Mathematics Solutions Solutions for Class 8 Maths Chapter 8 Quadrilateral : Constructions And Types are provided here with simple step-by-step explanations. These solutions for Quadrilateral : Constructions And Types are extremely popular among Class 8 students for Maths Quadrilateral : Constructions And Types Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Solutions Book of Class 8 Maths Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Solutions Solutions. All Mathematics Solutions Solutions for class Class 8 Maths are prepared by experts and are 100% accurate.

#### Question 1:

Construct the following quadrilaterals of given measures.
(1) In ☐ MORE, l(MO) = 5.8 cm, l(OR) = 4.4 cm, m∠M = 58$°$, m∠O = 105$°$, m∠R = 90$°$.
(2) Construct ☐ DEFG such that l(DE) = 4.5 cm, l(EF) = 6.5 cm, l(DG) = 5.5 cm, l(DF) = 7.2 cm, l(EG) = 7.8 cm.
(3) In ☐ ABCD l(AB) = 6.4 cm, l(BC) = 4.8 cm, m∠A = 70$°$, m∠B = 50$°$, m∠C = 140$°$.
(4) Construct ☐ LMNO such that l(LM) = l(LO) = 6 cm, l(ON) = l(NM) = 4.5 cm, l(OM) = 7.5 cm.

(1)
Steps of Construction:

Step 1:
Draw MO = 5.8 cm.

Step 2: Draw ∠MOX = 105$°$.

Step 3: With O as centre and radius 4.4 cm, draw an arc cutting ray OX at R.

Step 4: Draw ∠ORY = 90$°$.

Step 5: Draw ∠OMZ = 58$°$ such that ray MZ and ray RY intersect each other at E.

Here, MORE is the required quadrilateral.

(2)
Steps of Construction:

Step 1:
Draw DE = 4.5 cm.

Step 2: With D as centre and radius 7.2 cm, draw an arc.

Step 3: With E as centre and radius 6.5 cm, draw an arc cutting the previous arc at F.

Step 4: Join EF.

Step 5: With D as centre and radius 5.5 cm, draw an arc.

Step 6: With E as centre and radius 7.8 cm, draw an arc cutting the previous arc at G.

Step 7: Join DG and GF.

Here, DEFG is the required quadrilateral.

(3)
Steps of Construction:

Step 1:
Draw AB = 6.4 cm.

Step 2: Draw ∠ABX = 50$°$.

Step 3: With B as centre and radius 4.8 cm, draw an arc cutting ray BX at C.

Step 4: Draw ∠BCY = 140$°$.

Step 5: Draw ∠BAZ = 70$°$ such that ray AZ and ray CY intersect each other at D.

Here, ABCD is the required quadrilateral.

(4)
Steps of Construction:

Step 1:
Draw LM = 6 cm.

Step 2: With L as centre and radius 6 cm, draw an arc.

Step 3: With M as centre and radius 7.5 cm, draw an arc cutting the previous arc at O.

Step 4: Join OL.

Step 5: With O as centre and radius 4.5 cm, draw an arc.

Step 6: With M as centre and radius 4.5 cm, draw an arc cutting the previous arc at N.

Step 7: Join ON and MN.

Here, LMNO is the required quadrilateral.

#### Question 1:

Draw a rectangle ABCD such that l(AB) = 6.0 cm and l (BC) = 4.5 cm.

Steps of Construction:

Step 1:
Draw AB = 6 cm.

Step 2: Draw ∠ABX = 90º.

Step 3: With B as centre and radius 4.5 cm, draw an arc cutting the ray BX at C.

Step 4: With C as centre and radius 6 cm, draw an arc.

Step 5: With A as centre and radius 4.5 cm, draw an arc cutting the previous arc at D.

Step 6: Join AD and CD.

Here, ABCD is the required rectangle.

#### Question 2:

Draw a square WXYZ with side 5.2 cm.

Steps of Construction:

Step 1:
Draw WX = 5.2 cm.

Step 2: Draw ∠WXP = 90º.

Step 3: With X as centre and radius 5.2 cm, draw an arc cutting the ray XP at Y.

Step 4: With Y as centre and radius 5.2 cm, draw an arc.

Step 5: With W as centre and radius 5.2 cm, draw an arc cutting the previous arc at Z.

Step 6: Join YZ and WZ.

​Here, WXYZ is the required square.

#### Question 3:

Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75$°$.

Steps of Construction:

Step 1:
Draw KL = 4 cm.

Step 2: Draw ∠LKX = 75º.

Step 3: With K as centre and radius 4 cm, draw an arc cutting the ray KX at N.

Step 4: With N as centre and radius 4 cm, draw an arc.

Step 5: With L as centre and radius 4 cm, draw an arc cutting the previous arc at M.

Step 6: Join LM and NM.

​Here, KLMN is the required rhombus.

#### Question 4:

If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.

Suppose ABCD is a rectangle.

Here, segment AC is a diagonal and segment AD is one side of the rectangle ABCD.

l(AC) = 26 cm and l(AD) = 24 cm.

In right ∆ACD,

⇒ l(CD)= (26)2 − (24)2

⇒ l(CD)= 676 − 576 = 100

⇒ l(CD) = $\sqrt{100}$ = 10 cm

Thus, the other side of the rectangle is 10 cm.

#### Question 5:

Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.

ABCD is a rhombus.

Here, segment AC and segment BD are the diagonals of the rhombus ABCD.

l(AC) = 16 cm and l(BD) = 12 cm.

Diagonals of a rhombus are perpendicular bisectors of each other.

∴ m∠AOD = 90º

Also, l(OA) = $\frac{1}{2}$l(AC) = $\frac{1}{2}$ × 16 = 8 cm

l(OD) = $\frac{1}{2}$l(BD) = $\frac{1}{2}$ × 12 = 6 cm

In right ∆AOD,

⇒ l(AD)= 64 + 36 = 100

⇒ l(AD) $\sqrt{100}$ = 10 cm

All sides of a rhombus are equal.

∴ Perimeter of the rhombus ABCD = 4 × Side of a rhombus = 4 × 10 = 40 cm

Thus, the side and perimeter of the rhombus are 10 cm and 40 cm, respectively.

#### Question 6:

Find the length of diagonal of a square with side 8 cm.

Suppose ABCD is a square of side 8 cm.

Here, segment AC is a diagonal of square ABCD.

In ∆ABC,

l(AC)2l(AB)l(BC)2               (Pythagoras theorem)

⇒ l(AC)2 = (8)2 + (8)2

⇒ l(AC)= 64 + 64 = 128

⇒ l(AC) = $\sqrt{128}$ cm

Thus, the length of diagonal of square is $\sqrt{128}$ cm.

#### Question 7:

Measure of one angle of a rhombus is 50$°$. find the measures of remaining three angles.

Suppose ABCD is a rhombus.

Let m∠A = 50º.

Opposite angles of a rhombus are congruent.

∴ m∠C = m∠A = 50º and m∠B = m∠D

Now,

m∠A + m∠B  + m∠C + m∠D = 360º

∴ 50º + m∠B  + 50º + m∠D = 360º

⇒ m∠B + m∠D = 360º − 100º = 260º

⇒ 2 m∠B = 260º               (m∠B = m∠D)

m∠B = $\frac{260°}{2}$ = 130º

∴ m∠D = m∠B = 130º

Thus, the measures of the remaining angles of the rhombus are 130º, 50º and 130º.

#### Question 1:

Measures of opposite angles of a parallelogram are (3− 2)$°$ and (50 − x)$°$. Find the measure of its each angle.

Let ABCD be the parallelogram.

Suppose ∠A = (3− 2)º and ∠C = (50 − x)º.

We know that the opposite angles of a parallelogram are congruent.

∴ m∠A = m∠C

⇒ 3x − 2 = 50 − x

⇒ 3x + x = 50 + 2

⇒ 4x = 52

⇒ x = 13

∴ m∠A = (3− 2)º = (3 × 13 − 2)º= (39 − 2)​º= 37º

So, m∠C = m∠A = 37º

Also, the adjacent angles of a parallelogram are supplementary.

∴ m∠A + m∠D = 180º

⇒ 37º + m∠D = 180º

⇒ m∠D = 180º − 37º = 143º

Now,

m∠B = m∠D = 143º               (Opposite angles of a parallelogram are congruent)

Thus, the measure of the angles of the parallelogram are 37º, 143º, 37º and 143º.

#### Question 2:

Referring the adjacent figure of a parallelogram, write the answer of questions given below.

(1) If l(WZ) = 4.5 cm then l(XY) = ?
(2) If l(YZ) = 8.2 cm then l(XW) = ?
(3) If l(OX) = 2.5 cm then l(OZ) = ?
(4) If l(WO) = 3.3 cm then l(WY) = ?
(5) If m∠WZY = 120$°$ then m∠WXY = ? and m∠XWZ = ?

WXYZ is a parallelogram.

(1)
l(XY) = l(WZ) = 4.5 cm                 (Opposite sides of a parallelogram are congruent)

(2)
l(XW) = l(YZ) = 8.2 cm                 (Opposite sides of a parallelogram are congruent)

(3)
l(OZ) = l(OX) = 2.5 cm                  (Diagonals of parallelogram bisect each other)

(4)
l(WY) = 2 × l(WO) = 2 × 3.3 = 6.6 cm         (Diagonals of parallelogram bisect each other)

(5)
m∠WXY = m∠WZY = 120º      (Opposite angles of a parallelogram are congruent)

Now,

m∠WZY + m∠XWZ = 180º      (Adjacent angles of a parallelogram are supplementary)

⇒ 120º + m∠XWZ = 180º

⇒ m∠XWZ = 180º − 120º = 60º

#### Question 3:

Construct a parallelogram ABCD such that l(BC) = 7 cm, m∠ABC = 40$°$, l(AB) = 3 cm.

Steps of Construction:

Step 1: Draw AB = 3 cm.

Step 2: Draw ∠ABX = 40$°$.

Step 3: With B as centre and radius 7 cm, draw an arc cutting the ray BX at C.

Step 4: With C as centre and radius 3 cm, draw an arc.

Step 5: With A as centre and radius 7 cm, draw an arc cutting the previous arc at D.

Step 6: Join AD and CD.

Here, ABCD is the required parallelogram.

#### Question 4:

Ratio of consecutive angles of a quadrilateral is 1:2:3:4. Find the measure of its each angle. Write, with reason, what type of a quadrilateral it is.

Let m∠P : m∠Q : m∠R : m∠S = 1 : 2 : 3 : 4

So, m∠P = km∠Q = 2km∠R = 3k and m∠S = 4k, where k is some constant

Now,

m∠P + m∠Q  + m∠R + m∠S = 360º

+ 2k + 3k + 4k = 360º

⇒ 10k = 360º

⇒ k = 36º

∴ m∠P = 36º

m∠Q = 2k = 2 × 36º = 72º

m∠R = 3k = 3 × 36º = 108º

m∠S = 4= 4 × 36º = 144º

Now,  m∠P + m∠S = 36º + 144º = 180º

We know if two lines are intersected by a transversal such that the sum of interior angles on the same of the transversal are supplementary, then the two lines are parallel.

∴ Side PQ || Side SR

Also, m∠P + m∠Q = 36º + 72º = 108º ≠ 180º

So, side PS is not parallel to side QR.

In quadrilateral PQRS, only one pair of opposite sides is parallel. Therefore, quadrilateral PQRS is a trapezium.

#### Question 5:

Construct ☐ BARC such that l(BA) = l(BC) = 4.2 cm, l(AC) = 6.0 cm, l(AR) = l(CR) = 5.6 cm

Steps of Construction:

Step 1:
Draw BA = 4.2 cm.

Step 2: With B as centre and radius 4.2 cm, draw an arc.

Step 3: With A as centre and radius 6 cm, draw an arc cutting the previous arc at C.

Step 4: Join BC.

Step 5: With A as centre and radius 5.6 cm, draw an arc.

Step 6: With C as centre and radius 5.6 cm, draw an arc cutting the previous arc at R.

Step 7: Join AR and CR.

Here, BARC is the required quadrilateral.

#### Question 6:

Construct ☐ PQRS, such that l(PQ) = 3.5 cm, l(QR) = 5.6 cm, l(RS) = 3.5 cm, m∠Q = 110$°$, m∠R = 70$°$.
If it is given that ☐ PQRS is a parallelogram, which of the given information is unnecessary?

Steps of Construction:

Step 1:
Draw PQ = 3.5 cm.

Step 2: Draw ∠PQX = 110$°$.

Step 3: With Q as centre and radius 5.6 cm, draw an arc cutting ray QX at R.

Step 4: Draw ∠QRY = 70$°$.

Step 5: With R as centre and radius 3.5 cm, draw an arc cutting ray RY at S.

Step 6: Join PS.

Here, PQRS is the required quadrilateral.

If it is given that quadrilateral PQRS is a parallelogram, then the information l(RS) = 3.5 cm and m∠R = 70$°$ is unnecessary.

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