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#### Question 1:

In the given figure, $\angle$ACD is an exterior angle of $∆$ABC. $\angle$B = 40°, $\angle$A = 70°.

Find the measure of $\angle$ACD.

In $∆$ABC,
∠ACD = ∠A + ∠B   (Exterior angle property)
= 70 + 40
= 110
Hence, the measure of  $\angle$ACD is 110​.

#### Question 2:

In $∆$PQR, $\angle$P = 70°$\angle$Q = 65 ° then find  $\angle$R.

In $∆$PQR,
∠P + ∠Q + ∠R = 180   (Angle sum property)
⇒ 70 + 65 + ∠R = 180
⇒ 135 + ∠R = 180
⇒ ∠R = 180 − 135
= 45
Hence, the measure of  $\angle$R is 45.

#### Question 3:

The measures of angles of a triangle are x°, ( x$-$20)°, (x$-$40)°. Find the measure of each angle.

Let us suppose the angles ∠P, ∠Q, ∠Rof a $∆$PQR be x°, ($-$ 20)°, ($-$ 40)° respectively.
∠P + ∠Q + ∠R = 180   (Angle sum property)
x + ($-$ 20)° + ($-$ 40)° = 180
⇒ 3$-$ 60 = 180
⇒ 3= 240
​= 80
Therefore,
∠P = 80
∠R = (80 $-$ 20)°
= 60
∠R = (80 $-$ 40)°
= 40
Hence, the measure of each angle is 80, 60 and 40respectively.

#### Question 4:

The measure of one of the angles of a triangle is twice the measure of its smallest angle and the measure of the other is thrice the measure of the smallest angle. Find the measures of the three angles.

Let us suppose the angles of a $∆$PQR such that ∠P < ∠Q < ∠R.
A.T.Q,
∠Q = 2∠P
∠R = 2∠P
Now, ∠P + ∠Q + ∠R = 180   (Angle sum property)
⇒ ∠P + 2∠P + 3∠P = 180
⇒ 6∠P = 180
⇒ ∠P = 30
Therefore,
∠P = 30
∠R = 60
∠R = 90
Hence, the measure of each angle is 30, 60 and 90respectively.

#### Question 5:

In the given figure, measures of some angles are given. Using the measures find the values of x, y, z.

∠NEM + ∠NET =  180      (Linear angle property)
⇒ y + 100 = 180
⇒ y = 80
Also, ∠NME + ∠EMR =  180      (Linear angle property)
⇒ z + 140 = 180
⇒ z = 40
Now, In △NEM
∠N + ∠E + ∠M = 180   (Angle sum property)
⇒ x + y + z = 180
+ 80∘ + 40∘ = 180
⇒ + 120∘ = 180
⇒ x = 60
Hence, the values of xy and z are 60, 80 and 40respectively.

#### Question 6:

In the given figure, line AB || line DE. Find the measures of $\angle$DRE and $\angle$ARE using given measures of some angles.

AB || DE and AD is a transversal line.
∠BAR = ∠RDE = 70      (Alternate angles)
In △DER
∠D + ∠E + ∠DRE = 180   (Angle sum property)
⇒ 70 + 40 + ∠DRE = 180
⇒ 110 + ∠DRE = 180
⇒ ∠DRE = 70
Now, ∠ARE = ∠DRE + ∠RDE   (Exterior angle property)
= 70 + 40
= 110
Hence, the measures of ∠DRE and ∠ARE are 70∘ and 110respectively.

#### Question 7:

In $∆$ABC, bisectors of $\angle$A and $\angle$B intersect at point O. If $\angle$C = 70° . Find measure of $\angle$AOB.

If the bisectors of ∠X and ∠Y of a △XYZ intersect at point O, then $\angle \mathrm{XOY}=90°+\frac{1}{2}\angle \mathrm{XZY}$
$\therefore \angle \mathrm{AOB}=90°+\frac{1}{2}\angle \mathrm{ACB}\phantom{\rule{0ex}{0ex}}=90°+\frac{1}{2}\left(70°\right)$
= 90+ 35
= 125
Hence, the measure of ∠AOB is 125.

#### Question 8:

In the given Figure, line AB || line CD and line PQ is the transversal. Ray PT and ray QT are bisectors of $\angle$BPQ and $\angle$PQD respectively.
Prove that m $\angle$PTQ = 90 °.

AB || CD and PQ is a transversal line.
∠BPQ + ∠DQP = 180      (Angles on the same side of a transversal line are supplementary angles)
$⇒\frac{1}{2}\angle \mathrm{BPQ}+\frac{1}{2}\angle \mathrm{DQP}=\frac{180°}{2}$
⇒ ∠QPT + ∠PQT = 90
In △PQT
∠QPT + ∠PQT + ∠PTQ = 180   (Angle sum property)
⇒ 90 + ∠PTQ = 180
⇒ ∠PTQ = 90
Hence proved.

#### Question 9:

Using the information shown in figure, find the measures of $\angle$a, $\angle$b and $\angle$c.

c + 100 = 180                (Linear angle property)
⇒ c = 80
Now, b = 70∘     (Vertically opposite angles)
a + b + c = 180       (Angle sum property)
⇒ + 70∘ + 80∘ = 180
⇒ + 150∘ = 180
⇒ = 30
Hence, the values of ab and c are 30, 70 and 80respectively.

#### Question 10:

In the given figure, line DE ||   line GF ray EG and ray FG are bisectors of $\angle$DEF and $\angle$DFM respectively. Prove that,
(i)  $\angle$DEG = EDF    (ii)  EF = FG.

(i) Given: DE || GF
Now, $\angle$DEF = $\angle$GFM              (Corresponding angles as DM is a transversal line)
⇒ 2$\angle$DEG = $\angle$DFG         (Ray EG and ray FG are bisectors of $\angle$DEF and $\angle$DFM)
⇒ 2$\angle$DEG = $\angle$EDF          (∵ $\angle$EDF = $\angle$DFG, alternate angles as DF is a transversal line)
⇒ $\angle$DEG = EDF

(ii) Given: DE || GF
$\angle$DEG = $\angle$EGF              (Alternate angles as EG is a transversal line)
∴ $\angle$GEF = $\angle$EGF            (∵ $\angle$DEG = $\angle$GEF)
∴ EF = FG                         (Sides opposite to equal angles)

#### Question 1:

In each of the examples given below, a pair of triangles is shown. Equal parts of triangles in each pair are marked with the same signs. Observe the figures and state the test by which the triangles in each pair are congruent.

(i)

By . . . . . . . . . . test
$∆$ABC PQR

(ii)

By . . . . . . . . . . test
$∆$XYZ LMN

(iii)

By . . . . . . . . . . test
$∆$LMN PTR

(iv)

By . . . . . . . . . . test
$∆$LMN PTR

(i) SSC Test
(ii) SAS Test
(iii) ASA Test
(iv) Hypotenuse Side Test.

#### Question 2:

Observe the information shown in pairs of triangles given below. State the test by which the two triangles are congruent. Write the remaining congruent parts of the triangles.

(i)

From the information shown in the figure,
in $∆$ABC and  $∆$PQR
$\angle$ABC PQR
seg BC $\cong$ seg QR
$\angle$ACB$\cong$ $\angle$PRQ
ABC $\cong$ $∆$PQR ....... $\overline{){3243432}}$ test
BAC $\cong$ $\overline{){3243432}}$ .......corresponding angles of congruent triangles.

(ii)

From the information shown in the figure,
in $∆$PTQ and  $∆$STR
seg PT $\cong$ seg ST
$\angle$PTQ$\cong$ $\angle$STR...... vertically opposite angles
PTQ $\cong$ $∆$STR ....... $\overline{){3243432}}$ test

seg PQ $\cong$ $\overline{){3243432}}$  corresponding sides of congruent triangles

(i)
From the information shown in the figure,
in $∆$ABC and  $∆$PQR
$\angle$ABC PQR
seg BC $\cong$ seg QR
$\angle$ACB$\cong$ $\angle$PRQ
ABC $\cong$ $∆$PQR ....... $\overline{)\mathrm{ASA}}$ test
BAC $\cong$ $\overline{)\angle \mathrm{QPR}}$ .......corresponding angles of congruent triangles.

(ii)
From the information shown in the figure,
in $∆$PTQ and  $∆$STR
seg PT $\cong$ seg ST
$\angle$PTQ$\cong$ $\angle$STR...... vertically opposite angles
PTQ $\cong$ $∆$STR ....... $\overline{)\mathrm{SAS}}$ test

seg PQ $\cong$   corresponding sides of congruent triangles

#### Question 3:

From the information shown in the figure, state the test assuring the congruence of $∆$ ABC and $∆$ PQR. Write the remaining congruent parts of the triangles.

​In △ABC and △QPR
AB = PQ                        (Given)
BC = PR                        (Given)
∠A = ∠Q = 90             (Given)
By RHS test of congruency
△ABC ≅ △QPR
∠B = ∠P                (corresponding angles of congruent triangles)
∠C = ∠R               (corresponding angles of congruent triangles)
​AC = QR         (corresponding sides of congruent triangles)

#### Question 4:

As shown in the following figure, in  $∆$ LMN and  $∆$PNM, LM = PN, LN = PM. Write the test which assures the congruence of the two triangles. Write their remaining congruent parts.

​In △LMNand △PNM
LM = PN                         (Given)
LN = PM                        (Given)
MN = NM                      (Common)
By SSS test of congruency
△LMN ≅ △PNM
∠L = ∠P                (corresponding angles of congruent triangles)
∠LMN = ∠PNM   (corresponding angles of congruent triangles)
∠MNL = ∠NMP   (corresponding angles of congruent triangles)

#### Question 5:

In the given figure, seg AB $\cong$ seg CB and seg AD$\cong$seg CD.

Prove that $∆$ ABD$\cong$$∆$ CBD

In △ABD and △CBD
AB = CB                         (Given)
BD = BD                      (Common)
By SSS test of congruency
△ABD ≅ △CBD

#### Question 6:

In the given figure, $\angle$P $\cong$ $\angle$R seg PQ $\cong$ seg RQ

Prove that, $∆$ PQT $\cong$ $∆$ RQS

In △PQT and △RQS
∠P = ∠R                         (Given)
PQ = RQ                        (Given)
∠Q = ∠Q                      (Common)
By ASA test of congruency
△PQT ≅ △RQS

#### Question 1:

Find the values of x and y using the information shown in figure .

Find the measure of $\angle$ABD and m$\angle$ACD.

In △ABC, AB = AC
∴ ∠ABC = ∠ACB            (Angles opposite to equal sides are equal)
⇒ x = 50
In △BDC, DB = DC
∴ ∠DBC= ∠DCB            (Angles opposite to equal sides are equal)
⇒ y = 60∘
Now, ∠ABD = ∠ABC + ∠DBC
= 50 + 60
= 110
Also, ∠ACD = ∠ACB + ∠DCB
= 50 + 60
= 110
Hence, the values of x and y are 50∘ and 60respectively.

#### Question 2:

The length of hypotenuse of a right angled triangle is 15. Find the length of median of its hypotenuse.

In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse.
Therefore, the length of median of its hypotenuse is 7.5 units.

#### Question 3:

In $∆$PQR, $\angle$Q = 90° , PQ = 12, QR = 5 and QS is a median. Find l(QS).

In △PQR,
PR2 = PQ2 + QR2           (By Pythagoras theorem)
⇒ PR= 122 + 52
⇒ PR= 144 + 25
⇒ PR= 132
⇒ PR = 13
In a right angled triangle, the length of the median on its hypotenuse is half the length of the hypotenuse.
$\therefore l\left(\mathrm{QS}\right)=\frac{1}{2}\left(\mathrm{PR}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left(13\right)$
= 6.5 units.
Hence, the length of QR is 6.5 units.

#### Question 4:

In the given figure, P point G is the point of concurrence of the medians of $∆$ PQR . If GT = 2.5, find the lengths of PG and PT.

The point of concurrence of medians of a triangle divides each median in the ratio 2 : 1.
Let us suppose the length of PG and GT be 2x and x.
Therefore, the value of x is 2.5
Now, PG = 2x
= 2(2.5)
= 5 units
Now, PT = PG + GT
= 5 + 2.5
= 7.5 units
Hence, the lenght of PG and GT is 5 and 7.5units respectively.

#### Question 1:

In the given figure, point A is on the bisector of $\angle$XYZ. If AX = 2 cm then find AZ.

By Part I of the angle bisector theorem, every point on the bisector of an angle is equidistant from the sides of the angle.
∴ AZ = AX
= 2 cm
Hence, the length of AZ is 2 cm.

#### Question 2:

In the given figure, $\angle$RST = 56° , seg PT $\perp$ray ST, seg PR $\perp$ ray SR and seg PR$\cong$ seg PT Find the measure of $\angle$RSP. State the reason for your answer.

By Part II of the angle bisector theorem, any point equidistant from sides of an angle is on the bisector of the angle
∠TSP = ∠RSP
Now, ∠TSP + ∠RSP = ∠TSR
⇒ ∠RSP + ∠RSP = 56
⇒ 2∠RSP = 56
⇒ ∠RSP = 28
Hence, the measure of ∠RSP is 28.

#### Question 3:

In $∆$PQR, PQ = 10 cm, QR = 12 cm, PR = 8 cm. Find out the greatest and the smallest angle of the triangle.

In △PQR, QR and PR are the longest and smallest side.
The angle opposite to the longest side is the largest angle and the angle opposite to the smallest side is the smallest angle .
Hence, the greatest and the smallest angle of the triangle are ∠QPR and ∠PQR respectively.

#### Question 4:

In $∆$ FAN, $\angle$F = 80° $\angle$A = 40° . Find out the greatest and the smallest side of the triangle. State the reason.

In ∆FAN,
∠F + ∠A + ∠N = 180   (Angle sum property)
⇒ 80 + 40 + ∠N = 180
⇒ 120 + ∠N = 180
⇒ ∠N = 180 − 120
= 60
Now, in △FAN,  ∠F and ∠A are the largest and smallest angle.
The side opposite to the largest angle is the greatest side and the side opposite to the smallest angle is the smallest side .
Hence, the greatest and the smallest side of the triangle are AN and ​FN respectively.

#### Question 5:

Prove that an equilateral triangle is equiangular.

Consider an equilateral triangle ABC.

In △ABC, AB = BC
∴ ∠C = ∠A               ...(1)          (Angles opposite to equal sides)
In △ABC, AB = CA
∴ ∠C = ∠B               ...(2)          (Angles opposite to equal sides)
From (1) and (2), we get
∠A = ∠B =  ∠C

Hence, an equilateral triangle is equiangular.

#### Question 6:

Prove that, if the bisector of $\angle$BAC of $∆$ABC is perpendicular to side BC, then $∆$ABC is an isosceles triangle.

∠BDA = ∠CDA = 90                (Given)
By ASA test of congruency
∴ ∠B = ∠C        (corresponding angles of congruent triangles)
If two angles of a traingle are equal, then the traingle is said to be a isosceles triangle.

#### Question 7:

In the given figure, if seg PR$\cong$seg PQ, show that seg PS > seg PQ.

In △PRS, ∠PRS is an obtuse angle.
Since, a traingle can have maximum one obtuse angle.
Hence, ∠PRS is the greatest angle.
Therefore, the side opposite to ∠PRS i.e., PS is the longest side.
Now, seg PS > seg PR > seg RS
∴ seg PS > seg PQ                  [∵seg PR$\cong$seg PQ]

#### Question 8:

In the given figure, in $∆$ABC, seg AD and seg BE are altitudes and AE = BD. Prove that seg AD $\cong$seg BE

∠ADB = ∠BEA = 90                (Given)
BD = AE                                     (Given)
AB = BA                                     (Common)
By RHS test of congruency
∴ AD = BE        (corresponding angles of congruent triangles)
Hence proved.

#### Question 1:

If $∆$ XYZ ~  $∆$ LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.

Consider, $∆$ XYZ ~  $∆$ LMN

#### Question 2:

In $∆$ XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm, If  $∆$ XYZ ~ $∆$​ PQR  and PQ = 8 cm then find the lengths of remaining sides of $∆$PQR.

Consider, $∆$ XYZ ~  $∆$ PQR

Hence, the lengths of remaining sides of $∆$PQR are 12 and 10 cm.

#### Question 3:

Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same signs. Show the lengths of corresponding sides by numbers in proportion.

The pair of similar triangles is given below:

#### Question 1:

Choose the correct alternative answer for the following questions.

(i) If two sides of a triangle are 5 cm and 1.5 cm, the lenght of its third side cannot be . . . . . . . .
(A) 3.7 cm (B) 4.1 cm (C) 3.8 cm (D) 3.4 cm

(ii)   In $∆$PQR, If $\angle$ R >  $\angle$Q then . . . . . . . .

(A) QR > PR
(B) PQ > PR
(C) PQ < PR
(D) QR < PR

(iii) In $∆$TPQ, $\angle$ T = 65° , $\angle$ P = 95° which of the following is a true statement ?
(A) PQ < TP
(B) PQ < TQ
(C) TQ < TP < PQ
(D) PQ < TP < TQ

(i)
According to the triangle inequality "the sum of the lengths of any two sides of a triangle must be greater than or equal to the length of the third side".
Therefore, the lenght of the third side cannot be greater than 3.4 cm.
Hence, the correct option is (D).

(ii)  In $∆$PQR, If $\angle$ R >  $\angle$Q
∴ PQ > PR                       (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).

(iii) In $∆$TPQ, $\angle$ T = 65° , $\angle$ P = 95°
A triangle can have maximum one obtuse angle.
$\angle$T < $\angle$ P
PQ < TQ                         (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence, the correct option is (B).

#### Question 2:

$∆$ ABC is isosceles in which AB = AC. Seg BD and seg CE are medians. Show that BD = CE.

​In △ABC, AB = AC
∴ ∠B = ∠C                    (Angles opposite to equal sides)
Dividing both sides by 2, we get
$\frac{1}{2}\angle \mathrm{B}=\frac{1}{2}\angle \mathrm{C}$
⇒ ∠EBC = ∠DCB               ...(2)
In △BEC and △CDB
∠EBC = ∠DCB                         [From (2)]
BE = CD                                   (E and D are the mid points of AB & AC respectively and AB = AC)
BC = CB                                     (Common)
By SAS test of congruency
△BEC ≅ △CDB
∴ BD = CE        (corresponding sides of congruent triangles)

#### Question 3:

In$∆$PQR, If PQ > PR and bisectors of $\angle$Q and $\angle$R intersect at S. Show that SQ > SR.

In △PQR, PQ > PR
Now, the angle opposite to the greater side is greater than angle opposite to the smaller side.
∴ ∠R > ∠Q
Dividing both sides by 2, we get
$\frac{1}{2}\angle \mathrm{R}>\frac{1}{2}\angle \mathrm{Q}$
⇒ ∠SRQ > ∠SQR
⇒ SQ > SR       (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.

#### Question 4:

In the figure, point D and E are on side BC of $∆$ABC, such that BD = CE and AD = AE.
Show that $∆$ ABD$\cong$ $∆$ACE.

∴ ∠ADE = ∠AED                    (Angles opposite to equal sides)
Now,
∠AED + ∠AEC = 180            ...(2)
Subtracting (2) from (1), we get
In △ABD and △ACE
BD = CE                                   (Given)
By SAS test of congruency
△ABD ≅ △ACE

#### Question 5:

In the given figure, point S is any point on side QR of $∆$PQR Prove that : PQ + QR + RP > 2PS

In △PQS
PQ + QS > PS                     ...(1)
(Sum of two sides of a traingle is greater than the third side)
In △PRS
RP + RS > PS                     ...(2)
​                                                      (Sum of two sides of a traingle is greater than the third side)
Adding (1) and (2), we get
PQ + QS + RP + RS  > PS + PS
⇒ PQ + QR + RP > 2PS

#### Question 6:

In the given figure, bisector of  $\angle$BAC intersects side BC at point D. Prove that AB > BD

Given: AD is the bisector of ∠A
∴ ∠DAB = ∠DAC
Now, ∠BDA is the exterior angle of the ∆DAC
∴ ∠BDA > ∠DAC
⇒ ∠BDA > ∠DAB        (∵ ∠DAC = ∠DAB)
⇒ AB > BD                   (The sides opposite to the greater angle is greater than side opposite to the smaller angle)
Hence proved.

#### Question 7:

In the given figure, seg PT is the bisector of $\angle$QPR. A line through R intersects ray QP at point S. Prove that PS = PR

According to the figure, we have
PT || SR
If PT || SR and PS is a transversal line, then
$\angle$QPT = $\angle$PSR   (Corresponding angles)           ....(1)
If PT || SR and PR is a transversal line, then
$\angle$TPR = $\angle$PRS  (Alternate angles)              ....(2)
We are given that seg PT is the bisector of $\angle$QPR.
∴ $\angle$QPT = $\angle$TPR         ...(3)
From (1), (2) and (3)
$\angle$PSR = $\angle$PRS
The sides opposite to equal angles are also equal in a triangle.
∴ PS = PR

#### Question 8:

In the given figure, seg AD $\perp$ seg BC. seg AE is the bisector of $\angle$CAB and C - E - D. Prove that
$\angle$DAE =

In ∆ABC, since AE bisects ∠CAB, then
∠BAE = ∠CAE                         .......(1)
∠ADB + ∠DAB + ∠B = 180°   [Angle sum property]
⇒ 90° + ∠DAB + ∠B = 180°
⇒∠B = 90° − ∠DAB                    .....(2)
​⇒ $\angle$DAE =