RS Aggarwal 2020 2021 Solutions for Class 9 Maths Chapter 10 Quadrilaterals are provided here with simple step-by-step explanations. These solutions for Quadrilaterals are extremely popular among class 9 students for Maths Quadrilaterals Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the RS Aggarwal 2020 2021 Book of class 9 Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s RS Aggarwal 2020 2021 Solutions. All RS Aggarwal 2020 2021 Solutions for class 9 Maths are prepared by experts and are 100% accurate.
Page No 312:
Question 1:
Answer:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
So, the measure of the fourth angle is 120.
Page No 312:
Question 2:
Given: Three angles of a quadrilateral are 75°, 90° and 75°.
Let the fourth angle be x.
Using angle sum property of quadrilateral,
So, the measure of the fourth angle is 120.
Answer:
Let A = 2xââ.
Then B = (4x)â; C = (5x)â and D = (7x)â
Since the sum of the angles of a quadrilateral is 360o, we have:
2x + 4x + 5x + 7x = 360â
⇒ 18 x = 360â â
⇒ x = 20â
∴ A = 40â; B = 80â; C = 100â; D = 140â
Page No 312:
Question 3:
Let A = 2xââ.
Then B = (4x)â; C = (5x)â and D = (7x)â
Since the sum of the angles of a quadrilateral is 360o, we have:
2x + 4x + 5x + 7x = 360â
⇒ 18 x = 360â â
⇒ x = 20â
∴ A = 40â; B = 80â; C = 100â; D = 140â
Answer:
We have AB || DC.
A and D are the interior angles on the same side of transversal line AD, whereas B and C are the interior angles on the same side of transversal line BC.
Now, A + D = 180â
⇒ D = 180â − A
∴ D = 180â − 55â = 125â
Again , B + C = 180â
â⇒ C = 180â − B
∴ C = 180â − 70â = 110â
Page No 312:
Question 4:
We have AB || DC.
A and D are the interior angles on the same side of transversal line AD, whereas B and C are the interior angles on the same side of transversal line BC.
Now, A + D = 180â
⇒ D = 180â − A
∴ D = 180â − 55â = 125â
Again , B + C = 180â
â⇒ C = 180â − B
∴ C = 180â − 70â = 110â
Answer:
Given: ABCD is a square in which AB = BC = CD = DA. âEDC is an equilateral triangle in which ED = EC = DC and
EDC = DEC = âDCE = 60â.
To prove: AE = BE and DAE = 15ââ
Proof: In âADE and âBCE, we have:
AD = BC [Sides of a square]
ADE = BCE = 90â+ 60â = 150ââ
i.e., AE = BE
Now, ADE = 150â
DA = DC [Sides of a square]
DC = DE [Sides of an equilateral triangle]
So, DA = DE
âADE and âBCE are isosceles triangles.
i.e., DAE = DEA =
Page No 312:
Question 5:
Given: ABCD is a square in which AB = BC = CD = DA. âEDC is an equilateral triangle in which ED = EC = DC and
EDC = DEC = âDCE = 60â.
To prove: AE = BE and DAE = 15ââ
Proof: In âADE and âBCE, we have:
AD = BC [Sides of a square]
ADE = BCE = 90â+ 60â = 150ââ
i.e., AE = BE
Now, ADE = 150â
DA = DC [Sides of a square]
DC = DE [Sides of an equilateral triangle]
So, DA = DE
âADE and âBCE are isosceles triangles.
i.e., DAE = DEA =
Answer:
Given: A quadrilateral ABCD, in which BM â⊥ AC and DN ⊥ AC and BM = DN.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in âOND and âOMB, we have:
∠OND = ∠OMB (90o each)
∠DON = ∠ BOM (Vertically opposite angles)
Also, DN = BM (Given)
i.e., âOND ≅ âOMB (AAS congurence rule)
∴ OD = OB (CPCT)
âHence, AC bisects BD.
Page No 312:
Question 6:
Given: A quadrilateral ABCD, in which BM â⊥ AC and DN ⊥ AC and BM = DN.
To prove: AC bisects BD; or DO = BO
Proof:
Let AC and BD intersect at O.
Now, in âOND and âOMB, we have:
∠OND = ∠OMB (90o each)
∠DON = ∠ BOM (Vertically opposite angles)
Also, DN = BM (Given)
i.e., âOND ≅ âOMB (AAS congurence rule)
∴ OD = OB (CPCT)
âHence, AC bisects BD.
Answer:
âGiven: ABCD is a quadrilateral in which AB = AD and BC = DC
(i)
In âABC and âADC, we have:
AB = AD (Given)
AC is common.
i.e., âABC ≅ âADC (SSS congruence rule)
Thus, AC bisects ∠A and ∠ C.
(ii)
Now, in âABE and âADE, we have:
AB = AD (Given)â
∠BAE = ∠DAEâ (Proven above)
AE is common.
∴ âABE ≅ âADE (SAS congruence rule)
⇒ BE = DE (By CPCT)
(iii) âABC ≅ âADC (Proven above)
∴ ∠ABC = ∠ADâC (By CPCT)â
Page No 312:
Question 7:
âGiven: ABCD is a quadrilateral in which AB = AD and BC = DC
(i)
In âABC and âADC, we have:
AB = AD (Given)
AC is common.
i.e., âABC ≅ âADC (SSS congruence rule)
Thus, AC bisects ∠A and ∠ C.
(ii)
Now, in âABE and âADE, we have:
AB = AD (Given)â
∠BAE = ∠DAEâ (Proven above)
AE is common.
∴ âABE ≅ âADE (SAS congruence rule)
⇒ BE = DE (By CPCT)
(iii) âABC ≅ âADC (Proven above)
∴ ∠ABC = ∠ADâC (By CPCT)â
Answer:
Given: ABCD is a square and ∠PQR = 90°.
Also, PB = QC = DR
(i) We have:
BC = CD (Sides of square)
CQ = DR (Given)
BC = BQ + CQ
⇒ CQ = BC − BQ
∴ DR = BC − BQ ...(i)
Also, CD = RC+ DR
∴ DR = CD − RC = BC − RC ...(ii)
From (i) and (ii), we have:
BC − BQ = âBC − RC
∴ BQ = RC
(ii) In âRCQ and âQBP, we have:
PB = QC (Given)
BQ = RC (Proven above)
∠RCQ = ∠QBP (90o each)
i.e., âRCQ ≅ âQBP (SAS congruence rule)
∴ QR = PQ (By CPCT)
(iii) âRCQ ≅ âQBP and QR = PQ (Proven above)
∴ In âRPQ, ∠QPR = ∠QRP =
Page No 312:
Question 8:
Given: ABCD is a square and ∠PQR = 90°.
Also, PB = QC = DR
(i) We have:
BC = CD (Sides of square)
CQ = DR (Given)
BC = BQ + CQ
⇒ CQ = BC − BQ
∴ DR = BC − BQ ...(i)
Also, CD = RC+ DR
∴ DR = CD − RC = BC − RC ...(ii)
From (i) and (ii), we have:
BC − BQ = âBC − RC
∴ BQ = RC
(ii) In âRCQ and âQBP, we have:
PB = QC (Given)
BQ = RC (Proven above)
∠RCQ = ∠QBP (90o each)
i.e., âRCQ ≅ âQBP (SAS congruence rule)
∴ QR = PQ (By CPCT)
(iii) âRCQ ≅ âQBP and QR = PQ (Proven above)
∴ In âRPQ, ∠QPR = ∠QRP =
Answer:
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Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral.
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in âAOC, OA + OC > AC
⇒ OA + OB + OC + OD > AC + BD
Page No 313:
Question 9:
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Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral.
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in âAOC, OA + OC > AC
⇒ OA + OB + OC + OD > AC + BD
Answer:
Given: ABCD is a quadrilateral and AC is one of its diagonal.
(i) We know that the sum of any two sides of a triangle is greater than the third side.
In âABC, AB + BC > AC ...(1)
(ii) In âABC, we have :
â AB + BC > AC ...(1)
We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
In âACD, we have:â
AC > |DA − CD|â ...(2)
From (1) and (2), we have:
AB + BC > |DA − CD|â
⇒ AB + BC + CD > DA
(iii) In âABC, AB + BC > AC
âAdding these inequalities, we get:
2(AB + BC + CD + DA) > 2(AC + BD)
⇒ (AB + BC + CD + DA) > (AC + BD)
Page No 313:
Question 10:
Given: ABCD is a quadrilateral and AC is one of its diagonal.
(i) We know that the sum of any two sides of a triangle is greater than the third side.
In âABC, AB + BC > AC ...(1)
(ii) In âABC, we have :
â AB + BC > AC ...(1)
We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
In âACD, we have:â
AC > |DA − CD|â ...(2)
From (1) and (2), we have:
AB + BC > |DA − CD|â
⇒ AB + BC + CD > DA
(iii) In âABC, AB + BC > AC
âAdding these inequalities, we get:
2(AB + BC + CD + DA) > 2(AC + BD)
⇒ (AB + BC + CD + DA) > (AC + BD)
Answer:
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Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4 are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, âABD and âBCD.
In âABD, we have:
∠1 + ∠2 + â∠A = 180o ...(i)
In ââBCD, we have:
∠3 + ∠4 + ∠C = 180o ...(ii)
On adding (i) and (ii), we get:
⇒ ∠A + ∠C + ∠B + ∠D = 360o [ âµ ∠1 + ∠3 = ∠B; ∠4 + ∠2 = ∠D]
∴ ∠A + ∠C + ∠B + ∠D = 360o
Page No 328:
Question 1:
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Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4 are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, âABD and âBCD.
In âABD, we have:
∠1 + ∠2 + â∠A = 180o ...(i)
In ââBCD, we have:
∠3 + ∠4 + ∠C = 180o ...(ii)
On adding (i) and (ii), we get:
⇒ ∠A + ∠C + ∠B + ∠D = 360o [ âµ ∠1 + ∠3 = ∠B; ∠4 + ∠2 = ∠D]
∴ ∠A + ∠C + ∠B + ∠D = 360o
Answer:
ABCD is parallelogram and ∠A = 72°â.
We know that opposite angles of a parallelogram are equal.
∴∠A â= ∠C and ∠B â= ∠D â â
∴ ∠C = 72o
∠A and ∠B are adajcent angles.
i.e., ∠A â+ ∠Bâ = 180o
⇒ ∠B = 180o − ∠A
â⇒ ∠Bâ = 180o− 72o = 108o
∴â ∠Bâ =â ∠D = 108o
Hence, ∠Bâ =â ∠D = 108oâ and ∠Câ = 72oâ
Page No 328:
Question 2:
ABCD is parallelogram and ∠A = 72°â.
We know that opposite angles of a parallelogram are equal.
∴∠A â= ∠C and ∠B â= ∠D â â
∴ ∠C = 72o
∠A and ∠B are adajcent angles.
i.e., ∠A â+ ∠Bâ = 180o
⇒ ∠B = 180o − ∠A
â⇒ ∠Bâ = 180o− 72o = 108o
∴â ∠Bâ =â ∠D = 108o
Hence, ∠Bâ =â ∠D = 108oâ and ∠Câ = 72oâ
Answer:
Given: ABCD is parallelogram and ∠DAB = 80°â and ∠DBC = 60°
To find: Measure of ∠CDB and ∠ADB
In parallelogram ABCD, AD ||â BC
∴ ∠DBC = ∠ ADB = 60o (Alternate interior angles) ...(i)
As ∠DAB and ∠ADC are adajcent angles, ∠DAB â+ ∠ADCâ = 180o
∴ ∠ADC = 180o − ∠DAB
â ⇒∠ADCâ = 180o − 80o = 100o
Also, ∠ADCâ = ∠ADB + ∠CââDB
∴â ∠ADCâ =â 100o
⇒ ∠ADB + ∠CââDB = 100o ...(ii)
From (i) and (ii), we get:
60o + ∠CââDB = 100o
⇒ ∠CââDB = 100o − 60o = 40o
Hence, ∠CDBâ =â 40o and ∠ADBâ = 60oâ
â
Page No 329:
Question 3:
Given: ABCD is parallelogram and ∠DAB = 80°â and ∠DBC = 60°
To find: Measure of ∠CDB and ∠ADB
In parallelogram ABCD, AD ||â BC
∴ ∠DBC = ∠ ADB = 60o (Alternate interior angles) ...(i)
As ∠DAB and ∠ADC are adajcent angles, ∠DAB â+ ∠ADCâ = 180o
∴ ∠ADC = 180o − ∠DAB
â ⇒∠ADCâ = 180o − 80o = 100o
Also, ∠ADCâ = ∠ADB + ∠CââDB
∴â ∠ADCâ =â 100o
⇒ ∠ADB + ∠CââDB = 100o ...(ii)
From (i) and (ii), we get:
60o + ∠CââDB = 100o
⇒ ∠CââDB = 100o − 60o = 40o
Hence, ∠CDBâ =â 40o and ∠ADBâ = 60oâ
â
Answer:
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Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.
To prove: AD = 2CD
Proof:
Since, and AM is the transversal.
So, (Alternate interior angles)
But, (Given)
Therefore,
(Angles opposite to equal sides are equal.) ...(1)
Now, AB = CD (Opposite sides of a parallelogram are equal.)
(AB = BM and MC = BM)
Page No 329:
Question 4:
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Given: parallelogram ABCD, M is the midpoint of side BC and ∠BAM = ∠DAM.
To prove: AD = 2CD
Proof:
Since, and AM is the transversal.
So, (Alternate interior angles)
But, (Given)
Therefore,
(Angles opposite to equal sides are equal.) ...(1)
Now, AB = CD (Opposite sides of a parallelogram are equal.)
(AB = BM and MC = BM)
Answer:
ABCD is a parallelogram.
∴ â∠A = â∠C and ∠B =â ∠D (Opposite angles)
And â∠A + â∠B = 180o (Adjacent angles are supplementary)
∴ â∠B = 180o − ∠A
⇒ 180o − 60o = 120o ( âµ∠A = 60o)
∴ â∠A = â∠C = 60o and ∠B =â ∠Dâ = 120o
(i) In â APB, ∠âPAB = and ∠PBA =
∴â ∠âAPBâ = 180o − (30o + 60o) = 90o
(ii) In â ADP, ∠âPAD = 30o and ∠ADP = 120o
∴ ∠APB = 180o − (30o + 120o) = 30o
Thus, ∠âPAD = â∠APB = â30o
Hence, âADP is an isosceles triangle and AD = DP.
In â PBC, ∠â PBC = 60o, ∠â BPC = 180o − (90o +30o) = 60oand ∠â BCP = 60o (Opposite angle of ∠A)
∴ ∠ PBC = ∠â BPC = ∠â BCP
Hence, âPBC is an equilateral triangle and, therefore, PB = PC = BC.â
(iii) DC = DP + PC
From (ii), we have:
DC = AD + BC [AD = BC, opposite sides of a parallelogram]
⇒ DC = AD + AD
Page No 329:
Question 5:
ABCD is a parallelogram.
∴ â∠A = â∠C and ∠B =â ∠D (Opposite angles)
And â∠A + â∠B = 180o (Adjacent angles are supplementary)
∴ â∠B = 180o − ∠A
⇒ 180o − 60o = 120o ( âµ∠A = 60o)
∴ â∠A = â∠C = 60o and ∠B =â ∠Dâ = 120o
(i) In â APB, ∠âPAB = and ∠PBA =
∴â ∠âAPBâ = 180o − (30o + 60o) = 90o
(ii) In â ADP, ∠âPAD = 30o and ∠ADP = 120o
∴ ∠APB = 180o − (30o + 120o) = 30o
Thus, ∠âPAD = â∠APB = â30o
Hence, âADP is an isosceles triangle and AD = DP.
In â PBC, ∠â PBC = 60o, ∠â BPC = 180o − (90o +30o) = 60oand ∠â BCP = 60o (Opposite angle of ∠A)
∴ ∠ PBC = ∠â BPC = ∠â BCP
Hence, âPBC is an equilateral triangle and, therefore, PB = PC = BC.â
(iii) DC = DP + PC
From (ii), we have:
DC = AD + BC [AD = BC, opposite sides of a parallelogram]
⇒ DC = AD + AD
Answer:
ABCD is a parallelogram.
∴ AB â£â£â DC and BC ââ£â£â AD
(i) In âAOB, ∠BAO = 35°, â∠AOB = ∠COD = 105° (Vertically opposite angels)
∴ â∠ABO = 180o − (35o + 105o) = 40o
(ii)∠ODC and ∠ABO are alternate interior angles.
∴ ∠ODC = ∠ABO = 40o
(iii) ∠ACB = ∠âCAD = 40o (Alternate interior angles)
(iv) ∠CBD = ∠ABC − ∠ABD ...(i)
⇒∠CBD = 105o − ∠ABD (∠ABD = ∠ABO)
⇒∠CBD = 105o − 40o = 65o
Page No 329:
Question 6:
ABCD is a parallelogram.
∴ AB â£â£â DC and BC ââ£â£â AD
(i) In âAOB, ∠BAO = 35°, â∠AOB = ∠COD = 105° (Vertically opposite angels)
∴ â∠ABO = 180o − (35o + 105o) = 40o
(ii)∠ODC and ∠ABO are alternate interior angles.
∴ ∠ODC = ∠ABO = 40o
(iii) ∠ACB = ∠âCAD = 40o (Alternate interior angles)
(iv) ∠CBD = ∠ABC − ∠ABD ...(i)
⇒∠CBD = 105o − ∠ABD (∠ABD = ∠ABO)
⇒∠CBD = 105o − 40o = 65o
Answer:
ABCD is a parallelogram.
i.e., ∠A = ∠C and ∠B = ∠D (Opposite angles)
Also, ∠A + ∠B = 180o (Adjacent angles are supplementary) â
∴â (2x + 25)°â + (3x − 5)°â = 180
⇒ â5x +20 = 180
⇒â 5x = 160
⇒â x = 32o
∴â∠A = 2 ⨯ 32 + 25 = 89o and ∠B = 3 ⨯ 32 − 5 = 91o
Hence, x = 32o, ∠A = ∠C = 89o and ∠B = ∠D = 91o
Page No 329:
Question 7:
ABCD is a parallelogram.
i.e., ∠A = ∠C and ∠B = ∠D (Opposite angles)
Also, ∠A + ∠B = 180o (Adjacent angles are supplementary) â
∴â (2x + 25)°â + (3x − 5)°â = 180
⇒ â5x +20 = 180
⇒â 5x = 160
⇒â x = 32o
∴â∠A = 2 ⨯ 32 + 25 = 89o and ∠B = 3 ⨯ 32 − 5 = 91o
Hence, x = 32o, ∠A = ∠C = 89o and ∠B = ∠D = 91o
Answer:
Let ABCD be a parallelogram.
∴ ∠âA = ∠C and ∠B = ∠D (Opposite angles)
Let ∠A = xo and ∠B =
Now, ∠âA + ∠B = 180o (Adjacent angles are supplementary)
Hence, ∠A = ∠C = 100o; ∠B = ∠D = 80o
Page No 329:
Question 8:
Let ABCD be a parallelogram.
∴ ∠âA = ∠C and ∠B = ∠D (Opposite angles)
Let ∠A = xo and ∠B =
Now, ∠âA + ∠B = 180o (Adjacent angles are supplementary)
Hence, ∠A = ∠C = 100o; ∠B = ∠D = 80o
Answer:
Let ABCD be a parallelogram.
∴ ∠âA = ∠âC and ∠B = ∠D (Opposite angles)
Let ∠A be the smallest angle whose measure is xo.
∴â ∠âB = (2x − 30)o
Now, ∠âA + ∠B = 180o (Adjacent angles are supplementary)
⇒ x + 2x − 30o = 180o
⇒ 3x = 210o
⇒ x = 70o
∴ â∠âB = 2 ⨯ 70o − 30o = 110o
Hence, ∠A = ∠C = 70o; ∠B = ∠D = 110o
Page No 329:
Question 9:
Let ABCD be a parallelogram.
∴ ∠âA = ∠âC and ∠B = ∠D (Opposite angles)
Let ∠A be the smallest angle whose measure is xo.
∴â ∠âB = (2x − 30)o
Now, ∠âA + ∠B = 180o (Adjacent angles are supplementary)
⇒ x + 2x − 30o = 180o
⇒ 3x = 210o
⇒ x = 70o
∴ â∠âB = 2 ⨯ 70o − 30o = 110o
Hence, ∠A = ∠C = 70o; ∠B = ∠D = 110o
Answer:
ABCD is a parallelogram.
The opposite sides of a parallelogram are parallel and equal.
∴ AB = DC = 9.5 cm
Let BC = AD = x
∴â Perimeter of ABCD = AB + BC + CD + DA = 30 cm
⇒ 9.5 + x + 9.5 + x = 30
⇒ 19 + 2x = 30
⇒ 2x = 11
⇒ x = 5.5 cm
Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm
Page No 329:
Question 10:
ABCD is a parallelogram.
The opposite sides of a parallelogram are parallel and equal.
∴ AB = DC = 9.5 cm
Let BC = AD = x
∴â Perimeter of ABCD = AB + BC + CD + DA = 30 cm
⇒ 9.5 + x + 9.5 + x = 30
⇒ 19 + 2x = 30
⇒ 2x = 11
⇒ x = 5.5 cm
Hence, AB = DC = 9.5 cm and BC = DA = 5.5 cm
Answer:
ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
(i)â In âABC, ∠âBAC = ∠BCA =
i.e., x = 35o
Now, ∠âB + ∠C = 180o (Adjacent angles are supplementary) â
But ∠âCâ = x + y = 70o
⇒â y = 70o − x
⇒ây = 70o − 35o = 35o
Hence, x = 35o; y = 35o
(ii) The diagonals of a rhombus are perpendicular bisectors of each other.
So, in ââAOB, â∠OAB = 40o, ∠AOB = 90o and ∠ABO = 180o − (40o + 90o) = 50o
∴ âx = 50o
In ââABâD, AB = AD
So, ∠ABD = â∠ADB = â50o
Hence, x = 50o; y = 50o
â(iii) ∠âBAC = ∠âDCA (Alternate interior angles)â
i.e., x = 62o
In âBOC, ∠âBCO = 62o [In ââ ABC, AB = BC, so ∠âBAC = ∠âACB]
Also, ∠âBOC = 90o
∴ ∠âOBC = 180o − (90o + 62o) = 28o
Hence, x = 62o; y = 28o
Page No 329:
Question 11:
ABCD is a rhombus and a rhombus is also a parallelogram. A rhombus has four equal sides.
(i)â In âABC, ∠âBAC = ∠BCA =
i.e., x = 35o
Now, ∠âB + ∠C = 180o (Adjacent angles are supplementary) â
But ∠âCâ = x + y = 70o
⇒â y = 70o − x
⇒ây = 70o − 35o = 35o
Hence, x = 35o; y = 35o
(ii) The diagonals of a rhombus are perpendicular bisectors of each other.
So, in ââAOB, â∠OAB = 40o, ∠AOB = 90o and ∠ABO = 180o − (40o + 90o) = 50o
∴ âx = 50o
In ââABâD, AB = AD
So, ∠ABD = â∠ADB = â50o
Hence, x = 50o; y = 50o
â(iii) ∠âBAC = ∠âDCA (Alternate interior angles)â
i.e., x = 62o
In âBOC, ∠âBCO = 62o [In ââ ABC, AB = BC, so ∠âBAC = ∠âACB]
Also, ∠âBOC = 90o
∴ ∠âOBC = 180o − (90o + 62o) = 28o
Hence, x = 62o; y = 28o
Answer:
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Let ABCD be a rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â âAOB is a right angle triangle in which OA = AC/2 = 24/2 = 12 cm and OB = BD/2 = 18/2 = 9 cm.
Now, AB2= OA2 + OB2 [Pythagoras theorem]
⇒â AB2=â (12)2 + (9)2
⇒â AB2=â 144 + 81 = 225
⇒â AB=â 15 cm
Hence, the side of the rhombus is 15 cm.
Page No 329:
Question 12:
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Let ABCD be a rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 24 cm and BD = 18 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â âAOB is a right angle triangle in which OA = AC/2 = 24/2 = 12 cm and OB = BD/2 = 18/2 = 9 cm.
Now, AB2= OA2 + OB2 [Pythagoras theorem]
⇒â AB2=â (12)2 + (9)2
⇒â AB2=â 144 + 81 = 225
⇒â AB=â 15 cm
Hence, the side of the rhombus is 15 cm.
Answer:
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Let ABCD be a rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of ABCD. Let AC = x and BD = 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â âAOB is a right angle triangle, in which OA = AC 2 = x 2 and OB = BD 2 = 16 2 = 8 cm.
Now, AB2= OA2 + OB2 [Pythagoras theorem]
Hence, the other diagonal of the rhombus is 12 cm.
∴ Area of the rhombus =
Page No 330:
Question 13:
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Let ABCD be a rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of ABCD. Let AC = x and BD = 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â âAOB is a right angle triangle, in which OA = AC 2 = x 2 and OB = BD 2 = 16 2 = 8 cm.
Now, AB2= OA2 + OB2 [Pythagoras theorem]
Hence, the other diagonal of the rhombus is 12 cm.
∴ Area of the rhombus =
Answer:
(i) ABCD is a rectangle.
The diagonals of a rectangle are congruent and bisect each other. Therefore, inâ â AOB, we have:
OA = OB
∴â ∠âOAB = ∠âOBA = 35o
∴â x = 90o − 35o = 55o
And ∠AOB = 180o − (35o + 35o) = 110o
∴â y = ∠AOBâ = 110oâ [Vertically opposite angles]
Hence, x = 55o and y = 110oââ
(ii) In âAOB, we have:
OA = OB
Now, ∠âOAB = ∠OBA =
∴â y = ∠BAC = 35o [Interior alternate angles]
Also, x = 90o − y [ ââµ∠C = 90o = x + y ]
⇒â x = 90o − 35o = 55o
Hence, x = 55o and y = 35oââ
Page No 330:
Question 14:
(i) ABCD is a rectangle.
The diagonals of a rectangle are congruent and bisect each other. Therefore, inâ â AOB, we have:
OA = OB
∴â ∠âOAB = ∠âOBA = 35o
∴â x = 90o − 35o = 55o
And ∠AOB = 180o − (35o + 35o) = 110o
∴â y = ∠AOBâ = 110oâ [Vertically opposite angles]
Hence, x = 55o and y = 110oââ
(ii) In âAOB, we have:
OA = OB
Now, ∠âOAB = ∠OBA =
∴â y = ∠BAC = 35o [Interior alternate angles]
Also, x = 90o − y [ ââµ∠C = 90o = x + y ]
⇒â x = 90o − 35o = 55o
Hence, x = 55o and y = 35oââ
Answer:
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Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB
(Common side)
(Given)
(Given)
(By SAS congruence Criteria)
(CPCT)
Also, (Sides of rhombus are equal)
Thus, is an equilateral triangle.
Therefore,
(Opposite angles of rhombus are equal)
(Adjacent angles of rhombus are supplementary.)
Hence, the angles of rhombus are .
Page No 330:
Question 15:
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Given: ABCD is a rhombus, DE is altitude which bisects AB i.e. AE = EB
(Common side)
(Given)
(Given)
(By SAS congruence Criteria)
(CPCT)
Also, (Sides of rhombus are equal)
Thus, is an equilateral triangle.
Therefore,
(Opposite angles of rhombus are equal)
(Adjacent angles of rhombus are supplementary.)
Hence, the angles of rhombus are .
Answer:
The angles of a square are bisected by the diagonals.
∴ ∠âOBX = 45o [âµ∠âABC = 90o and BD bisects ∠âABCâ]
And ∠âBOX = ∠COD = 80o [Vertically opposite angles]
∴â In âBOX, we have:
∠AXO = ∠OBX + â∠BOX [Exterior angle of âBOX]
⇒â ∠AXO = 45o + 80o = 125o
∴ âx =125o
Page No 330:
Question 16:
The angles of a square are bisected by the diagonals.
∴ ∠âOBX = 45o [âµ∠âABC = 90o and BD bisects ∠âABCâ]
And ∠âBOX = ∠COD = 80o [Vertically opposite angles]
∴â In âBOX, we have:
∠AXO = ∠OBX + â∠BOX [Exterior angle of âBOX]
⇒â ∠AXO = 45o + 80o = 125o
∴ âx =125o
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: A rhombus ABCD.
To prove: Diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Proof:
In ,
(Sides of rhombus are equal.)
(Angles opposite to equal sides are equal.) ...(1)
Now,
(Opposite sides of rhombus are parallel.)
AC is transversal.
So, (Alternate interior angles) ...(2)
From (1) and (2), we get
Thus, AC bisects .
Similarly,
Since, and AC is transversal.
So, (Alternate interior angles) ...(3)
From (1) and (3), we get
Thus, AC bisects ∠C.
Hence, AC bisects
In ,
(Sides of rhombus are equal.)
(Angles opposite to equal sides are equal.) ...(4)
Now,
(Opposite sides of rhombus are parallel.)
BD is transversal.
So, (Alternate interior angles) ...(5)
From (4) and (5), we get
Thus, DB bisects .
Similarly,
Since, and BD is transversal.
So, (Alternate interior angles) ...(6)
From (4) and (6), we get
Thus, BD bisects ∠B.
Hence, BD bisects
Page No 330:
Question 17:
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Given: A rhombus ABCD.
To prove: Diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.
Proof:
In ,
(Sides of rhombus are equal.)
(Angles opposite to equal sides are equal.) ...(1)
Now,
(Opposite sides of rhombus are parallel.)
AC is transversal.
So, (Alternate interior angles) ...(2)
From (1) and (2), we get
Thus, AC bisects .
Similarly,
Since, and AC is transversal.
So, (Alternate interior angles) ...(3)
From (1) and (3), we get
Thus, AC bisects ∠C.
Hence, AC bisects
In ,
(Sides of rhombus are equal.)
(Angles opposite to equal sides are equal.) ...(4)
Now,
(Opposite sides of rhombus are parallel.)
BD is transversal.
So, (Alternate interior angles) ...(5)
From (4) and (5), we get
Thus, DB bisects .
Similarly,
Since, and BD is transversal.
So, (Alternate interior angles) ...(6)
From (4) and (6), we get
Thus, BD bisects ∠B.
Hence, BD bisects
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In a parallelogram ABCD, AM = CN.
To prove: AC and MN bisect each other.
Construction: Join AN and MC.
Proof:
Since, ABCD is a parallelogram.
Also, AM = CN (Given)
Thus, AMCN is a parallelogram.
Since, diagonals of a parallelogram bisect each other.
Hence, AC and MN bisect each other.
Page No 330:
Question 18:
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Given: In a parallelogram ABCD, AM = CN.
To prove: AC and MN bisect each other.
Construction: Join AN and MC.
Proof:
Since, ABCD is a parallelogram.
Also, AM = CN (Given)
Thus, AMCN is a parallelogram.
Since, diagonals of a parallelogram bisect each other.
Hence, AC and MN bisect each other.
Answer:
We have:
∠B = ∠D [Opposite angles of parallelogram ABCD]
AD = BC and AB = DC [Opposite sides of parallelogram ABCD]
Also, AD || BC and AB|| DC
It is given that .
∴ âAP = CQ [âµ AD = BC]
In ââDPC and ââBQA, we have:
AB = CD, ∠B = ∠D and DP = QB [âµDP = â AD and QB = BC]
i.e., ââDPC ≅ ââBQA
∴â PC = QA
Thus, in quadrilatreal AQCP, we have:
AP = CQ ...(i)
PC = QA ...(ii)
∴ âAQCP is a parallelogram.
Page No 330:
Question 19:
We have:
∠B = ∠D [Opposite angles of parallelogram ABCD]
AD = BC and AB = DC [Opposite sides of parallelogram ABCD]
Also, AD || BC and AB|| DC
It is given that .
∴ âAP = CQ [âµ AD = BC]
In ââDPC and ââBQA, we have:
AB = CD, ∠B = ∠D and DP = QB [âµDP = â AD and QB = BC]
i.e., ââDPC ≅ ââBQA
∴â PC = QA
Thus, in quadrilatreal AQCP, we have:
AP = CQ ...(i)
PC = QA ...(ii)
∴ âAQCP is a parallelogram.
Answer:
In ââODF and ââOBE, we have:
OD = OB (Diagonals bisects each other)
∠DOF = ∠BOE (Vertically opposite angles)
∠FDO = ∠OBE (Alternate interior angles)
i.e., ââODF ≅ ââOBE
∴â OF = OE (CPCT)
Hence, proved.
Page No 330:
Question 20:
In ââODF and ââOBE, we have:
OD = OB (Diagonals bisects each other)
∠DOF = ∠BOE (Vertically opposite angles)
∠FDO = ∠OBE (Alternate interior angles)
i.e., ââODF ≅ ââOBE
∴â OF = OE (CPCT)
Hence, proved.
Answer:
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âGiven: In parallelogram ABCD, DP⊥ AB, AQ ⊥ BC and ∠PDQ = 60°
In quadrilateral DPBQ, by angle sum property, we have
Therefore,
Now,
(Opposite angles of a parallelogram are equal.)
(Adjacent angles of a parallelogram are supplementary.)
Also,
(Opposite angles of a parallleogram are equal.)
So, the angles of a parallelogram are
Page No 330:
Question 21:
![](/img/site_content/ask-answer/loader.gif)
âGiven: In parallelogram ABCD, DP⊥ AB, AQ ⊥ BC and ∠PDQ = 60°
In quadrilateral DPBQ, by angle sum property, we have
Therefore,
Now,
(Opposite angles of a parallelogram are equal.)
(Adjacent angles of a parallelogram are supplementary.)
Also,
(Opposite angles of a parallleogram are equal.)
So, the angles of a parallelogram are
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In rectangle ABCD, AC bisects ∠A, i.e. ∠1 = ∠2 and AC bisects ∠C, i.e. ∠3 = ∠4.
To prove:
(i) ABCD is a square,
(ii) diagonal BD bisects ∠B as well as ∠D.
Proof:
(i)
Since, (Opposite sides of a rectangle are parallel.)
So, (Alternate interior angles)
But, (Given)
So,
In
Since,
So, (Sides opposite to equal angles are equal.)
But these are adjacent sides of the rectangle ABCD.
Hence, ABCD is a square.
(ii)
Since, the diagonals of a square bisects its angles.
So, diagonals BD bisects ∠B as well as ∠D.
Page No 331:
Question 22:
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Given: In rectangle ABCD, AC bisects ∠A, i.e. ∠1 = ∠2 and AC bisects ∠C, i.e. ∠3 = ∠4.
To prove:
(i) ABCD is a square,
(ii) diagonal BD bisects ∠B as well as ∠D.
Proof:
(i)
Since, (Opposite sides of a rectangle are parallel.)
So, (Alternate interior angles)
But, (Given)
So,
In
Since,
So, (Sides opposite to equal angles are equal.)
But these are adjacent sides of the rectangle ABCD.
Hence, ABCD is a square.
(ii)
Since, the diagonals of a square bisects its angles.
So, diagonals BD bisects ∠B as well as ∠D.
Answer:
DC = BE (âµ DC = AB)
∠COD = ∠BOE (Vertically opposite angles)
∠OCD = ∠OBE ( Alternate interior angles)
i.e., ââODC ≅ ââOEB
⇒ OC = OB (CPCT)
We know that BC = OC + OB.
∴ ED bisects BC.
Page No 331:
Question 23:
DC = BE (âµ DC = AB)
∠COD = ∠BOE (Vertically opposite angles)
∠OCD = ∠OBE ( Alternate interior angles)
i.e., ââODC ≅ ââOEB
⇒ OC = OB (CPCT)
We know that BC = OC + OB.
∴ ED bisects BC.
Answer:
Given: ABCD is a parallelogram.
BE = CE (E is the mid point of BC)
DE and AB when produced meet at F.
To prove: AF = 2AB
âProof:
In parallelogram ABCD, we have:
AB || DC
∠DCE = ∠EBF (Alternate interior angles)
In âDCE and âBFE, we have:
∠DCE = ∠EBF (Proved above)
Also, BE = CE (Given)
∴ âDCE ≅â âBFE (By ASA congruence rule)
∴ DC = BF (CPCT)
But DC = AB, as ABCD is a parallelogram.
∴ DC = AB = BF ...(i)
Now, AF = AB + BF ...(ii)
From (i), we get:
AF = AB + AB = 2AB
Hence, proved.
Page No 331:
Question 24:
Given: ABCD is a parallelogram.
BE = CE (E is the mid point of BC)
DE and AB when produced meet at F.
To prove: AF = 2AB
âProof:
In parallelogram ABCD, we have:
AB || DC
∠DCE = ∠EBF (Alternate interior angles)
In âDCE and âBFE, we have:
∠DCE = ∠EBF (Proved above)
Also, BE = CE (Given)
∴ âDCE ≅â âBFE (By ASA congruence rule)
∴ DC = BF (CPCT)
But DC = AB, as ABCD is a parallelogram.
∴ DC = AB = BF ...(i)
Now, AF = AB + BF ...(ii)
From (i), we get:
AF = AB + AB = 2AB
Hence, proved.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: l || m and the bisectors of interior angles intersect at B and D.
To prove: ABCD is a rectangle.
Proof:
Since, l || m (Given)
So, (Alternate interior angles)
but, these are a pair of alternate interior angles for AB and DC.
Similarly,
So, ABCD is a parallelogram.
Also,
(Linear pair)
But, this an angle of the parallleogram ABCD.
Hence, ABCD is a rectangle.
Page No 331:
Question 25:
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Given: l || m and the bisectors of interior angles intersect at B and D.
To prove: ABCD is a rectangle.
Proof:
Since, l || m (Given)
So, (Alternate interior angles)
but, these are a pair of alternate interior angles for AB and DC.
Similarly,
So, ABCD is a parallelogram.
Also,
(Linear pair)
But, this an angle of the parallleogram ABCD.
Hence, ABCD is a rectangle.
Answer:
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âGiven: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.
Proof:
In square ABCD,
AB = BC = CD = DA (All sides of a square are equal.)
And, AK = BL = CM = DN (Given)
So, AB AK = BC BL = CD CM = DA DN
KB = CL = DM = AN ...(1)
In and ,
(Each angle of a square is a right angle.)
(Given)
[From (1)]
So, by SAS congruence criteria,
(CPCT) ...(2)
Similarly,
MN = NK and (CPCT) ...(3)
MN = JM and (CPCT) ...(4)
ML = LK and (CPCT) ...(5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML ...(6)
And,
Now,
![](/img/site_content/ask-answer/loader.gif)
In ,
Let
So, (Exterior angles equals sum of interior opposite angles.)
Similarly,
...(7)
Using (6) and (7), we get
All sides of quadrikateral KLMN are equal and all angles are 90.
So, KLMN is a square.
Page No 331:
Question 26:
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âGiven: In square ABCD, AK = BL = CM = DN.
To prove: KLMN is a square.
Proof:
In square ABCD,
AB = BC = CD = DA (All sides of a square are equal.)
And, AK = BL = CM = DN (Given)
So, AB AK = BC BL = CD CM = DA DN
KB = CL = DM = AN ...(1)
In and ,
(Each angle of a square is a right angle.)
(Given)
[From (1)]
So, by SAS congruence criteria,
(CPCT) ...(2)
Similarly,
MN = NK and (CPCT) ...(3)
MN = JM and (CPCT) ...(4)
ML = LK and (CPCT) ...(5)
From (2), (3), (4) and (5), we get
NK = KL = MN = ML ...(6)
And,
Now,
![](/img/site_content/ask-answer/loader.gif)
In ,
Let
So, (Exterior angles equals sum of interior opposite angles.)
Similarly,
...(7)
Using (6) and (7), we get
All sides of quadrikateral KLMN are equal and all angles are 90.
So, KLMN is a square.
Answer:
BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(i)
Similarly, BC || AR and AB || CR.
i.e., BCRA is a parallelogram.
∴ BC = AR ...(ii)
But QR = QA + AR
From (i) and (ii), we get:
QR = BC + BC
⇒ QR = 2BCâ
∴ BC =
Page No 331:
Question 27:
BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(i)
Similarly, BC || AR and AB || CR.
i.e., BCRA is a parallelogram.
∴ BC = AR ...(ii)
But QR = QA + AR
From (i) and (ii), we get:
QR = BC + BC
⇒ QR = 2BCâ
∴ BC =
Answer:
Perimeter of ââABC = AB + BC + CA ...(i)
Perimeter of âPQR =â PQ + QR + PR ...(ii)
BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(iii)
Similarly, BC || AR and AB || CR
i.e., BCRA is a parallelogram.
∴ BC = AR ...(iv)
But, QR = QA + AR
From (iii) and (iv), we get:
⇒ QR = BC + BC
⇒ QR = 2BCâ
∴ BC =
Similarly, CA = â and AB = â
i.e., Perimeter of ââABC = (Perimeter of ââPQR)
∴ Perimeter of âPQR = 2 ⨯ Perimeter of âABC
Page No 345:
Question 1:
Perimeter of ââABC = AB + BC + CA ...(i)
Perimeter of âPQR =â PQ + QR + PR ...(ii)
BC || QA and CA || QB
i.e., BCQA is a parallelogram.
∴ BC = QA ...(iii)
Similarly, BC || AR and AB || CR
i.e., BCRA is a parallelogram.
∴ BC = AR ...(iv)
But, QR = QA + AR
From (iii) and (iv), we get:
⇒ QR = BC + BC
⇒ QR = 2BCâ
∴ BC =
Similarly, CA = â and AB = â
i.e., Perimeter of ââABC = (Perimeter of ââPQR)
∴ Perimeter of âPQR = 2 ⨯ Perimeter of âABC
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
(i) PQ || AC and PQ = AC
(ii) PQ || SR
(iii) PQRS is a parallelogram.
Proof:
(i)
In ,
Since, P and Q are the mid points of sides AB and BC, respectively. (Given)
(Using mid-point theorem.)
(ii)
In ,
Since, S and R are the mid-points of AD and DC, respectively. (Given)
(Using mid-point theorem.) ...(1)
From (i) and (1), we get
PQ || SR
(iii)
From (i) and (ii), we get
So, PQ and SR are parallel and equal.
Hence, PQRS is a parallelogram.
Page No 345:
Question 2:
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Given: In quadrilateral ABCD, P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA.
To prove:
(i) PQ || AC and PQ = AC
(ii) PQ || SR
(iii) PQRS is a parallelogram.
Proof:
(i)
In ,
Since, P and Q are the mid points of sides AB and BC, respectively. (Given)
(Using mid-point theorem.)
(ii)
In ,
Since, S and R are the mid-points of AD and DC, respectively. (Given)
(Using mid-point theorem.) ...(1)
From (i) and (1), we get
PQ || SR
(iii)
From (i) and (ii), we get
So, PQ and SR are parallel and equal.
Hence, PQRS is a parallelogram.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In an isosceles right âABC, CMPN is a square.
To prove: P bisects the hypotenuse AB i.e., AP = PB.
Proof:
In square CMPN,
∴ CM = MP = PN = CN (All sides are equal.)
Also, âABC is an isosceles with AC = BC.
⇒ AN + NC = CM + MB
⇒ AN = MB (âµ CN = CM) ...(i)
Now,
In âANP and âPMB,
AN = MB [From (i)]
∠ANP = ∠PMB = 90°
PN = PM (Sides of square CMPN)
∴ By SAS congruence criteria,
âANP ≅ âBMP
Hence, AP = PB (By CPCT)
Page No 346:
Question 3:
![](/img/site_content/ask-answer/loader.gif)
Given: In an isosceles right âABC, CMPN is a square.
To prove: P bisects the hypotenuse AB i.e., AP = PB.
Proof:
In square CMPN,
∴ CM = MP = PN = CN (All sides are equal.)
Also, âABC is an isosceles with AC = BC.
⇒ AN + NC = CM + MB
⇒ AN = MB (âµ CN = CM) ...(i)
Now,
In âANP and âPMB,
AN = MB [From (i)]
∠ANP = ∠PMB = 90°
PN = PM (Sides of square CMPN)
∴ By SAS congruence criteria,
âANP ≅ âBMP
Hence, AP = PB (By CPCT)
Answer:
In parallelogram ABCD, we have:
AD || BC and AB || DC
AB = AE + BE and DC = DF + FC
â∴â AD || EFâ
Similarly, âBEFC is also a parallelogram.
∴â EF || BC
∴â AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.
These lines AD, EF and BCâ are also cut by the transversal AB at A, E and B, respectively such thatâ AE = BE. â
Similarly, they are also cut byâ GH.
∴ GP = PH (By intercept theorem)
Page No 346:
Question 4:
In parallelogram ABCD, we have:
AD || BC and AB || DC
AB = AE + BE and DC = DF + FC
â∴â AD || EFâ
Similarly, âBEFC is also a parallelogram.
∴â EF || BC
∴â AD || EF || BC
Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.
These lines AD, EF and BCâ are also cut by the transversal AB at A, E and B, respectively such thatâ AE = BE. â
Similarly, they are also cut byâ GH.
∴ GP = PH (By intercept theorem)
Answer:
Given: A parallelogram ABCD
To prove: MN is bisected at O
Proof:
In OAM and OCN,
OA = OC (Diagonals of parallelogram bisect each other)
∠AOM = ∠CON (Vertically opposite angles)
∠MAO = ∠OCN (Alternate interior angles)
By ASA congruence criteria,
(CPCT)
Hence, MN is bisected at O.
Page No 346:
Question 5:
Given: A parallelogram ABCD
To prove: MN is bisected at O
Proof:
In OAM and OCN,
OA = OC (Diagonals of parallelogram bisect each other)
∠AOM = ∠CON (Vertically opposite angles)
∠MAO = ∠OCN (Alternate interior angles)
By ASA congruence criteria,
(CPCT)
Hence, MN is bisected at O.
Answer:
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Given: In trapezium PQRS, PQ || SR, M is the midpoint of PS and MN || PQ.
To prove: N is the midpoint of QR.
Construction: Join QS.
Proof:
In âSPQ,
Since, M is the mid-point of SP and MO || PQ.
Therefore, O is the mid-point of SQ. (By Mid-point theorem)
Similarly, in âSRQ,
Since, O is the mid-point of SQ and ON || SR (SR || PQ and MN || PQ)
Therefore, N is the mid-point of QR. (By Mid-point theorem)
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Question 6:
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Given: In trapezium PQRS, PQ || SR, M is the midpoint of PS and MN || PQ.
To prove: N is the midpoint of QR.
Construction: Join QS.
Proof:
In âSPQ,
Since, M is the mid-point of SP and MO || PQ.
Therefore, O is the mid-point of SQ. (By Mid-point theorem)
Similarly, in âSRQ,
Since, O is the mid-point of SQ and ON || SR (SR || PQ and MN || PQ)
Therefore, N is the mid-point of QR. (By Mid-point theorem)
Answer:
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Given: In parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠SPQ meets SR at M.
Let ∠SPQ = 2x.
⇒ ∠SRQ = 2x and ∠TPQ = x.
Also, PQ ⥠SR
⇒ ∠TMR = ∠TPQ = x.
In â³TMR, ∠SRQ is an exterior angle.
⇒ ∠SRQ = ∠TMR + ∠MTR
⇒ 2x = x + ∠MTR
⇒ ∠MTR = x
⇒ â³TPQ is an isosceles triangle.
⇒ TQ = PQ = 12 cm
Now,
RT = TQ − QR
= TQ − PS
= 12 − 9
= 3 cm
Page No 346:
Question 7:
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Given: In parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of ∠SPQ meets SR at M.
Let ∠SPQ = 2x.
⇒ ∠SRQ = 2x and ∠TPQ = x.
Also, PQ ⥠SR
⇒ ∠TMR = ∠TPQ = x.
In â³TMR, ∠SRQ is an exterior angle.
⇒ ∠SRQ = ∠TMR + ∠MTR
⇒ 2x = x + ∠MTR
⇒ ∠MTR = x
⇒ â³TPQ is an isosceles triangle.
⇒ TQ = PQ = 12 cm
Now,
RT = TQ − QR
= TQ − PS
= 12 − 9
= 3 cm
Answer:
Given: AB || DC, AP = PD and BQ = CQ
(i) In âQCD and âQBE, we have:
∠DQC = â∠BQE (Vertically opposite angles)
BQ = CQ (P is the midpoints)
∴ âQCD ≅ âQBE
âHence, DQ = QE (CPCT)
(ii) Now, in âADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
⇒ PQ || AB || DC
â⇒â AB || PR || DC
(iii) PQ, AB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC.
â Similarly, lines PQ, AB and DC are also cut by AC at R.
∴ AR = RC (By intercept theorem)
Page No 346:
Question 8:
Given: AB || DC, AP = PD and BQ = CQ
(i) In âQCD and âQBE, we have:
∠DQC = â∠BQE (Vertically opposite angles)
BQ = CQ (P is the midpoints)
∴ âQCD ≅ âQBE
âHence, DQ = QE (CPCT)
(ii) Now, in âADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
⇒ PQ || AB || DC
â⇒â AB || PR || DC
(iii) PQ, AB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC.
â Similarly, lines PQ, AB and DC are also cut by AC at R.
∴ AR = RC (By intercept theorem)
Answer:
AD is a median of âABC.
∴ BD = DC
We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Here, in âABC, D is the mid point of BC and DE || BA (given). Then DE bisects AC.
i.e., AE = EC
∴ E is the midpoint of AC.
⇒ BE is the median of âABC.
Page No 346:
Question 9:
AD is a median of âABC.
∴ BD = DC
We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Here, in âABC, D is the mid point of BC and DE || BA (given). Then DE bisects AC.
i.e., AE = EC
∴ E is the midpoint of AC.
⇒ BE is the median of âABC.
Answer:
In âABC, we have:
AC = AE + EC ...(i)
AE = EC ...(ii) [BE is the median of âABC]
∴ AC = 2EC ...(iii)
âIn âBEC, DF || BE.
∴ EF = CF (By midpoint theorem, as D is the midpoint of BC)
But EC = EF + CF
⇒ EC =â 2 ⨯ âCF ...(iv)
From (iii) and (iv), we get:
AC = 2 ⨯ (2 ⨯â CF)
∴â CF =
Page No 346:
Question 10:
In âABC, we have:
AC = AE + EC ...(i)
AE = EC ...(ii) [BE is the median of âABC]
∴ AC = 2EC ...(iii)
âIn âBEC, DF || BE.
∴ EF = CF (By midpoint theorem, as D is the midpoint of BC)
But EC = EF + CF
⇒ EC =â 2 ⨯ âCF ...(iv)
From (iii) and (iv), we get:
AC = 2 ⨯ (2 ⨯â CF)
∴â CF =
Answer:
âABC is shown below. D, E and F are the midpoints of sides AB, BC and CA, respectively.
As, D and E are the mid points of sides AB, and BC of â ABC.
∴ DE â£â£ AC (By midpoint theorem)
Similarly, DF ââ£â£ BC and EF ââ£â£ AB.
Therefore, ADEF, BDFE and DFCE are all parallelograms.
Now, DE is the diagonal of the parallelogram BDFE.
∴ ââBDE ≅ ââFED
Simiilarly, DF is the âdiagonal of the parallelogram ADEF.
And, EF is the âdiagonal of the parallelogram DFCE.
So, all the four triangles are congruent.
Page No 346:
Question 11:
âABC is shown below. D, E and F are the midpoints of sides AB, BC and CA, respectively.
As, D and E are the mid points of sides AB, and BC of â ABC.
∴ DE â£â£ AC (By midpoint theorem)
Similarly, DF ââ£â£ BC and EF ââ£â£ AB.
Therefore, ADEF, BDFE and DFCE are all parallelograms.
Now, DE is the diagonal of the parallelogram BDFE.
∴ ââBDE ≅ ââFED
Simiilarly, DF is the âdiagonal of the parallelogram ADEF.
And, EF is the âdiagonal of the parallelogram DFCE.
So, all the four triangles are congruent.
Answer:
â ABC is shown below. D, E and F are the midpoints of sides BC, CA and AB, respectively.
As F and E are the mid points of sides AB and AC of â ABC.
∴ FE â£â£ BC (By mid point theorem)
Similarly, DE ââ£â£ FB and FD ââ£â£ AC.
Therefore, AFDE, BDEF and DCEF are all parallelograms.
In parallelogram AFDE, we have:
∠A = ∠EDF (Opposite angles are equal)
In parallelogram BDEF, we have:
∠B = ∠DEF (Opposite angles are equal)
In parallelogram DCEF, we have:
∠ C = ∠â DFE (Opposite angles are equal)
Page No 347:
Question 12:
â ABC is shown below. D, E and F are the midpoints of sides BC, CA and AB, respectively.
As F and E are the mid points of sides AB and AC of â ABC.
∴ FE â£â£ BC (By mid point theorem)
Similarly, DE ââ£â£ FB and FD ââ£â£ AC.
Therefore, AFDE, BDEF and DCEF are all parallelograms.
In parallelogram AFDE, we have:
∠A = ∠EDF (Opposite angles are equal)
In parallelogram BDEF, we have:
∠B = ∠DEF (Opposite angles are equal)
In parallelogram DCEF, we have:
∠ C = ∠â DFE (Opposite angles are equal)
Answer:
Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join AC, a diagonal of the rectangle.
In â ABC, we have:
∴ PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in â DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Now, PQ â£â£ AC and SR â£â£ AC
⇒ PQ â£â£ SR
So, PQRS is a parallelogram.
Now, in âSAP and âQBP, we have:
AS = BQ
∠A = ∠B = 90o
AP = BP
i.e., âSAP ≅ âQBPâ
∴ PS = PQ ...(ii)
Similarly, âSDR ≅ âQCRâ
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.
Page No 347:
Question 13:
Let ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join AC, a diagonal of the rectangle.
In â ABC, we have:
∴ PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in â DAC, the points S and R are the mid points of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Now, PQ â£â£ AC and SR â£â£ AC
⇒ PQ â£â£ SR
So, PQRS is a parallelogram.
Now, in âSAP and âQBP, we have:
AS = BQ
∠A = ∠B = 90o
AP = BP
i.e., âSAP ≅ âQBPâ
∴ PS = PQ ...(ii)
Similarly, âSDR ≅ âQCRâ
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PQ = SR = RQ
Hence, PQRS is a rhombus.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In â ABC, we have:
PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in âDAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Now, PQ â£â£ AC and SR â£â£ AC ⇒ PQ â£â£ SR
So, PQRS is a parallelogram.
We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90o
âNow, RQâ£â£ DB
⇒RE â£â£ FO
Also, SRâ£â£ AC
⇒FR â£â£ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90oâ (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o.
∴â PQRS is a rectangle.
Page No 347:
Question 14:
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Let ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals, AC and BD.
In â ABC, we have:
PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in âDAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Now, PQ â£â£ AC and SR â£â£ AC ⇒ PQ â£â£ SR
So, PQRS is a parallelogram.
We know that the diagonals of a rhombus bisect each other at right angles.
∴ ∠EOF = 90o
âNow, RQâ£â£ DB
⇒RE â£â£ FO
Also, SRâ£â£ AC
⇒FR â£â£ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90oâ (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o.
∴â PQRS is a rectangle.
Answer:
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Let ABCD be a square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let O be the intersection point of AC and BD.
In â ABC, we have:
∴ PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in âDAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Now, PQ â£â£ AC and SR â£â£ AC ⇒ PQ â£â£ SR
So, PQRS is a parallelogram.
Now, in âSAP and âQBP, we have:
AS = BQ
∠A = ∠B = 90o
AP = BP
i.e., âSAP ≅ âQBPâ
∴ PS = PQ ...(ii)
Similarly, âSDR ≅ âRCQâ
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PS = SR = RQ ...(iv)
We know that the diagonals of a square bisect each other at right angles.
∴ ∠âEOF = 90o
âNow, RQ â£â£ DB
⇒RE â£â£ FO
Also, SR â£â£ AC
⇒FR â£â£ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90oâ (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o and PQ = PS = SR = RQ.
∴â PQRS is a square.
Page No 347:
Question 15:
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Let ABCD be a square and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.
Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let O be the intersection point of AC and BD.
In â ABC, we have:
∴ PQ â£â£ AC and PQ = AC [By midpoint theorem]
Again, in âDAC, the points S and R are the midpoints of AD and DC, respectively.
∴ SR â£â£ AC and SR = AC [By midpoint theorem]
Now, PQ â£â£ AC and SR â£â£ AC ⇒ PQ â£â£ SR
So, PQRS is a parallelogram.
Now, in âSAP and âQBP, we have:
AS = BQ
∠A = ∠B = 90o
AP = BP
i.e., âSAP ≅ âQBPâ
∴ PS = PQ ...(ii)
Similarly, âSDR ≅ âRCQâ
∴ SR = RQ ...(iii)
From (i), (ii) and (iii), we have:
PQ = PS = SR = RQ ...(iv)
We know that the diagonals of a square bisect each other at right angles.
∴ ∠âEOF = 90o
âNow, RQ â£â£ DB
⇒RE â£â£ FO
Also, SR â£â£ AC
⇒FR â£â£ OE
∴ OERF is a parallelogram.
So, ∠FRE = ∠EOF = 90oâ (Opposite angles are equal)
Thus, PQRS is a parallelogram with ∠R = 90o and PQ = PS = SR = RQ.
∴â PQRS is a square.
Answer:
Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA, respectively.
Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.
In ΔABD, S and P are the midpoints of AD and AB, respectively.
∴â SP || BD and SP = BD ... (i) (By midpoint theorem)
Similarly in Δ BCD, we have:
QR || BD and QR = BD ... (ii) (By midpoint theorem)
From equations (i) and (ii), we get:
SP || BD || QR
∴ SP || QR and SP = QR [Each equal to BD]
In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.
∴ SPQR is a parallelogram.
We know that the diagonals of a parallelogram bisect each other.
∴ PR and QS bisect each other.
Page No 347:
Question 16:
Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of sides AB, BC, CD, and DA, respectively.
Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.
In ΔABD, S and P are the midpoints of AD and AB, respectively.
∴â SP || BD and SP = BD ... (i) (By midpoint theorem)
Similarly in Δ BCD, we have:
QR || BD and QR = BD ... (ii) (By midpoint theorem)
From equations (i) and (ii), we get:
SP || BD || QR
∴ SP || QR and SP = QR [Each equal to BD]
In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.
∴ SPQR is a parallelogram.
We know that the diagonals of a parallelogram bisect each other.
∴ PR and QS bisect each other.
Answer:
![](/img/site_content/ask-answer/loader.gif)
To prove: KLMN is a rhombus.
Proof:
In âADC,
Since, K and L are the mid-points of sides AD and CD, respectively.
So, KL || AC and KL = AC ...(1)
Similarly, in âABC,
Since, M and N are the mid-points of sides BC and AB, respectively.
So, NM || AC and NM = AC ...(2)
From (1) and (2), we get
But this a pair of opposite sides of the quadrilateral KLMN.
So, KLMN is a parallelogram.
Now, in âABD,
Since, K and N are the mid-points of sides AD and AB, respectively.
So, KN || BD and KN = BD ...(3)
But BD = AC (Given)
BD = AC
KN = NM [From (2) and (3)]
But these are a pair of adjacent sides of the parallelogram KLMN.
Hence, KLMN is a rhombus.
Page No 347:
Question 17:
![](/img/site_content/ask-answer/loader.gif)
To prove: KLMN is a rhombus.
Proof:
In âADC,
Since, K and L are the mid-points of sides AD and CD, respectively.
So, KL || AC and KL = AC ...(1)
Similarly, in âABC,
Since, M and N are the mid-points of sides BC and AB, respectively.
So, NM || AC and NM = AC ...(2)
From (1) and (2), we get
But this a pair of opposite sides of the quadrilateral KLMN.
So, KLMN is a parallelogram.
Now, in âABD,
Since, K and N are the mid-points of sides AD and AB, respectively.
So, KN || BD and KN = BD ...(3)
But BD = AC (Given)
BD = AC
KN = NM [From (2) and (3)]
But these are a pair of adjacent sides of the parallelogram KLMN.
Hence, KLMN is a rhombus.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In quadrilateral ABCD, AC BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.
To prove: PQRS is a rectangle.
Proof:
In ΔABC, P and Q are mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = AC (Mid-point theorem) ...(1)
Similarly, in ΔACD,
So, R and S are mid-points of sides CD and AD, respectively.
∴ SR || AC and SR = AC (Mid-point theorem) ...(2)
From (1) and (2), we get
PQ || SR and PQ = SR
But this is a pair of opposite sides of the quadrilateral PQRS,
So, PQRS is parallelogram.
Now, in ΔBCD, Q and R are mid-points of BC and CD, respectively.
∴ QR || BD and QR = BD (Mid-point theorem) ...(3)
From (2) and (3), we get
SR || AC and QR || BD
But, AC ⊥ BD (Given)
∴ RS ⊥ QR
Hence, PQRS is a rectangle.
Page No 347:
Question 18:
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Given: In quadrilateral ABCD, AC BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.
To prove: PQRS is a rectangle.
Proof:
In ΔABC, P and Q are mid-points of AB and BC, respectively.
∴ PQ || AC and PQ = AC (Mid-point theorem) ...(1)
Similarly, in ΔACD,
So, R and S are mid-points of sides CD and AD, respectively.
∴ SR || AC and SR = AC (Mid-point theorem) ...(2)
From (1) and (2), we get
PQ || SR and PQ = SR
But this is a pair of opposite sides of the quadrilateral PQRS,
So, PQRS is parallelogram.
Now, in ΔBCD, Q and R are mid-points of BC and CD, respectively.
∴ QR || BD and QR = BD (Mid-point theorem) ...(3)
From (2) and (3), we get
SR || AC and QR || BD
But, AC ⊥ BD (Given)
∴ RS ⊥ QR
Hence, PQRS is a rectangle.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In quadrilateral ABCD, AC = BD and AC ⊥ BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.
To prove: PQRS is a square.
Construction: Join AC and BD.
Proof:
In ΔABC,
P and Q are mid-points of AB and BC, respectively.
PQ || AC and PQ = AC (Mid-point theorem) ...(1)
Similarly, in ΔACD,
R and S are mid-points of sides CD and AD, respectively.
SR || AC and SR = AC (Mid-point theorem) ...(2)
From (1) and (2), we get
PQ || SR and PQ = SR
But this a pair of opposite sides of the quadrilateral PQRS.
So, PQRS is parallelogram.
Now, in ΔBCD,
Q and R are mid-points of sides BC and CD, respectively.
QR || BD and QR = BD (Mid-point theorem) ...(3)
From (2) and (3), we get
RS || AC and QR || BD
But, AC ⊥ BD (Given)
∴ RS ⊥ QR
But this a pair of adjacent sides of the parallelogram PQRS.
So, PQRS is a rectangle.
Again, AC = BD (Given)
AC = BD
RS = QR [From (2) and (3)]
But this a pair of adjacent sides of the rectangle PQRS.
Hence, PQRS is a square.
Page No 349:
Question 1:
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Given: In quadrilateral ABCD, AC = BD and AC ⊥ BD. P, Q, R and S are the mid-points of AB, BC, CD and AD, respectively.
To prove: PQRS is a square.
Construction: Join AC and BD.
Proof:
In ΔABC,
P and Q are mid-points of AB and BC, respectively.
PQ || AC and PQ = AC (Mid-point theorem) ...(1)
Similarly, in ΔACD,
R and S are mid-points of sides CD and AD, respectively.
SR || AC and SR = AC (Mid-point theorem) ...(2)
From (1) and (2), we get
PQ || SR and PQ = SR
But this a pair of opposite sides of the quadrilateral PQRS.
So, PQRS is parallelogram.
Now, in ΔBCD,
Q and R are mid-points of sides BC and CD, respectively.
QR || BD and QR = BD (Mid-point theorem) ...(3)
From (2) and (3), we get
RS || AC and QR || BD
But, AC ⊥ BD (Given)
∴ RS ⊥ QR
But this a pair of adjacent sides of the parallelogram PQRS.
So, PQRS is a rectangle.
Again, AC = BD (Given)
AC = BD
RS = QR [From (2) and (3)]
But this a pair of adjacent sides of the rectangle PQRS.
Hence, PQRS is a square.
Answer:
(b) 73°â
Explanation:
Let the measure of the fourth angle be xo.
Since the sum of the angles of a quadrilateral is 360o, we have:
80o + 95o + 112o + x = 360o
⇒ 287o + x = 360o â
⇒ x = 73o
Hence, the measure of the fourth angle is 73o.
Page No 349:
Question 2:
(b) 73°â
Explanation:
Let the measure of the fourth angle be xo.
Since the sum of the angles of a quadrilateral is 360o, we have:
80o + 95o + 112o + x = 360o
⇒ 287o + x = 360o â
⇒ x = 73o
Hence, the measure of the fourth angle is 73o.
Answer:
(b) 60°â
Explanation:
Let ∠A = 3xâ, ∠B = 4x, ∠C = 5x and ∠D = 6x.
Since the sum of the angles of a quadrilateral is 360o, we have:
3x + 4x + 5x + 6x = 360o
⇒ 18x = 360o â
⇒ x = 20o
∴ ∠A = 60oâ, ∠B = 80o, ∠C = 100o and ∠D = 120o
âHence, the smallest angle is 60.
Page No 349:
Question 3:
(b) 60°â
Explanation:
Let ∠A = 3xâ, ∠B = 4x, ∠C = 5x and ∠D = 6x.
Since the sum of the angles of a quadrilateral is 360o, we have:
3x + 4x + 5x + 6x = 360o
⇒ 18x = 360o â
⇒ x = 20o
∴ ∠A = 60oâ, ∠B = 80o, ∠C = 100o and ∠D = 120o
âHence, the smallest angle is 60.
Answer:
(c) 45°
Explanation:
∠B = 180o − ∠A
⇒ ∠B = 180o − 75o = 105o
âNow, ∠B =â ∠ABD + ∠CBD
⇒ââ 105oâ = ∠ABD + 60o
⇒ ∠ABDâ = 105o − 60o = 45o
⇒ ∠ABD = â∠BDCâ = 45o (Alternate angles)
Page No 349:
Question 4:
(c) 45°
Explanation:
∠B = 180o − ∠A
⇒ ∠B = 180o − 75o = 105o
âNow, ∠B =â ∠ABD + ∠CBD
⇒ââ 105oâ = ∠ABD + 60o
⇒ ∠ABDâ = 105o − 60o = 45o
⇒ ∠ABD = â∠BDCâ = 45o (Alternate angles)
Answer:
![](/img/site_content/ask-answer/loader.gif)
We know that diagonals of rhombus bisect each other at 90°.
Then, in ΔBOC,
90° + 50° + ∠OBC = 180° (Angle sum property of triangle)
∠OBC = 180° 140°
∠OBC = 40°
But ∠OBC = ∠ADB (Alternate interior angles)
Thus, ∠ADB = 40°
Hence, the corerct option is (a).
Page No 349:
Question 5:
![](/img/site_content/ask-answer/loader.gif)
We know that diagonals of rhombus bisect each other at 90°.
Then, in ΔBOC,
90° + 50° + ∠OBC = 180° (Angle sum property of triangle)
∠OBC = 180° 140°
∠OBC = 40°
But ∠OBC = ∠ADB (Alternate interior angles)
Thus, ∠ADB = 40°
Hence, the corerct option is (a).
Answer:
(d) Rectangle.
The diagonals of a rectangle are equal.
Page No 349:
Question 6:
(d) Rectangle.
The diagonals of a rectangle are equal.
Answer:
(d) rhombus
The diagonals of a rhombus bisect each other at right angles.
Page No 350:
Question 7:
(d) rhombus
The diagonals of a rhombus bisect each other at right angles.
Answer:
(a) 10 cm
Explanation:
Let ABCD be the rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 16 cm and BD = 12 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â ââAOB is a right angle triangle, in which OA = AC /2 = 16/2 = 8 cm and OB = BD/2 = 12/2 = 6 cm.
Now, AB2= OA2 + OB2 [Pythagoras theorem]
⇒ âAB2 = (8)2 + (6)2
⇒â AB2 =â 64 + 36 = 100
⇒â AB =â 10 cm
Hence, the side of the rhombus is 10 cm.
Page No 350:
Question 8:
(a) 10 cm
Explanation:
Let ABCD be the rhombus.
∴ AB = BC = CD = DA
Here, AC and BD are the diagonals of ABCD, where AC = 16 cm and BD = 12 cm.
Let the diagonals intersect each other at O.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â ââAOB is a right angle triangle, in which OA = AC /2 = 16/2 = 8 cm and OB = BD/2 = 12/2 = 6 cm.
Now, AB2= OA2 + OB2 [Pythagoras theorem]
⇒ âAB2 = (8)2 + (6)2
⇒â AB2 =â 64 + 36 = 100
⇒â AB =â 10 cm
Hence, the side of the rhombus is 10 cm.
Answer:
(b) 12 cm
Explanation:
Let ABCD be the rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of the rhombus.
Let AC be x and BD be 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â âAOB is a right angle triangle in which OA = AC 2 = x
Now, AB2= OA2 + OB2 [Pythagoras theorem]
Page No 350:
Question 9:
(b) 12 cm
Explanation:
Let ABCD be the rhombus.
∴ AB = BC = CD = DA = 10 cm
Let AC and BD be the diagonals of the rhombus.
Let AC be x and BD be 16 cm and O be the intersection point of the diagonals.
We know that the diagonals of a rhombus are perpendicular bisectors of each other.
∴â âAOB is a right angle triangle in which OA = AC 2 = x
Now, AB2= OA2 + OB2 [Pythagoras theorem]
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In rectangle ABCD, ∠OAD = 35°.
Since, ∠BAD = 90°
∠OAB = 90° 35° = 55°
In ΔOAB,
Since, OA = OB (Diagonals of a rectangle are equal and bisect each other)
∠OAB = ∠OBA = 55° (Angles opposite to equal sides are equal)
Now, in ΔODA,
55° + 55° + ∠DOA = 180° (Angle sum property of a triangle)
∠DOA = 180° 110°
∠DOA = 70°
Thus, the acute angle between the diagonals is 70°.
Hence, the correct option is (b).
Page No 350:
Question 10:
![](/img/site_content/ask-answer/loader.gif)
Given: In rectangle ABCD, ∠OAD = 35°.
Since, ∠BAD = 90°
∠OAB = 90° 35° = 55°
In ΔOAB,
Since, OA = OB (Diagonals of a rectangle are equal and bisect each other)
∠OAB = ∠OBA = 55° (Angles opposite to equal sides are equal)
Now, in ΔODA,
55° + 55° + ∠DOA = 180° (Angle sum property of a triangle)
∠DOA = 180° 110°
∠DOA = 70°
Thus, the acute angle between the diagonals is 70°.
Hence, the correct option is (b).
Answer:
(c) Rectangle
Explanation:
∠A = ∠B
âThen ∠A + ∠B = 180o
⇒ 2∠A = 180o
⇒ ∠Aâ = 90o
⇒ ∠Aâ =â ∠Bâ =â∠Câ =ââ∠D = 90o
∴â The parallelogram is a rectangle.
Page No 350:
Question 11:
(c) Rectangle
Explanation:
∠A = ∠B
âThen ∠A + ∠B = 180o
⇒ 2∠A = 180o
⇒ ∠Aâ = 90o
⇒ ∠Aâ =â ∠Bâ =â∠Câ =ââ∠D = 90o
∴â The parallelogram is a rectangle.
Answer:
(b) 50o
ââExplanation:
∠C = 70o and ∠D = 30o
âThen ∠A + ∠B = 360o - (70 +30)o = 260o
∴ (∠A +∠B) =(260o) = 130o
In ââ AOB, we have:
∠AOBâ = 180o - [(∠A +∠B)â]
⇒ ∠AOBâ = â180 - 130 = 50o
Page No 350:
Question 12:
(b) 50o
ââExplanation:
∠C = 70o and ∠D = 30o
âThen ∠A + ∠B = 360o - (70 +30)o = 260o
∴ (∠A +∠B) =(260o) = 130o
In ââ AOB, we have:
∠AOBâ = 180o - [(∠A +∠B)â]
⇒ ∠AOBâ = â180 - 130 = 50o
Answer:
(d) 90°
Explanation:
Sum of two adjacent angles = 180o
Now, sum of angle bisectors of two adjacent angles =
∴ Intersection angle of bisectors of two adjacent angles = 180o − 90o = 90o
Page No 350:
Question 13:
(d) 90°
Explanation:
Sum of two adjacent angles = 180o
Now, sum of angle bisectors of two adjacent angles =
∴ Intersection angle of bisectors of two adjacent angles = 180o − 90o = 90o
Answer:
(c) Rectangle
The bisectors of the angles of a parallelogram encloses a rectangle.
Page No 350:
Question 14:
(c) Rectangle
The bisectors of the angles of a parallelogram encloses a rectangle.
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In quadrilateral ABCD, AS, BQ, CQ and DS are angle bisectors of angles A, B, C and D, respectively.
∠QPS = ∠APB (Vertically opposite angles) ...(1)
In APB,
∠APB + ∠PAB + ∠ABP = 180° (Angle sum property of triangle.)
∠APB + ∠A + ∠B = 180°
∠APB = 180° – (∠A + ∠B) ...(2)
From (1) and (2), we get
∠QPS = 180° – (∠A + ∠B) ...(3)
Similarly, ∠QRS = 180° – (∠C + ∠D) ...(4)
From (3) and (4), we get
∠QPS + ∠QRS = 360° – (∠A + ∠B + ∠C + ∠D)
= 360° – (360°)
= 360° – 180°
= 180°
So, PQRS is a quadrilateral whose opposite angles are supplementary.
Hence, the correct option is (d).
Page No 350:
Question 15:
![](/img/site_content/ask-answer/loader.gif)
Given: In quadrilateral ABCD, AS, BQ, CQ and DS are angle bisectors of angles A, B, C and D, respectively.
∠QPS = ∠APB (Vertically opposite angles) ...(1)
In APB,
∠APB + ∠PAB + ∠ABP = 180° (Angle sum property of triangle.)
∠APB + ∠A + ∠B = 180°
∠APB = 180° – (∠A + ∠B) ...(2)
From (1) and (2), we get
∠QPS = 180° – (∠A + ∠B) ...(3)
Similarly, ∠QRS = 180° – (∠C + ∠D) ...(4)
From (3) and (4), we get
∠QPS + ∠QRS = 360° – (∠A + ∠B + ∠C + ∠D)
= 360° – (360°)
= 360° – 180°
= 180°
So, PQRS is a quadrilateral whose opposite angles are supplementary.
Hence, the correct option is (d).
Answer:
(d) parallelogram
The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.
Page No 350:
Question 16:
(d) parallelogram
The figure formed by joining the mid points of the adjacent sides of a quadrilateral is a parallelogram.
Answer:
(b) Square
The figure formed by joining the mid points of the adjacent sides of a square is a square.
Page No 350:
Question 17:
(b) Square
The figure formed by joining the mid points of the adjacent sides of a square is a square.
Answer:
(d) parallelogram.
The figure made by joining the mid points of the adjacent sides of a parallelogram is a parallelogram.
Page No 350:
Question 18:
(d) parallelogram.
The figure made by joining the mid points of the adjacent sides of a parallelogram is a parallelogram.
Answer:
(a) rhombus
The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.
Page No 351:
Question 19:
(a) rhombus
The figure formed by joining the mid points of the adjacent sides of a rectangle is a rhombus.
Answer:
(c) Rectangle
The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.
Page No 351:
Question 20:
(c) Rectangle
The figure formed by joining the mid points of the adjacent sides of a rhombus is a rectangle.
Answer:
Since,
The quadrilateral formed by joining the mid-points of the sides of a parallelogram is parallelogram ,
The quadrilateral formed by joining the mid-points of the sides of a rectangle is rhombus,
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus, and
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular to each other is rectangle.
Hence, the correct option is (d).
Page No 351:
Question 21:
Since,
The quadrilateral formed by joining the mid-points of the sides of a parallelogram is parallelogram ,
The quadrilateral formed by joining the mid-points of the sides of a rectangle is rhombus,
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus, and
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular to each other is rectangle.
Hence, the correct option is (d).
Answer:
Given:
The quadrilateral ABCD is a rhombus.
So, the sides AB, BC, CD and AD are equal.
Now, in we have
(Using mid-point theorem) ...(1)
Similarly, in
...(2)
As, BC = DC
[From (1) and (2)]
So, QS = PR
Thus, the diagonals of PQRS are equal.
Hence, the correct option is (c).
Page No 351:
Question 22:
Given:
The quadrilateral ABCD is a rhombus.
So, the sides AB, BC, CD and AD are equal.
Now, in we have
(Using mid-point theorem) ...(1)
Similarly, in
...(2)
As, BC = DC
[From (1) and (2)]
So, QS = PR
Thus, the diagonals of PQRS are equal.
Hence, the correct option is (c).
Answer:
Since,
The quadrilateral formed by joining the mid-points of the sides of a rhombus is rectangle,
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus,
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular is rectangle, and
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal and perpendicular is square.
Hence, the correct option is (d).
Page No 351:
Question 23:
Since,
The quadrilateral formed by joining the mid-points of the sides of a rhombus is rectangle,
The quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal is rhombus,
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals perpendicular is rectangle, and
âThe quadrilateral formed by joining the mid-points of the sides of a quadrilateral with diagonals equal and perpendicular is square.
Hence, the correct option is (d).
Answer:
(c) 72°â
Explanation:
Let ABCD be a parallelogram.
∴ ∠A = ∠C and ∠B = ∠D (Opposite angles)
Let ∠A = x and ∠âB =
∴ ∠âA + ∠B = 180o (Adjacent angles are supplementary)
⇒ x+ x= 180o
⇒
⇒ x = 108o
∴ ∠B = (108o) = 72o
Hence, ∠A = ∠C = 108o and ∠B = ∠D = 72o
Page No 351:
Question 24:
(c) 72°â
Explanation:
Let ABCD be a parallelogram.
∴ ∠A = ∠C and ∠B = ∠D (Opposite angles)
Let ∠A = x and ∠âB =
∴ ∠âA + ∠B = 180o (Adjacent angles are supplementary)
⇒ x+ x= 180o
⇒
⇒ x = 108o
∴ ∠B = (108o) = 72o
Hence, ∠A = ∠C = 108o and ∠B = ∠D = 72o
Answer:
(c)112°â
Explanation:
Let ABCD is a parallelogram.
∴ ∠âA = ∠C and ∠âB = ∠D (Opposite angles)
Let ∠A be the smallest angle whose measure is x.
∴â∠B = (2x − 24)o
Now, ∠âA + ∠B = 180o (Adjacent angles are supplementary)
⇒ x + 2x − 24o = 180o
⇒ 3x = 204o
⇒ x = 68o
∴â∠âB = 2 ⨯ 68o − 24o = 112o
Hence, ∠A = ∠C = 68o and ∠B = ∠D = 112o
Page No 351:
Question 25:
(c)112°â
Explanation:
Let ABCD is a parallelogram.
∴ ∠âA = ∠C and ∠âB = ∠D (Opposite angles)
Let ∠A be the smallest angle whose measure is x.
∴â∠B = (2x − 24)o
Now, ∠âA + ∠B = 180o (Adjacent angles are supplementary)
⇒ x + 2x − 24o = 180o
⇒ 3x = 204o
⇒ x = 68o
∴â∠âB = 2 ⨯ 68o − 24o = 112o
Hence, ∠A = ∠C = 68o and ∠B = ∠D = 112o
Answer:
(c) Trapezium
Explanation:
Let the angles be (3x), (7x), (6x) and (4x)â.
Then 3x + 7x + 6x + 4x = 360o
∴ x = 18o
âThus, the angles are 3 ⨯18o = 54o, â7 ⨯ 18o = 126o, â6 ⨯ 18o = 108o and â4 ⨯18o = 72o.
But 54o + 126o = 180o and 72o + 108o = 180o
∴ ABCD is a trapezium.
Page No 351:
Question 26:
(c) Trapezium
Explanation:
Let the angles be (3x), (7x), (6x) and (4x)â.
Then 3x + 7x + 6x + 4x = 360o
∴ x = 18o
âThus, the angles are 3 ⨯18o = 54o, â7 ⨯ 18o = 126o, â6 ⨯ 18o = 108o and â4 ⨯18o = 72o.
But 54o + 126o = 180o and 72o + 108o = 180o
∴ ABCD is a trapezium.
Answer:
(c) Opposite angles are bisected by the diagonals.
Page No 351:
Question 27:
(c) Opposite angles are bisected by the diagonals.
Answer:
(c) Rectangle
If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a rectangle.
Page No 352:
Question 28:
(c) Rectangle
If APB and CQD are two parallel lines, then the bisectors of ∠APQ, ∠BPQ, ∠CQP and ∠PQD enclose a rectangle.
Answer:
(c)â 60°
Explanation:
∠BAD = ∠BCD = 75o [Opposite angles are equal]
In â BCD, ∠ C = 75o
∴ â∠CBD = 180o− (75o + 45o) = 60o
Page No 352:
Question 29:
(c)â 60°
Explanation:
∠BAD = ∠BCD = 75o [Opposite angles are equal]
In â BCD, ∠ C = 75o
∴ â∠CBD = 180o− (75o + 45o) = 60o
Answer:
(c) A < B
Explanation:
Let h be the height of parallelogram.
Then clearly, h < b
∴ âA = a ⨯â h < a ⨯ b = B
Hence, A < B
Page No 352:
Question 30:
(c) A < B
Explanation:
Let h be the height of parallelogram.
Then clearly, h < b
∴ âA = a ⨯â h < a ⨯ b = B
Hence, A < B
Answer:
(b) AF = 2 AB
Explanation:
Iân parallelogram ABCD, we have:
AB || DC
∠DCE = ∠â EBF (Alternate interior angles)
In â DCE and â â BFE, we have:
∠DCE = ∠ EBF (Proved above)
BE = CE ( Given)
i.e., â DCE ≅â â BFE (By ASA congruence rule)
∴ DC = BF (CPCT)
But DC= AB, as ABCD is a parallelogram.
∴ DC = AB = BF ...(i)
Now, AF = AB + BF ...(ii)
From (i), we get:
∴ AF = AB + AB = 2AB
Page No 352:
Question 31:
(b) AF = 2 AB
Explanation:
Iân parallelogram ABCD, we have:
AB || DC
∠DCE = ∠â EBF (Alternate interior angles)
In â DCE and â â BFE, we have:
∠DCE = ∠ EBF (Proved above)
BE = CE ( Given)
i.e., â DCE ≅â â BFE (By ASA congruence rule)
∴ DC = BF (CPCT)
But DC= AB, as ABCD is a parallelogram.
∴ DC = AB = BF ...(i)
Now, AF = AB + BF ...(ii)
From (i), we get:
∴ AF = AB + AB = 2AB
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: In âABC, M, N, D and E are the mid-points of BP, CP, AB and AC, respectivley.
In âABP,
D and M are the mid-points of AB,and BP, respectively. (Given)
BM = AP and BM || AP (Mid-point theorem) ...(i)
Again, in âACP,
E and N are the mid-points of AC,and CP, respectively. (Given)
EN = AP and EN || AP (Mid-point theorem) ...(ii)
From (i) and (ii), we get
BM = EN and BM || EN
But this a pair of opposite sides of the quadrilateral DENM.
So, DENM is a parallelgram.
Hence, the correct option is (b).
Page No 352:
Question 32:
![](/img/site_content/ask-answer/loader.gif)
Given: In âABC, M, N, D and E are the mid-points of BP, CP, AB and AC, respectivley.
In âABP,
D and M are the mid-points of AB,and BP, respectively. (Given)
BM = AP and BM || AP (Mid-point theorem) ...(i)
Again, in âACP,
E and N are the mid-points of AC,and CP, respectively. (Given)
EN = AP and EN || AP (Mid-point theorem) ...(ii)
From (i) and (ii), we get
BM = EN and BM || EN
But this a pair of opposite sides of the quadrilateral DENM.
So, DENM is a parallelgram.
Hence, the correct option is (b).
Answer:
(b)
Explanation:
Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in â DAB, E is the midpoint of AD and EM || AB.
∴ M is the mid point of BD and EM =
Similarly, M is the mid point of BD and MF || DC.
i.e., F is the midpoint of BC and MF =
Page No 352:
Question 33:
(b)
Explanation:
Suppose ABCD is a trapezium.
Draw EF parallel to AB.
Join BD to cut EF at M.
Now, in â DAB, E is the midpoint of AD and EM || AB.
∴ M is the mid point of BD and EM =
Similarly, M is the mid point of BD and MF || DC.
i.e., F is the midpoint of BC and MF =
Answer:
(d)
Explanation:
Join CF and produce it to cut AB at G.
Then âCDF ≅ âGBF [âµ DF = BF, ∠âDCF = ∠âBGF and ∠âCDF = ∠âGBF]
∴ CD = GB
Thus, in ââCAG, the points E and F are the mid points of AC and CG, respectively.
∴ EF = =
Page No 352:
Question 34:
(d)
Explanation:
Join CF and produce it to cut AB at G.
Then âCDF ≅ âGBF [âµ DF = BF, ∠âDCF = ∠âBGF and ∠âCDF = ∠âGBF]
∴ CD = GB
Thus, in ââCAG, the points E and F are the mid points of AC and CG, respectively.
∴ EF = =
Answer:
(c) â90°
Explanation:
∠B = ∠D
⇒ ∠B = ∠Dâ
⇒â ∠ADB = ∠âABD
∴ âABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of âABD.
∴ AM ⊥ BD and, hence, â∠AMB =ââ 90°
Page No 352:
Question 35:
(c) â90°
Explanation:
∠B = ∠D
⇒ ∠B = ∠Dâ
⇒â ∠ADB = ∠âABD
∴ âABD is an isosceles triangle and M is the midpoint of BD. We can also say that M is the median of âABD.
∴ AM ⊥ BD and, hence, â∠AMB =ââ 90°
Answer:
(c) âAC2 + BD2 = 4AB2
Explanation:
We know that the diagonals of a rhombus bisect each other at right angles.
Here, OA = AC, OB = BD and ∠âAOB â= 90°
Now, AB2= OA2 + OB2 = (AC)2 + (BD)2
∴ 4AB2 = (AC2 + BD2)
Page No 353:
Question 36:
(c) âAC2 + BD2 = 4AB2
Explanation:
We know that the diagonals of a rhombus bisect each other at right angles.
Here, OA = AC, OB = BD and ∠âAOB â= 90°
Now, AB2= OA2 + OB2 = (AC)2 + (BD)2
∴ 4AB2 = (AC2 + BD2)
Answer:
(c) BC2 + AD2 + 2AB.CD
Explanation:
Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
∴â DEFC is a parallelogram and EF = CD.
In âABC, ∠B is acute.
∴ AC2= BC2 + AB2 - 2AB.AE
In âABD, ∠A is acute.â
∴ âBD2 = AD2 + AB2 - 2AB.AF
∴ âAC2 + BD2 = (BC2 + AD2) + (AB2+ AB2 ) - 2AB(AE + BF)
= (BC2 + AD2) + 2AB(AB - AE - BF) [âµ AB = AE + EF + FB and AB - AE = BE]
= (BC2 + AD2) + 2AB(BE - BF)
= (BC2 + AD2) + 2AB.EF
∴ AC2 + BD2 = â(BC2 + AD2) + 2AB.CD
Page No 353:
Question 37:
(c) BC2 + AD2 + 2AB.CD
Explanation:
Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
∴â DEFC is a parallelogram and EF = CD.
In âABC, ∠B is acute.
∴ AC2= BC2 + AB2 - 2AB.AE
In âABD, ∠A is acute.â
∴ âBD2 = AD2 + AB2 - 2AB.AF
∴ âAC2 + BD2 = (BC2 + AD2) + (AB2+ AB2 ) - 2AB(AE + BF)
= (BC2 + AD2) + 2AB(AB - AE - BF) [âµ AB = AE + EF + FB and AB - AE = BE]
= (BC2 + AD2) + 2AB(BE - BF)
= (BC2 + AD2) + 2AB.EF
∴ AC2 + BD2 = â(BC2 + AD2) + 2AB.CD
Answer:
(d) 1:1
Area of a parallelogram = base ⨯ height
If both parallelograms stands on the same base and between the same parallels, then their heights are the same.
So, their areas will also be the same.
Page No 353:
Question 38:
(d) 1:1
Area of a parallelogram = base ⨯ height
If both parallelograms stands on the same base and between the same parallels, then their heights are the same.
So, their areas will also be the same.
Answer:
(b) â
AC
Explanation:
Let G be the mid point of FC. Join DG.
âIn âBCF, D is the mid point of BC and G is the mid point of FC.
∴ DG || BF
⇒ DG || EF
âIn â ADG, E is the mid point of AD and EF || DG.
i.e., F is the mid point of AG.
Now, AF = FG = GC [âµ G is the mid point of FC]
∴ AF =â
ACâ
Page No 353:
Question 39:
(b) â
AC
Explanation:
Let G be the mid point of FC. Join DG.
âIn âBCF, D is the mid point of BC and G is the mid point of FC.
∴ DG || BF
⇒ DG || EF
âIn â ADG, E is the mid point of AD and EF || DG.
i.e., F is the mid point of AG.
Now, AF = FG = GC [âµ G is the mid point of FC]
∴ AF =â
ACâ
Answer:
(a) 40°
Explanation:
∠OAD = â∠OCB = 30o (Alternate interior angles)
∠AOB + ∠BOC = 180o (Linear pair of angles)
∴ ∠BOC = 180o − 70o = 110o (∠ AOB = 70o)
In âBOC, we have:
∠OBC = 180o − (110o + 30o) = 40o
∴ â∠DBC = 40o
Page No 353:
Question 40:
(a) 40°
Explanation:
∠OAD = â∠OCB = 30o (Alternate interior angles)
∠AOB + ∠BOC = 180o (Linear pair of angles)
∴ ∠BOC = 180o − 70o = 110o (∠ AOB = 70o)
In âBOC, we have:
∠OBC = 180o − (110o + 30o) = 40o
∴ â∠DBC = 40o
Answer:
(c) I and II
Statements I and II are true. Statement III is false, as the triangle formed by joining the midpoints of the sides of an isosceles triangle is an isosceles triangle.
Page No 353:
Question 41:
(c) I and II
Statements I and II are true. Statement III is false, as the triangle formed by joining the midpoints of the sides of an isosceles triangle is an isosceles triangle.
Answer:
(b) II and III
Page No 353:
Question 42:
(b) II and III
Answer:
Since, the opposite angles of the quadrilateral PQRS are equal.
Quadrilateral PQRS is a parallelogram.
(Opposite sides of a parallelogram are equal.)
Page No 354:
Question 43:
Since, the opposite angles of the quadrilateral PQRS are equal.
Quadrilateral PQRS is a parallelogram.
(Opposite sides of a parallelogram are equal.)
Answer:
No, the statement is incorrect because the diagonals of a parallelogram bisect each other.
Page No 354:
Question 44:
No, the statement is incorrect because the diagonals of a parallelogram bisect each other.
Answer:
Since, in quadrilateral PQRS, ∠P + ∠S = 180°.
i.e. the sum of adjacent angles is 180.
So, PQRS is a parallelogram.
Page No 354:
Question 45:
Since, in quadrilateral PQRS, ∠P + ∠S = 180°.
i.e. the sum of adjacent angles is 180.
So, PQRS is a parallelogram.
Answer:
No, the statement is false because if all the four angles of a quadrilateral are less than 90, then the sum of all four angles will be less than 360.
Page No 354:
Question 46:
No, the statement is false because if all the four angles of a quadrilateral are less than 90, then the sum of all four angles will be less than 360.
Answer:
Yes, the statement is true because all the angles of a quadrilateral such as rectangle and square are right angles.
Page No 354:
Question 47:
Yes, the statement is true because all the angles of a quadrilateral such as rectangle and square are right angles.
Answer:
No, the statement is false because if all angles are greater than 90, then the sum of four obtuse angles will be greater than 360.
Page No 354:
Question 48:
No, the statement is false because if all angles are greater than 90, then the sum of four obtuse angles will be greater than 360.
Answer:
Since, the sum of all angles (i.e. 70° + 115° + 60° + 120° = 365).
So, we cannot form a quadrilateral with these angles.
Page No 354:
Question 49:
Since, the sum of all angles (i.e. 70° + 115° + 60° + 120° = 365).
So, we cannot form a quadrilateral with these angles.
Answer:
We know that, sum of all angles in a quadrilateral is 360°.
Let each angle of the quadrilateral be x.
x + x + x + x = 360°
⇒ 4x = 360°
⇒ x = 90°
⇒ All angles of the quadrilateral are 90°.
Hence, given quadrilateral is a rectangle.
Page No 354:
Question 50:
We know that, sum of all angles in a quadrilateral is 360°.
Let each angle of the quadrilateral be x.
x + x + x + x = 360°
⇒ 4x = 360°
⇒ x = 90°
⇒ All angles of the quadrilateral are 90°.
Hence, given quadrilateral is a rectangle.
Answer:
In ABC,
Since, D and E are respectively the mid-points of sides AB and BC. (Given)
So, DE = AC (Uing mid-point theorem)
But AC = 3.6 cm (Given)
DE = (3.6)
or, DE = 1.8 cm
Hence, the length of DE is 1.8 cm.
Page No 354:
Question 51:
In ABC,
Since, D and E are respectively the mid-points of sides AB and BC. (Given)
So, DE = AC (Uing mid-point theorem)
But AC = 3.6 cm (Given)
DE = (3.6)
or, DE = 1.8 cm
Hence, the length of DE is 1.8 cm.
Answer:
Since, the diagonals PR and QS of quadrilateral PQRS bisect each other. (Given)
So, PQRS is a parallelogram.
Now, (Adjacent angles are supplementary.)
Page No 354:
Question 52:
Since, the diagonals PR and QS of quadrilateral PQRS bisect each other. (Given)
So, PQRS is a parallelogram.
Now, (Adjacent angles are supplementary.)
Answer:
![](/img/site_content/ask-answer/loader.gif)
Given: Parallelograms BDEF and AFDE.
Since, BF = DE (Opposite sides of parallelogram BDEF) ...(i)
And, AF = DE (Opposite sides of parallelogram AFDE) ...(ii)
From (i) and (ii), we get
AF = FB
Page No 354:
Question 53:
![](/img/site_content/ask-answer/loader.gif)
Given: Parallelograms BDEF and AFDE.
Since, BF = DE (Opposite sides of parallelogram BDEF) ...(i)
And, AF = DE (Opposite sides of parallelogram AFDE) ...(ii)
From (i) and (ii), we get
AF = FB
Answer:
We know that if the diagonals of a âquadrilateral bisects each other, then it is a parallelogram.
∴ I gives the answer.
If the diagonals of a quadrilateral are equal, then it is not necessarily a âparallelogram.
∴ II does not give the answer.
Page No 355:
Question 54:
We know that if the diagonals of a âquadrilateral bisects each other, then it is a parallelogram.
∴ I gives the answer.
If the diagonals of a quadrilateral are equal, then it is not necessarily a âparallelogram.
∴ II does not give the answer.
Answer:
Clearly, I alone is not sufficient to answer the given question.
Also, II alone is not sufficient to answer the given question.â
However, both I and II together will give the answer.
∴ Hence, the correct answer is (c).
Page No 355:
Question 55:
Clearly, I alone is not sufficient to answer the given question.
Also, II alone is not sufficient to answer the given question.â
However, both I and II together will give the answer.
∴ Hence, the correct answer is (c).
Answer:
When the diagonals of a parallelogram are equal, it is either a rectangle or a square.
Also, if the diagonals intersects at a right angle, then it is a square.
∴ Both I and II together will give the answer.
Hence, the correct answer is (c).
Page No 355:
Question 56:
When the diagonals of a parallelogram are equal, it is either a rectangle or a square.
Also, if the diagonals intersects at a right angle, then it is a square.
∴ Both I and II together will give the answer.
Hence, the correct answer is (c).
Answer:
Page No 355:
Question 57:
Answer:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Explanation:
Fourth angle = 360o − (130o + 70o + 60o) = 100o
Clearly, reason (R) and assertion (A) are both true and the reason gives the assertion.
Hence, the correct answer is (a).
Page No 355:
Question 58:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
Explanation:
Fourth angle = 360o − (130o + 70o + 60o) = 100o
Clearly, reason (R) and assertion (A) are both true and the reason gives the assertion.
Hence, the correct answer is (a).
Answer:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
âExplanation:
Clearly, reason (R) and assertion (A) are both true and reason (R) gives assertion (A).
Hence, the correct answer is (a).
Page No 355:
Question 59:
(a) Both Assertion and Reason are true and Reason is a correct explanation of Assertion.
âExplanation:
Clearly, reason (R) and assertion (A) are both true and reason (R) gives assertion (A).
Hence, the correct answer is (a).
Answer:
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
âExplanation:
Clearly, reason (R) and assertion (A) are both true.
Page No 356:
Question 60:
(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
âExplanation:
Clearly, reason (R) and assertion (A) are both true.
Answer:
(d) Assertion is false and Reason is true.
âExplanation:
We can easily prove reason (R). So, reason (R) is true.
Page No 356:
Question 61:
(d) Assertion is false and Reason is true.
âExplanation:
We can easily prove reason (R). So, reason (R) is true.
Answer:
â(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
âExplanation:
Clearly, assertion (A) is true.
We can easily prove reason (R). So, (R) is also true.
Page No 356:
Question 62:
â(b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion.
âExplanation:
Clearly, assertion (A) is true.
We can easily prove reason (R). So, (R) is also true.
Answer:
(a) - (q), (b) - (r), (c) - (s), (d) - (p)
Page No 356:
Question 63:
(a) - (q), (b) - (r), (c) - (s), (d) - (p)
Answer:
â(a) - (r), (b) - (s), (c) - (p), (d) - (q)
Explanation:
(a) PQ = (AB+ CD) = (17) = 8.5 cm
(b) OR = (PR) = â(13) = 6.5 cm
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