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#### Page No 159:

#### Question 1:

**In the given question, four options are given, out of which only one is correct. Choose the correct one.**

The sides of a triangle have lengths (in cm) 10, 6.5 and *a*, where a is a whole number. The minimum value that *a* can take is

(a) 6

(b) 5

(c) 3

(d) 4

#### Answer:

The sides of a triangle have lengths (in cm) 10, 6.5 and *a*, where a is a whole number. The minimum value that *a* can take is

The sum of the lengths of two sides of the triangle should be greater than the third side.

So, *a* + 6.5 > 10

⇒a > 3.5 > 3

So, the minimum value of *a* should be 3.

#### Page No 159:

#### Question 2:

**In the given question, four options are given, out of which only one is correct. Choose the correct one.**

Triangle DEF of in the given figure is a right triangle with ∠E = 90°.

What type of angles are ∠D and ∠F?

(a) They are equal angles

(b) They form a pair of adjacent angles

(c) They are complementary angles

(d) They are supplementary angles

#### Answer:

Triangle DEF in the given figure is a right triangle with ∠E = 90°

Now, ∠D + ∠E + ∠F = 180°

⇒ ∠D + 90° + ∠F = 180°

⇒ â€‹∠D + ∠F = 180° - 90° = 90°

So, they are complementary angles.

#### Page No 159:

#### Question 3:

**In the given question, four options are given, out of which only one is correct. Choose the correct one.**

In the given figure, PQ = PS. The value of *x* is

(a) 35°

(b) 45°

(c) 55°

(d) 70°

#### Answer:

In $\u25b3\mathrm{PQS}$,

PQ = PS (Given)

⇒$\angle \mathrm{PQS}=\angle \mathrm{PSQ}=70\xb0$ (âˆµ Angles corresponding to equal sides are equal)

In $\u25b3\mathrm{PRS}$,

$\angle \mathrm{PSQ}=\angle \mathrm{SPQ}+\angle \mathrm{SRQ}\phantom{\rule{0ex}{0ex}}\Rightarrow 70\xb0=x+25\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=70\xb0-25\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=45\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the correct answer is option (b).

#### Page No 159:

#### Question 4:

In a right-angled triangle, the angles other than the right angle are

(a) obtuse

(b) right

(c) acute

(d) straight

#### Answer:

In a right-angled triangle, one of the angles is a right angle, i.e., of 90°.

So, the sum of the other two angles is 90°.

Therefore, the other two angles are less than 90° which means they are acute angles.

Thus, in a right-angled triangle, the angles other than the right angle are acute angles.

â€‹Hence, the correct answer is option (c).

#### Page No 159:

#### Question 5:

In an isosceles triangle, one angle is 70°. The other two angles are of

(i) 55° and 55°

(ii) 70° and 40°

(iii) any measure

In the given option(s) which of the above statement(s) are true?

(a) (i) only

(b) (ii) only

(c) (iii) only

(d) (i) and (ii)

#### Answer:

In an isosceles triangle, one angle is 70°.

The other two angles are of**(i)** 55° and 55°

By Angle Sum Property of Triangle,

70° + 55° + 55° = 180°

⇒ 180° = 180°

Thus, statement (i) is true.**(ii)** 70° and 40°

By Angle Sum Property of Triangle,

70° + 70° + 40° = 180°

⇒ 180° = 180°

Thus, statement (ii) is true.**(iii)** any measure

Not possible as in an isosceles triangle two angles are equal.

Thus, statement (iii) is false.

So, in the given option(s) the statement(s) (i) and (ii) are true.

â€‹â€‹Hence, the correct answer is option (d).

#### Page No 160:

#### Question 6:

In a triangle, one angle is of 90°. Then

(i) The other two angles are of 45° each

(ii) In remaining two angles, one angle is 90° and other is 45°

(iii) Remaining two angles are complementary

In the given option(s) which is true?

(a) (i) only

(b) (ii) only

(c) (iii) only

(d) (i) and (ii)

#### Answer:

In a triangle, one angle is of 90°. Then**(i)** The other two angles are of 45° each

By Angle Sum Property of Triangle,

90° + 45° + 45° = 180°

⇒ 180° = 180°

But this is not a trivial solution.

Thus, statement (i) is false.**(ii)** In remaining two angles, one angle is 90° and other is 45°

By Angle Sum Property of Triangle,

90° + 90° + 45° = 180°

⇒ 225° ≠ 180°

Thus, statement (ii) is false.**(iii)** Remaining two angles are complementary

Let the remaining two angles be *x* and *y* such that they are complementary angles.

⇒ *x* + *y* = 90°

Now,

90° + *x* + *y* = 180°

⇒ 90° + 90° + = 180°

⇒ 180° = 180°

This is a trivia; solution.

Thus, statement (iii) is true.

So, in the given option(s) the statement (iii) is true.

â€‹â€‹Hence, the correct answer is option (c).

#### Page No 160:

#### Question 7:

Lengths of sides of a triangle are 3 cm, 4 cm and 5 cm. The triangle is

(a) Obtuse angled triangle

(b) Acute-angled triangle

(c) Right-angled triangle

(d) An Isosceles right triangle

#### Answer:

Lengths of the sides of a triangle are 3 cm, 4 cm and 5 cm.

According to Pythagoras' theorem,

${5}^{2}={3}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25=9+16\phantom{\rule{0ex}{0ex}}\Rightarrow 25=25\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{LHS}=\mathrm{RHS}$

Since these sides satisfy the Pythagoras theorem, therefore it is right angled triangle.

Hence, the correct answer is option (c).

#### Page No 160:

#### Question 8:

In the given figure, PB = PD. The value of *x* is

(a) 85°

(b) 90°

(c) 25°

(d) 35°

#### Answer:

In $\u2206\mathrm{PBD}$,

$\angle \mathrm{PBE}+\angle \mathrm{PBD}=180\xb0\left(\because \mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 120\xb0+\angle \mathrm{PBD}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{PBD}=60\xb0.....\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

$\Rightarrow \angle \mathrm{PBD}=\angle \mathrm{PDB}\left(\because \mathrm{Isosceles}\mathrm{triangle}\mathrm{property}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{PDB}=60\xb0\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Now,

In $\u2206\mathrm{PDC}$,

$\angle \mathrm{PDB}=\angle \mathrm{DCP}+\angle \mathrm{DPC}\left(\because \mathrm{Exterior}\mathrm{angle}\mathrm{property}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0=x+35\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0-35\xb0=x\phantom{\rule{0ex}{0ex}}\Rightarrow 25\xb0=x$

Hence, the correct answer is option (c).

#### Page No 160:

#### Question 9:

In âˆ†PQR,

(a) PQ – QR > PR

(b) PQ + QR < PR

(c) PQ – QR< PR

(d) PQ + PR< QR

#### Answer:

Since the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

∴ In âˆ†PQR,

QR + PR > PQ

⇒ PR > PQ − QR

⇒ PQ − QR < PR

Hence, the correct answer is option (c).

#### Page No 160:

#### Question 10:

In âˆ† ABC,

(a) AB + BC > AC

(b) AB + BC < AC

(c) AB + AC < BC

(d) AC + BC < AB

#### Answer:

Since the sum of the lengths of any two sides of a triangle is always greater than the length of the third side.

∴ In âˆ†ABC,

AB + BC > AC

Hence, the correct answer is option (a).

#### Page No 161:

#### Question 11:

The top of a broken tree touches the ground at a distance of 12 m from its base. If the tree is broken at a height of 5 m from the ground then the actual height of the tree is

(a) 25 m

(b) 13 m

(c) 18 m

(d) 17 m

#### Answer:

Let AB be the given that tree of height* h* m, which is broken at D which is 12 m away from its base and the height of the remaining part, i.e., CB = 5 m.

$\mathrm{In}\u25b3\mathrm{BCD}\phantom{\rule{0ex}{0ex}}{\mathrm{DC}}^{2}={\mathrm{BC}}^{2}+{\mathrm{BD}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{DC}}^{2}={5}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{DC}}^{2}=25+144\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{DC}}^{2}=169\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{DC}}^{2}={13}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{DC}=13$

Since AC and CD both represents the same broken part of the tree.

∴ CA = DC = 13 m

Now, AB = BC + CA

⇒* h* = 5 + 13

⇒* **h* = 18 m

Hence, the correct answer is option (c).

#### Page No 161:

#### Question 12:

The triangle ABC formed by AB = 5 cm, BC = 8 cm, AC = 4 cm is

(a) an isosceles triangle only

(b) a scalene triangle only

(c) an isosceles right triangle

(d) scalene as well as a right triangle

#### Answer:

Given: The triangle ABC formed by AB = 5 cm, BC = 8 cm, AC = 4 cm.

Here,

${\mathrm{AB}}^{2}+{\mathrm{AC}}^{2}={\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {5}^{2}+{4}^{2}={8}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 25+16=64\phantom{\rule{0ex}{0ex}}\Rightarrow 41\ne 64$

Also, AB ≠ BC ≠ CA

⇒ Triangle ABC is an isosceles triangle.

Thus, the triangle ABC formed by AB = 5 cm, BC = 8 cm, AC = 4 cm is an isosceles triangle only.

Hence, the correct answer is option (a).

#### Page No 161:

#### Question 13:

Two trees 7 m and 4 m high stand upright on a ground. If their bases (roots) are 4 m apart, then the distance between their tops is

(a) 3 m

(b) 5 m

(c) 4 m

(d) 11 m

#### Answer:

Let BE be the smaller tree and AD be the bigger tree.

AB is the distance between their tops.

So. BE = CD = 4 cm

ED = 4 cm and â€‹AD = 7 cm (Given)

⇒ AC = 3 cm

Also, BC = ED = 4 cm

$\mathrm{In}\u25b3\mathrm{ABC},\phantom{\rule{0ex}{0ex}}{\mathrm{AB}}^{2}={\mathrm{AC}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AB}}^{2}={3}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AB}}^{2}=9+16\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AB}}^{2}=25\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AB}=5\mathrm{m}\phantom{\rule{0ex}{0ex}}$

Thus, the distance between their tops is 5 m.

Hence, the correct answer is option (b).

#### Page No 161:

#### Question 14:

If in an isosceles triangle, each of the base angles is 40°, then the triangle is

(a) Right-angled triangle

(b) Acute angled triangle

(c) Obtuse angled triangle

(d) Isosceles right-angled triangle

#### Answer:

Given: An isosceles triangle, each of the base angles is 40°.

Consider $\u25b3\mathrm{ABC}$ with $\angle \mathrm{B}\mathrm{and}\angle \mathrm{C}$ as base angles each of 40°.

In $\u25b3\mathrm{ABC}$

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}+40\xb0+40\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}+80\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}=100\xb0$

Here, $\angle \mathrm{A}=100\xb0>90\xb0$

Thus, the triangle is Obtuse angled triangle.

Hence, the correct answer is option (c).

#### Page No 161:

#### Question 15:

If two angles of a triangle are 60° each, then the triangle is

(a) Isosceles but not equilateral

(b) Scalene

(c) Equilateral

(d) Right-angled

#### Answer:

Given: Two angles of a triangle are 60°.

Consider $\u25b3\mathrm{ABC}$ with $\angle \mathrm{B}\mathrm{and}\angle \mathrm{C}$ each of 60°.

In $\u25b3\mathrm{ABC}$

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}+60\xb0+60\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}+120\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}=60\xb0$

Here, $\angle \mathrm{A}=\angle \mathrm{B}=\angle \mathrm{C}=60\xb0$

Thus, the triangle is an equilateral triangle.

Hence, the correct answer is option (c).

#### Page No 161:

#### Question 16:

The perimeter of the rectangle whose length is 60 cm and a diagonal is 61 cm is

(a) 120 cm

(b) 122 cm

(c) 71 cm

(d) 142 cm

#### Answer:

Given: Length of rectangle = 60 cm and diagonal of rectangle = 61 cm

Let the breadth of the rectangle be *x* cm.

Here, AC = 61 cm and AB = 60 cm

In $\u25b3\mathrm{ABC}$, by Pythagoras Theorem

${\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(61\right)}^{2}={\left(60\right)}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 3721=3600+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 121={\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {11}^{2}={\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BC}=11$

$\begin{array}{rcl}\therefore \mathrm{Perimeter}\mathrm{of}\mathrm{triangle}& =& 2\times \left(\mathrm{length}+\mathrm{breadth}\right)\\ & =& 2\times \left(\mathrm{AB}+\mathrm{BC}\right)\\ & =& 2\times \left(60+11\right)\\ & =& 2\times 71\\ & =& 142\mathrm{cm}\end{array}$

Hence, the correct answer is option (d).

#### Page No 161:

#### Question 17:

In âˆ†PQR, if PQ = QR and ∠Q = 100°, then ∠R is equal to

(a) 40°

(b) 80°

(c) 120°

(d) 50°

#### Answer:

Given: In âˆ†PQR, PQ = QR and ∠Q = 100°

In âˆ†PQR,

PQ = QR

$\Rightarrow \angle \mathrm{P}=\angle \mathrm{R}$ .....(1) (âˆµ Angles corresponding to equal sides are equal)

By Angle Sum Property of Triangle

$\Rightarrow \angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 100\xb0+2\angle \mathrm{R}=180\xb0\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{R}=80\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{R}=40\xb0$

Hence, the correct answer is option (a).

#### Page No 161:

#### Question 18:

Which of the following statements is not correct?

(a) The sum of any two sides of a triangle is greater than the third side

(b) A triangle can have all its angles acute

(c) A right-angled triangle cannot be equilateral

(d) Difference of any two sides of a triangle is greater than the third side

#### Answer:

The difference of the length of any two sides of a triangle is always smaller than the length of the third side.

Thus, statements in option (d) is not correct.

Hence, the correct answer is option (d).

#### Page No 161:

#### Question 19:

In the given figure, BC = CA and ∠A = 40°. Then, ∠ACD is equal to

(a) 40°

(b) 80°

(c) 120°

(d) 60°

#### Answer:

Given: BC = CA and ∠A = 40°

In $\u25b3\mathrm{ABC}$,

BC = CA

$\Rightarrow \angle \mathrm{CAB}=\angle \mathrm{CBA}\left(\because \mathrm{Angles}\mathrm{corresponding}\mathrm{to}\mathrm{equal}\mathrm{sides}\mathrm{are}\mathrm{equal}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{CAB}=40\xb0$

By Exterior Angle Property of Triangle

$\angle \mathrm{CAB}+\angle \mathrm{CBA}=\angle \mathrm{ACD}\phantom{\rule{0ex}{0ex}}\Rightarrow 40\xb0+40\xb0=\angle \mathrm{ACD}\phantom{\rule{0ex}{0ex}}\Rightarrow 80\xb0=\angle \mathrm{ACD}$

Hence, the correct answer is option (b).

#### Page No 162:

#### Question 20:

The length of two sides of a triangle are 7 cm and 9 cm. The length of the third side may lie between

(a) 1 cm and 10 cm

(b) 2 cm and 8 cm

(c) 3 cm and 16 cm

(d) 1 cm and 16 cm

#### Answer:

Given: The length of two sides of a triangle are 7 cm and 9 cm.

Consider $\u25b3\mathrm{ABC}$ with AB = 7 cm and BC = 9 cm.

Since the sum of any two sides of a triangle is greater than the third side.

∴ AB + BC > CA

⇒ 7 + 9 > CA

⇒â€‹ 16 cm > CA .....(1)

Since the difference of the length of any two sides of a triangle is always smaller than the length of the third side.

∴ BC − AB < CA

⇒ 9 − 7 < CA

â€‹⇒â€‹ 2 cm < CA .....(2)

From (1) and (2)

2 cm < CA < 16 cm

Thus, the length of the third side may lie between 2 cm and 16 cm.**Disclaimer:** Correct answer is not there in the options.

#### Page No 162:

#### Question 21:

From in the given figure, the value of *x* is

(a) 75°

(b) 90°

(c) 120°

(d) 60°

#### Answer:

In $\u25b3\mathrm{ABC}$, by Exterior Angle Property

$\angle \mathrm{CAB}+\angle \mathrm{CBA}=\angle \mathrm{ACD}\phantom{\rule{0ex}{0ex}}\Rightarrow 25\xb0+35\xb0=\angle \mathrm{ACD}\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0=\angle \mathrm{ACD}$

In $\u25b3\mathrm{PDC}$, by Exterior Angle Property

$\angle \mathrm{PDC}+\angle \mathrm{PCD}=\angle \mathrm{APD}\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0+60\xb0=\angle \mathrm{APD}\phantom{\rule{0ex}{0ex}}\Rightarrow 120\xb0=\angle \mathrm{APD}$

Hence, the correct answer is option (c).

#### Page No 162:

#### Question 22:

In the given figure, the value of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F is

(a) 190°

(b) 540°

(c) 360°

(d) 180°

#### Answer:

In $\u25b3\mathrm{ABC}$, by Angle Sum Property of Triangle

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0$ .....(1)

In $\u25b3\mathrm{DEF}$, by Angle Sum Property of Triangle

$\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=180\xb0$ .....(2)

Adding (1) and (2) on both sides, we get

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}+\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=360\xb0$

Hence, the correct answer is option (c).

#### Page No 162:

#### Question 23:

In the given figure, PQ = PR, RS = RQ and ST || QR. If the exterior angle RPU is 140°, then the measure of angle TSR is

(a) 55°

(b) 40°

(c) 50°

(d) 45°

#### Answer:

Given: PQ = PR, RS = RQ, ST || QR and exterior angle RPU is 140°.

In $\u25b3\mathrm{PQR}$,

PQ = PR

$\Rightarrow \angle \mathrm{PRQ}=\angle \mathrm{PQR}$ .....(1) (Isosceles triangle property)

In $\u25b3\mathrm{PQR}$, by Exterior angle property

$\angle \mathrm{UPT}=\angle \mathrm{PQR}+\angle \mathrm{PRQ}\phantom{\rule{0ex}{0ex}}\Rightarrow 140\xb0=2\angle \mathrm{PQR}\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 70\xb0=\angle \mathrm{PQR}.....\left(2\right)$

In $\u25b3\mathrm{SQR}$,

RS = RQ

$\Rightarrow \angle \mathrm{PRQ}=\angle \mathrm{RSQ}$ .....(3) (Isosceles triangle property)

$\Rightarrow \angle \mathrm{RSQ}=70\xb0$

ST || QR

$\Rightarrow \angle \mathrm{TSP}=\angle \mathrm{RQS}\left(\mathrm{Corresponding}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{TSP}=70\xb0$

$\angle \mathrm{RSQ}+\angle \mathrm{RST}+\angle \mathrm{TSP}=180\xb0\left(\mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 70\xb0+\angle \mathrm{RST}+70\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 140\xb0+\angle \mathrm{RST}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{RST}=40\xb0$

Hence, the correct answer is option (b).

#### Page No 162:

#### Question 24:

In the given figure, ∠BAC = 90°, AD ⊥ BC and ∠BAD = 50°, then ∠ACD is

(a) 50°

(b) 40°

(c) 70°

(d) 60°

#### Answer:

Given: ∠BAC = 90°, AD ⊥ BC and ∠BAD = 50°

∠BAC = 90°

⇒ ∠BAD + ∠DAC = 90°

⇒ 50° + ∠DAC = 90°

⇒ ∠DAC = 40°

In âˆ†ADC,

∠DAC + ∠ACD + ∠ADC = 180°

⇒ 40° + ∠ACD + 90° = 180°

⇒ ∠ACD + 130° = 180°

⇒ ∠ACD = 50°

Hence, the correct answer is option (a).

#### Page No 162:

#### Question 25:

If one angle of a triangle is equal to the sum of the other two angles, the triangle is

(a) obtuse

(b) acute

(c) right

(d) equilateral

#### Answer:

Let A, B and C be the angles of the triangle. Then, one angle of a triangle is equal to the sum of the other two angles,

i.e. ∠A = ∠B + ∠C …(1)

By Angle Sum Property of Triangle

∠A + ∠B + ∠C = 180°

⇒ ∠A + ∠A = 180°

⇒ â€‹2∠A = 180°

⇒ â€‹∠A = 90°

Thus, $\u25b3\mathrm{ABC}$ is right angled triangle.

Hence, the correct answer is option (c).

#### Page No 162:

#### Question 26:

If the exterior angle of a triangle is 130° and its interior opposite angles are equal, then measure of each interior opposite angle is

(a) 55°

(b) 65°

(c) 50°

(d) 60°

#### Answer:

Given: The exterior angle of a triangle is 130°

Let the interior angle be *x*.

Given that, interior opposite angles are equal.

Since the measure of any exterior angle is equal to the sum of two opposite interior angles.

$x+x=130\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2x=130\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=65\xb0\phantom{\rule{0ex}{0ex}}$

Hence, the correct answer is option (b).

#### Page No 163:

#### Question 27:

If one of the angles of a triangle is 110°, then the angle between the bisectors of the other two angles is

(a) 70°

(b) 110°

(c) 35°

(d) 145°

#### Answer:

Given: one of the angles of a triangle is 110°

Consider $\u25b3\mathrm{ABC},\mathrm{with}\angle \mathrm{A}=110\xb0$

In, by Angle Sum Property of Triangle

In, by Angle Sum Property of Triangle

Hence, the correct answer is option (d).

#### Page No 163:

#### Question 28:

In âˆ†ABC, AD is the bisector of ∠A meeting BC at D, CF ⊥ AB and E is the mid-point of AC. Then median of the triangle is

(a) AD

(b) BE

(c) FC

(d) DE

#### Answer:

Given; In âˆ†ABC, AD is the bisector of ∠A meeting BC at D, CF ⊥ AB and E is the mid-point of AC.

Since the median of a triangle bisects the opposite sides.

As, E is the mid-point of AC ⇒ AE = EC

Thus, BE is the median of the âˆ†ABC.

Hence, the correct answer is option (b).

#### Page No 163:

#### Question 29:

In âˆ†PQR, if ∠P = 60°, and ∠Q = 40°, then the exterior angle formed by producing QR is equal to

(a) 60°

(b) 120°

(c) 100°

(d) 80°

#### Answer:

Given: In âˆ†PQR, if ∠P = 60°, and ∠Q = 40°

Since the measure of the exterior angle is equal to the sum of the opposite two interior angles.

In âˆ†PQR

$\angle \mathrm{P}+\angle \mathrm{Q}=x\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0+40\xb0=x\phantom{\rule{0ex}{0ex}}\Rightarrow 100\xb0=x$

Hence, the correct answer is option (c).

#### Page No 163:

#### Question 30:

Which of the following triplets cannot be the angles of a triangle?

(a) 67°, 51°, 62°

(b) 70°, 83°, 27°

(c) 90°, 70°, 20°

(d) 40°, 132°, 18°

#### Answer:

The sum of the interior angles of a triangle is 180°.

Now, verifing the given triplets:

(a) 67°+ 51°+ 62° = 180°

(b) 70° + 83° + 27° = 180°

(c) 90° + 70° + 20° = 180°

(d) 40° + 132°+ 18° = 190°

Thus, option (d) cannot be the angles of a triangle.

Hence, the correct answer is option (d).

#### Page No 163:

#### Question 31:

Which of the following can be the length of the third side of a triangle whose two sides measure 18 cm and 14 cm?

(a) 4 cm

(b) 3 cm

(c) 5 cm

(d) 32 cm

#### Answer:

Since the sum of any two sides of a triangle is always greater than the third side. Hence, option (c) satisfies the given condition.

(a) 18 + 14 > 4, 18 + 4 > 14, 4 + 14 = 18 ⇒ Not possible

(b) 18 + 14 > 3, 18 + 3 > 14, 3 + 14 < 18 ⇒ Not possible

(c) 18 + 14 > 5, 18 + 5 > 14, 5 + 14 > 18 ⇒ Possible

(d) 18 + 14 = 32 ⇒ Not possible

Thus, the length of the third side of a triangle whose two sides measure 18 cm and 14 cm is 5 cm.

Hence, the correct answer is option (c).

#### Page No 163:

#### Question 32:

How many altitudes does a triangle have?

(a) 1

(b) 3

(c) 6

(d) 9

#### Answer:

A triangle has 3 altitudes

Hence, the correct answer is option (b).

#### Page No 163:

#### Question 33:

If we join a vertex to a point on opposite side which divides that side in the ratio 1:1, then what is the special name of that line segment?

(a) Median

(b) Angle bisector

(c) Altitude

(d) Hypotenuse

#### Answer:

Consider ΔABC in which AD divides BC in the ratio 1:1.

$\mathrm{BD}:\mathrm{DC}=1:1\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{\mathrm{BD}}{\mathrm{DC}}=\frac{1}{1}\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{BD}=\mathrm{DC}$

⇒ D is the mid-point of BC.

Thus, AD is the mdian of the ΔABC.

Hence, the correct answer is option (a).

#### Page No 163:

#### Question 34:

The measures of ∠*x* and ∠*y* in the given figure are respectively

(a) 30°, 60°

(b) 40°, 40°

(c) 70°, 70°

(d) 70°, 60°

#### Answer:

In $\u25b3\mathrm{PQR}$, by Exterior angle property

$\angle \mathrm{RPQ}+\angle \mathrm{PQR}=120\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle x+50\xb0=120\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle x=70\xb0$

$\angle \mathrm{PRQ}+\angle y=180\xb0\left(\mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 120\xb0+\angle y=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle y=60\xb0$

Hence, the correct answer is option (d).

#### Page No 163:

#### Question 35:

If length of two sides of a triangle are 6 cm and 10 cm, then the length of the third side can be

(a) 3 cm

(b) 4 cm

(c) 2 cm

(d) 6 cm

#### Answer:

Given: The length of two sides of a triangle are 6 cm and 10 cm.

Let the length of the third side be *x *cm.

Since the sum of any two sides of a triangle is always greater than the third side.

∴ *x* < 6 +10

⇒ *x* < 16 .....(1)

Also, the difference of any two sides of a triangle is always lesser than the third side.

∴ *x* > 10 − 6

⇒ *x* > 4 .....(2)

From (1) and (2), we get

4 < *x* < 16

Thus, the length of the third side of the triangle is 6 cm.

Hence, the correct answer is option (d).

#### Page No 163:

#### Question 36:

In a right-angled triangle ABC, if angle B = 90°, BC = 3 cm and AC = 5 cm, then the length of side AB is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

#### Answer:

Given: In a right-angled triangle ABC, if angle B = 90°, BC = 3 cm, and AC = 5 cm.

Hence, the correct answer is option (b).

#### Page No 164:

#### Question 37:

In a right-angled triangle ABC, if angle B = 90°, then which of the following is true?

(a) AB^{2} = BC^{2} + AC^{2}

(b) AC^{2} = AB^{2} + BC^{2}

(c) AB = BC + AC

(d) AC = AB + BC

#### Answer:

According to Pythagoras theorem

$\mathrm{In}\u25b3\mathrm{ABC},\mathrm{by}\mathrm{Pythagoras}\mathrm{Theorem}\phantom{\rule{0ex}{0ex}}{\left(\mathrm{Hypotenuse}\right)}^{2}={\left(\mathrm{Base}\right)}^{2}+{\left(\mathrm{Perpendicular}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={\mathrm{BC}}^{2}+{\mathrm{AB}}^{2}$

Hence, the correct answer is option (b).

#### Page No 164:

#### Question 38:

Which of the following figures will have it’s altitude outside the triangle?

#### Answer:

Since the perpendicular line segment from a vertex of a triangle to its opposite side is called the altitude of the triangle.

Thus, in option (d) the figure will have the altitude outside the triangle.

Hence, the correct answer is option (d).

#### Page No 164:

#### Question 39:

In the given figure, if AB || CD, then

(a) ∠2 = ∠3

(b) ∠1 = ∠4

(c) ∠4 = ∠1 + ∠2

(d) ∠1 + ∠2 = ∠3 + ∠4

#### Answer:

Given: AB || CD

$\therefore \angle 1=\angle 3\left(\mathrm{Alternate}\mathrm{interior}\mathrm{angles}\right)$

$\mathrm{In}\u25b3\mathrm{ABC},\mathrm{by}\mathrm{Exterior}\mathrm{angle}\mathrm{proprty}\phantom{\rule{0ex}{0ex}}\angle 1+\angle 2=\left(\angle 3+\angle 4\right)$

Hence, the correct answer is option (d).

#### Page No 165:

#### Question 40:

In âˆ†ABC, ∠A = 100°, AD bisects ∠A and AD⊥BC. Then, ∠B is equal to

(a) 80°

(b) 20°

(c) 40°

(d) 30°

#### Answer:

Given: In âˆ†ABC, ∠A = 100°, AD bisects ∠A and AD ⊥ BC.

Since AD bisects ∠A.

$\therefore \angle \mathrm{BAD}=\angle \mathrm{CAD}=50\xb0$

Since AD ⊥ BC.

$\therefore \angle \mathrm{ADB}=\angle \mathrm{ADC}=90\xb0$

Now, in $\u25b3\mathrm{ABD}$, by Angle Sum Property of triangle

$\angle \mathrm{ABD}+\angle \mathrm{BDA}+\angle \mathrm{DAB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ABD}+90\xb0+50\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ABD}+140\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ABD}=40\xb0$

Thus, ∠B = 40°

Hence, the correct answer is option (c).

#### Page No 165:

#### Question 41:

In âˆ†ABC, ∠A = 50°, ∠B = 70° and bisector of ∠C meets AB in D . Measure of $\angle $ADC is.

(a) 50°

(b) 100°

(c) 30°

(d) 70°

#### Answer:

Given: In âˆ†ABC, ∠A = 50°, ∠B = 70°, and the bisector of ∠C meet AB in D.

In âˆ†ADC, by angle sum property of the triangle

$\angle \mathrm{ADC}+\angle \mathrm{DAC}+\angle \mathrm{ACD}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ADC}+50\xb0+\angle \mathrm{ACD}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ACD}=130\xb0-\angle \mathrm{ADC}$

In âˆ†BDC, by exterior angle property of the triangle

$\angle \mathrm{ADC}=\angle \mathrm{DBC}+\angle \mathrm{BCD}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ADC}=70\xb0+130\xb0-\angle \mathrm{ADC}\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{ADC}=200\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ADC}=100\xb0$

Hence, the correct answer is option (b).

#### Page No 165:

#### Question 42:

If for âˆ†ABC and âˆ†DEF, the correspondence CAB ↔ EDF gives a congruence, then which of the following is not true?

(a) AC = DE

(b) AB = EF

(c) ∠A =∠D

(d) ∠C =∠E

#### Answer:

Given: In âˆ†ABC and âˆ†DEF, the correspondence CAB ↔ EDF gives a congruence.

Two figures are said to be congruent, if the trace copy of figure 1 fits exactly on that of figure 2.

So,

CA = ED

AB = DF

CB = EF

$\angle \mathrm{C}=\angle \mathrm{E}\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{D}\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}=\angle \mathrm{F}$

Hence, the correct answer is option (b).

#### Page No 165:

#### Question 43:

In the given figure, M is the mid-point of both AC and BD. Then

(a) ∠1 = ∠2

(b) ∠1 = ∠4

(c) ∠2 = ∠4

(d) ∠1 = ∠3

#### Answer:

Given: M is the mid-point of both AC and BD.

$\mathrm{In}\u25b3\mathrm{AMB}\mathrm{and}\u25b3\mathrm{CMD}\phantom{\rule{0ex}{0ex}}\mathrm{AM}=\mathrm{CM}\left(\because \mathrm{M}\mathrm{is}\mathrm{the}\mathrm{midpoint}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{AMB}=\angle \mathrm{CMD}\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{anbles}\right)\phantom{\rule{0ex}{0ex}}\mathrm{BM}=\mathrm{DM}\left(\because \mathrm{M}\mathrm{is}\mathrm{the}\mathrm{midpoint}\right)\phantom{\rule{0ex}{0ex}}\therefore \u25b3\mathrm{AMB}\cong \u25b3\mathrm{CMD}\left(\mathrm{By}\mathrm{SAS}\mathrm{congruence}\mathrm{criterion}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow \angle 1=\angle 4\left(\mathrm{By}\mathrm{CPCT}\right)\phantom{\rule{0ex}{0ex}}$

Hence, the correct answer is option (b).

#### Page No 165:

#### Question 44:

If D is the mid-point of the side BC in âˆ†ABC where AB = AC, then ∠ADC is

(a) 60°

(b) 45°

(c) 120°

(d) 90°

#### Answer:

Given: D is the mid-point of the side BC in âˆ†ABC where AB = AC.

$\mathrm{In}\u25b3\mathrm{ADB}\mathrm{and}\u25b3\mathrm{ADC}\phantom{\rule{0ex}{0ex}}\mathrm{BD}=\mathrm{DC}\left(\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{midpoint}\right)\phantom{\rule{0ex}{0ex}}\mathrm{AB}=\mathrm{AC}\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{AD}=\mathrm{AD}\left(\mathrm{Common}\mathrm{side}\right)$

By SSS congruency criterion

$\u25b3\mathrm{ADB}\cong \mathrm{ACD}\phantom{\rule{0ex}{0ex}}\angle \mathrm{ADB}=\angle \mathrm{ADC}\left(\mathrm{By}\mathrm{CPCT}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{ADB}+\angle \mathrm{ADC}=180\xb0\left(\mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{ADC}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ADC}=90\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Hence, the correct answer is option (d).

#### Page No 166:

#### Question 45:

Two triangles are congruent, if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them of the other triangle. This is known as the

(a) RHS congruence criterion

(b) ASA congruence criterion

(c) SAS congruence criterion

(d) AAA congruence criterion

#### Answer:

Under ASA congruence criterion, two triangles are congruent, if two angles and the side included between them in one of the triangles are equal to the two angles and the side included between them in the other triangle.

Hence, the correct answer is option (b).

#### Page No 166:

#### Question 46:

By which congruency criterion, the two triangles in the given figure are congruent?

(a) RHS (b) ASA

(c) SSS (d) SAS

#### Answer:

$\mathrm{In}\u25b3\mathrm{PRQ}\mathrm{and}\u25b3\mathrm{PSQ}\phantom{\rule{0ex}{0ex}}\mathrm{PR}=\mathrm{PS}\left(=\mathrm{a}\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\mathrm{RQ}=\mathrm{SQ}\left(=\mathrm{b}\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\mathrm{PQ}=\mathrm{PQ}\left(\mathrm{common}\right)\phantom{\rule{0ex}{0ex}}\therefore \u25b3\mathrm{PRQ}\cong \u25b3\mathrm{PSQ}\left(\mathrm{SSS}\mathrm{congruency}\mathrm{criterion}\right)\phantom{\rule{0ex}{0ex}}$

Hence, the correct answer is option (c).

#### Page No 166:

#### Question 47:

By which of the following criterion two triangles cannot be proved congruent?

(a) AAA

(b) SSS

(c) SAS

(d) ASA

#### Answer:

AAA is not a congruency criterion, because if all three angles of two triangles are equal; this does not imply that both the triangles fit exactly on each other.

Hence, the correct answer is option (a).

#### Page No 166:

#### Question 48:

If âˆ†PQR is congruent to âˆ†STU , then what is the length of TU?

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) cannot be determined

#### Answer:

Given: âˆ†PQR ≅ âˆ†STU

QR = TU (By CPCT)

⇒ 6 cm = TU

Hence, the correct answer is option (b).

#### Page No 166:

#### Question 49:

If âˆ†ABC and âˆ†DBC are on the same base BC, AB = DC and AC = DB , then which of the following gives a congruence relationship?

(a) âˆ† ABC ≅ âˆ† DBC

(b) âˆ† ABC ≅ âˆ†CBD

(c) âˆ† ABC ≅ âˆ† DCB

(d) âˆ† ABC ≅ âˆ†BCD

#### Answer:

Given: âˆ†ABC and âˆ†DBC are on the same base BC, AB = DC and AC = DB

In âˆ†ABC and âˆ†DBC,

AB = DC (Given)

AC = DB (Given)

BC = CB (Common)

∴âˆ†ABC ≅ âˆ†DCB (By SSS congruency)

Hence, the correct answer is option (c).

#### Page No 167:

#### Question 50:

**Fill in the blanks to make the statements true.**

The ___________ triangle always has altitude outside itself.

#### Answer:

The __ obtuse__ angled triangle always has altitude outside itself.

#### Page No 167:

#### Question 51:

**Fill in the blanks to make the statements true.**

The sum of an exterior angle of a triangle and its adjacent angle is always_____________ .

#### Answer:

The sum of an exterior angle of a triangle and its adjacent angle is always,** 180° because they form a linear pair**.

#### Page No 167:

#### Question 52:

**Fill in the blanks to make the statements true.**

The longest side of a right angled triangle is called its ______________.

#### Answer:

The longest side of a right angled triangle is called its __ Hypotenuse__.

#### Page No 167:

#### Question 53:

**Fill in the blanks to make the statements true.**

Median is also called___________ in an equilateral triangle.

#### Answer:

Median is also called__ an altitude __in an equilateral triangle.

#### Page No 167:

#### Question 54:

**Fill in the blanks to make the statements true.**

Measures of each of the angles of an equilateral triangle is _____________ .

#### Answer:

Measures of each of the angles of an equilateral triangle is __ 60°__.

As all the angles in an equilateral triangle are equal.

Let

*x*be the angle of the equilateral triangle.

$x+x+x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=60\xb0$

#### Page No 167:

#### Question 55:

**Fill in the blanks to make the statements true.**

In an isosceles triangle, two angles are always__________ .

#### Answer:

In an isosceles triangle, two angles are always __ equal__.

Since, if two sides are equal, then the angles opposite them are equal.

#### Page No 167:

#### Question 56:

**Fill in the blanks to make the statements true.**

In an isosceles triangle, angles opposite to equal sides are______________ .

#### Answer:

In an isosceles triangle, angles opposite to equal sides are ** equal**.

Because if two angles are equal then the sides opposite to them are also equal.

#### Page No 167:

#### Question 57:

**Fill in the blanks to make the statements true.**

If one angle of a triangle is equal to the sum of other two, then the measure of that angle is _________________ .

#### Answer:

Let the angles of a triangle be *a*, *b*, and *c*.

According to the question

$a=b+c$ .....(1)

By angle sum property of triangle

$a+b+c=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow a+a=180\xb0\left[\mathrm{From}\left(1\right)\right]\phantom{\rule{0ex}{0ex}}\Rightarrow 2a=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow a=90\xb0$

If one angle of a triangle is equal to the sum of other two, then the measure of that angle is __ 90°__.

#### Page No 167:

#### Question 58:

**Fill in the blanks to make the statements true.**

Every triangle has at least _____________ acute angle (s).

#### Answer:

Every triangle has at least__ two __acute angle (s).

#### Page No 167:

#### Question 59:

**Fill in the blanks to make the statements true.**

Two line segments are congruent, if they are of_____________lengths.

#### Answer:

Two line segments are congruent, if they are of ** equal **lengths.

#### Page No 167:

#### Question 60:

**Fill in the blanks to make the statements true.**

Two angles are said to be_______________, if they have equal measures.

#### Answer:

Two angles are said to be** congruent**, if they have equal measures.

#### Page No 167:

#### Question 61:

**Fill in the blanks to make the statements true.**

Two rectangles are congruent, if they have same _______________and______________.

#### Answer:

Two rectangles are congruent, if they have same __ length __and

__.__

**breadth**#### Page No 167:

#### Question 62:

**Fill in the blanks to make the statements true.**

Two squares are congruent, if they have same________________ .

#### Answer:

Two squares are congruent, if they have same ** side**.

#### Page No 167:

#### Question 63:

**Fill in the blanks to make the statements true.**

If âˆ†PQR and âˆ†XYZ are congruent under the correspondence QPR ↔ XYZ, then

(i) ∠R = _____________

(ii) QR =_____________

(iii) ∠P = ____________

(iv) QP =_____________

(v) ∠Q = _____________

(vi) RP =_____________

#### Answer:

Given: âˆ†PQR and âˆ†XYZ are congruent under the correspondence QPR ↔ XYZ, then

(i) ∠R = ∠Z

(ii) QR = XZ

(iii) ∠P = ∠Y

(iv) QP = XY

(v) ∠Q = ∠X

(vi) RP = ZY

#### Page No 168:

#### Question 64:

**Fill in the blanks to make the statements true.**

In the given figure, âˆ†PQR ≅ âˆ†___________

#### Answer:

$\mathrm{In}\u25b3\mathrm{PQR}\mathrm{and}\u25b3\mathrm{XZY}\phantom{\rule{0ex}{0ex}}\mathrm{PQ}=\mathrm{XZ}\left(=3.5\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{Q}=\angle \mathrm{Z}\left(\mathrm{Each}45\xb0\right)\phantom{\rule{0ex}{0ex}}\mathrm{QR}=\mathrm{YZ}\left(=5\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{PQR}\cong \u25b3\mathrm{XZY}\left(\mathrm{By}\mathrm{SAS}\mathrm{Congruency}\right)$

In the given figure, âˆ†PQR ≅ **âˆ†XZY **

#### Page No 168:

#### Question 65:

**Fill in the blanks to make the statements true.**

In the given figure, âˆ†PQR≅ âˆ†_____________

#### Answer:

$\mathrm{In}\u25b3\mathrm{PSR}\mathrm{and}\u25b3\mathrm{RQP}\phantom{\rule{0ex}{0ex}}\mathrm{PS}=\mathrm{QR}\left(=4.1\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{P}=\angle \mathrm{R}\left(\mathrm{Each}45\xb0\right)\phantom{\rule{0ex}{0ex}}\mathrm{PR}=\mathrm{PR}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{PSR}\cong \u25b3\mathrm{RQP}\left(\mathrm{By}\mathrm{SAS}\mathrm{Congruency}\right)$

#### Page No 168:

#### Question 66:

**Fill in the blanks to make the statements true.**

In the given figure, âˆ† ____________≅ âˆ† PQR

#### Answer:

$\mathrm{In}\u25b3\mathrm{PQR}\mathrm{and}\u25b3\mathrm{DRQ}\phantom{\rule{0ex}{0ex}}\angle \mathrm{PQR}=\angle \mathrm{DRQ}\left(\mathrm{Each}70\xb0\right)\phantom{\rule{0ex}{0ex}}\mathrm{QR}=\mathrm{QR}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{PRQ}=\angle \mathrm{DQR}\left(\mathrm{Each}40\xb0\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{PQR}\cong \u25b3\mathrm{DRQ}\left(\mathrm{By}\mathrm{ASA}\mathrm{Congruency}\right)$

In the given figure, __ âˆ†DRQ__ ≅ âˆ† PQR

#### Page No 168:

#### Question 67:

**Fill in the blanks to make the statements true.**

In the given figure, âˆ† ARO ≅ âˆ† _____________

#### Answer:

$\mathrm{In}\u25b3\mathrm{ARO}\mathrm{and}\u25b3\mathrm{PQO}\phantom{\rule{0ex}{0ex}}\angle \mathrm{AOR}=\mathrm{POQ}\left(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{ARO}=\angle \mathrm{PQO}=55\xb0\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{AO}=\mathrm{PO}\left(=2.5\mathrm{cm}\right)\phantom{\rule{0ex}{0ex}}\u25b3\mathrm{ARO}\cong \u25b3\mathrm{PQO}\left(\mathrm{By}\mathrm{ASS}\mathrm{Congruency}\mathrm{criterion}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

In the given figure, âˆ† ARO ≅ **âˆ† PQO**

#### Page No 168:

#### Question 68:

**Fill in the blanks to make the statements true.**

In the given figure, AB = AD and ∠BAC = ∠DAC. Then

(i) âˆ† ___________ ≅ âˆ†ABC.

(ii) BC =________________

(iii) ∠BCA =______________ .

(iv) Line segment AC bisects _____________ and ____________.

#### Answer:

$\u25b3\mathrm{ABC}\mathrm{and}\u25b3\mathrm{ADC}\phantom{\rule{0ex}{0ex}}\mathrm{AB}=\mathrm{AD}\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{BAC}=\angle \mathrm{DAC}\left(\mathrm{Given}\right)\phantom{\rule{0ex}{0ex}}\mathrm{AC}=\mathrm{AC}\left(\mathrm{Common}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{i}\right)\u25b3\overline{)\mathbf{ABC}}\cong \u25b3\mathrm{ADC}\phantom{\rule{0ex}{0ex}}\left(\mathrm{ii}\right)\mathrm{BC}=\overline{)\mathbf{DC}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iii}\right)\angle \mathrm{BCA}=\overline{)\mathbf{\angle}\mathbf{DCA}}\phantom{\rule{0ex}{0ex}}\left(\mathrm{iv}\right)\mathrm{Line}\mathrm{segment}\mathrm{AC}\mathrm{bisects}\overline{)\mathbf{\angle}\mathbf{BAC}}\mathbf{}\mathrm{and}\mathbf{}\overline{)\mathbf{\angle}\mathbf{BCD}}$

#### Page No 169:

#### Question 69:

**Fill in the blanks to make the statements true.**

In the given figure,

(i) ∠TPQ = ∠ _____ + ∠ _____

(ii) ∠UQR = ∠ _____ + ∠ _____

(iii) ∠PRS = ∠ _____ + ∠ _____

#### Answer:

According to the exterior angle property, the measure of an exterior angle is equal to the sum of the two opposite interior angles.

(i) ∠TPQ= ** ∠PQR** +

__∠PRQ__(ii) ∠UQR=

**+**

__∠QRP__

__∠QPR__(iii) ∠PRS =

**+**

__∠RPQ__

__∠RQP__#### Page No 169:

#### Question 70:

**State whether the statement is True or False.**

In a triangle, sum of squares of two sides is equal to the square of the third side.

#### Answer:

False

Only in a right angled triangle, the sum of two shorter sides is equal to the square of the third side.

#### Page No 169:

#### Question 71:

**State whether the statement is True or False.**

Sum of two sides of a triangle is greater than or equal to the third side.

#### Answer:

False

Sum of two sides of a triangle is greater than the third side.

#### Page No 169:

#### Question 72:

**State whether the statement is True or False.**

The difference between the lengths of any two sides of a triangle is smaller than the length of third side.

#### Answer:

True

The difference between the lengths of any two sides of a triangle is smaller than the length of third side.

#### Page No 169:

#### Question 73:

**State whether the statement is True or False.**

In âˆ†ABC, AB = 3.5 cm, AC = 5 cm, BC = 6 cm and in âˆ†PQR, PR= 3.5 cm, PQ = 5 cm, RQ = 6 cm. Then âˆ†ABC ≅ âˆ†PQR.

#### Answer:

In âˆ†ABC and âˆ†PQR

AB = PR (= 3.5 cm)

AC = PQ (=5 cm)

BC = RQ (= 6 cm)

∴ âˆ†ABC ≅ âˆ†PQR (By SSS Congruency criterion)

Thus, the given statement is False.

#### Page No 170:

#### Question 74:

**State whether the statement is True or False.**

Sum of any two angles of a triangle is always greater than the third angle.

#### Answer:

False

It is not necessary that sum of any two angles of a triangle is always greater than the third angle, e.g. Let the angles of a triangle be 20°, 50° and 110°, respectively.

Hence, 20° + 50° = 70°, which is less than 110°.

#### Page No 170:

#### Question 75:

**State whether the statement is True or False.**

The sum of the measures of three angles of a triangle is greater than 180°.

#### Answer:

False

The sum of the measures of three angles of a triangle is always equal to 180°.

#### Page No 170:

#### Question 76:

**State whether the statement is True or False.**

It is possible to have a right-angled equilateral triangle.

#### Answer:

False

In a right angled triangle, one angle is equal to 90° and in equilateral triangle, all angles are equal to 60°.

#### Page No 170:

#### Question 77:

**State whether the statement is True or False.**

If M is the mid-point of a line segment AB, then we can say that AM and MB are congruent.

#### Answer:

True.

Given: M is the mid-point of a line segment AB.

∴ AM = MB

When two lines are of same length they are congruent to each other.

#### Page No 170:

#### Question 78:

**State whether the statement is True or False.**

It is possible to have a triangle in which two of the angles are right angles.

#### Answer:

False

Because if in a triangle two angles are right angles, then by the Angle Sum Property of the triangle the third angle = 180° – (90° + 90°) = 0°, which is not possible.

#### Page No 170:

#### Question 79:

**State whether the statement is True or False.**

It is possible to have a triangle in which two of the angles are obtuse.

#### Answer:

False

Obtuse angles are angles which are greater than 90°.

Thus, the sum of two obtuse angles will be greater than 180°, which is impossible as the sum of all angles of a triangle is 180°.

#### Page No 170:

#### Question 80:

**State whether the statement is True or False.**

It is possible to have a triangle in which two angles are acute.

#### Answer:

True

In a triangle, at least two angles must be acute angle.

#### Page No 170:

#### Question 81:

**State whether the statement is True or False.**

It is possible to have a triangle in which each angle is less than 60°.

#### Answer:

False.

The sum of all angles in a triangle is equal to 180°.

Thus, all three angles can never be less than 60°.

#### Page No 170:

#### Question 82:

**State whether the statement is True or False.**

It is possible to have a triangle in which each angle is greater than 60°.

#### Answer:

False

If all the angles are greater than 60° in a triangle, then the sum of all three angles will exceed 180°, which cannot be possible in the case of a triangle.

#### Page No 170:

#### Question 83:

**State whether the statement is True or False.**

It is possible to have a triangle in which each angle is equal to 60°.

#### Answer:

True

The triangle in which each angle is equal to 60° is called an equilateral triangle.

#### Page No 170:

#### Question 84:

**State whether the statement is True or False.**

A right-angled triangle may have all sides equal.

#### Answer:

False

Hypotenuse is always the greater than the other two sides of the right angled triangle.

#### Page No 170:

#### Question 85:

**State whether the statement is True or False.**

If two angles of a triangle are equal, the third angle is also equal to each of the other two angles.

#### Answer:

False

In an isosceles triangle, always two angles are equal and not the third one.

#### Page No 170:

#### Question 86:

**State whether the statement is True or False.**

In the given figure, two triangles are congruent by RHS.

#### Answer:

False

$\mathrm{In}\u25b3\mathrm{XYZ},\angle \mathrm{X}=90\xb0.\phantom{\rule{0ex}{0ex}}\mathrm{And}\mathrm{in}\u25b3\mathrm{ABC},\angle \mathrm{A}\mathrm{can}\mathrm{be}\mathrm{of}\mathrm{any}\mathrm{measure}.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Thus},\u25b3\mathrm{XYZ}\ncong \u25b3\mathrm{ABC}\phantom{\rule{0ex}{0ex}}$

#### Page No 170:

#### Question 87:

**State whether the statement is True or False.**

The congruent figures super impose each other completely.

#### Answer:

True

Because congruent figures have the same shape and same size. Thus, they superimpose on each other completely.

#### Page No 170:

#### Question 88:

**State whether the statement is True or False.**

A one rupee coin is congruent to a five rupee coin.

#### Answer:

False

Because a five-rupee coin is bigger in size than a one-rupee coin. So, they are not congruent with each other.

#### Page No 170:

#### Question 89:

**State whether the statement is True or False.**

The top and bottom faces of a kaleidoscope are congruent.

#### Answer:

True

Because they are of the same size and shape.

#### Page No 170:

#### Question 90:

**State whether the statement is True or False.**

Two acute angles are congruent.

#### Answer:

False

Because the measure of two acute angles may be different as it's not necessarily to be equal.

#### Page No 170:

#### Question 91:

**State whether the statement is True or False.**

Two right angles are congruent.

#### Answer:

True

Since, the measure of right angles is always same.

#### Page No 170:

#### Question 92:

**State whether the statement is True or False.**

Two figures are congruent, if they have the same shape.

#### Answer:

False

Because two figures are congruent if they have the same shape and same size.

#### Page No 170:

#### Question 93:

**State whether the statement is True or False.**

If the areas of two squares is same, they are congruent.

#### Answer:

True

Because two squares will have the same areas only if their sides are equal and squares with the same sides will superimpose each other.

#### Page No 170:

#### Question 94:

**State whether the statement is True or False.**

If the areas of two rectangles are same, they are congruent.

#### Answer:

False

Because rectangles with different lengths and breadths may have equal areas. But, they will not superimpose on each other.

#### Page No 170:

#### Question 95:

**State whether the statement is True or False.**

If the areas of two circles are the same, they are congruent.

#### Answer:

True

Because areas of two circles will be equal only if their radii are equal and circles with the same radii will superimpose on each other.

#### Page No 170:

#### Question 96:

**State whether the statement is True or False.**

Two squares having same perimeter are congruent.

#### Answer:

True

Because if two squares have the same perimeter, then their sides will be equal. Hence, the squares will superimpose on each other.

#### Page No 170:

#### Question 97:

**State whether the statement is True or False.**

Two circles having same circumference are congruent.

#### Answer:

True

If two circles have the same circumference, then their radii will be equal. Thus, the circles will superimpose on each other.

#### Page No 171:

#### Question 98:

**State whether the statement is True or False.**

If three angles of two triangles are equal, triangles are congruent.

#### Answer:

False

Consider two equilateral triangles ΔABC and ΔDEF with sides 4 cm and 7 cm respectively.

Since both the triangles have same angles but their size is different because of different side lengths.

Thus, they are not congruent.

#### Page No 171:

#### Question 99:

**State whether the statement is True or False.**

If two legs of a right triangle are equal to two legs of another right triangle, then the right triangles are congruent.

#### Answer:

True

If two legs of a right-angled triangle are equal to two legs of another right-angled triangle, then their third leg will also be equal. Thus, they will have the same shape and same size.

#### Page No 171:

#### Question 100:

**State whether the statement is True or False.**

If two sides and one angle of a triangle are equal to the two sides and angle of another triangle, then the two triangles are congruent.

#### Answer:

False

Because if two sides and the angle included between them of the other triangle, then the two triangles will be congruent.

#### Page No 171:

#### Question 101:

**State whether the statement is True or False.**

If two triangles are congruent, then the corresponding angles are equal.

#### Answer:

True

Because if two triangles are congruent, then the pairs of all their corresponding sides and pairs of all their corresponding angles are equal.

#### Page No 171:

#### Question 102:

**State whether the statement is True or False.**

If two angles and a side of a triangle are equal to two angles and a side of another triangle, then the triangles are congruent.

#### Answer:

False

Because if two angles and the side included between them of a triangle are equal to two angles and included a side between them of the other triangle, then triangles are congruent.

#### Page No 171:

#### Question 103:

**State whether the statement is True or False.**

If the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle, then the triangles are congruent.

#### Answer:

False

Because if the hypotenuse and a side of one of the triangles are equal to the hypotenuse and one of the sides of the other triangle then the right-angled triangles are congruent.

#### Page No 171:

#### Question 104:

**State whether the statement is True or False.**

If hypotenuse and an acute angle of one right triangle are equal to the hypotenuse and an acute angle of another right triangle, then the triangles are congruent.

#### Answer:

True

Because if an acute angle of one right triangle is equal to the acute angle of another right triangle. Then the remaining angle in both the triangles must also be equal according to the Angle Sum Property of the triangle.

Since the hypotenuse of one right triangle is equal to the hypotenuse of another right triangle and the two angles on the vertices of the hypotenuse of both triangles are also equal to each other. So, by ASA congruency both the triangles will be congruent to each other.

#### Page No 171:

#### Question 105:

**State whether the statement is True or False.**

AAS congruence criterion is same as ASA congruence criterion.

#### Answer:

False

In ASA congruence criterion, the side ‘S’ included between the two angles of the triangle. In AAS congruence criterion, side ‘S’ is not included between two angles.

#### Page No 171:

#### Question 106:

**State whether the statement is True or False.**

In the given figure, AD ⊥ BC and AD is the bisector of angle BAC. Then, âˆ†ABD ≅ âˆ†ACD by RHS.

#### Answer:

False

In âˆ†ABD and âˆ†ACD

$\angle $BAD = $\angle $DAC (âˆµAD is the bisector of $\angle $BAC)

AD = AD (Common)

$\angle \mathrm{ADB}=\angle \mathrm{ADC}$ (âˆµAD ⊥ BC)

∴ âˆ†ABD ≅ âˆ†ACD by ASA Congruency

#### Page No 171:

#### Question 107:

The measure of three angles of a triangle are in the ratio 5 : 3 : 1. Find the measures of these angles.

#### Answer:

Given: The measure of three angles of a triangle are in the ratio 5 : 3 : 1.

Let the common ratio be *x*.

∴ The measure of the three angles of the triangle be 5*x*, 3*x* and *x*.

By the Angle Sum Property of the triangle,

5*x + *3*x* + *x *=180°

⇒ 8*x *=180°

⇒ *x *= 22.5°

∴ 5*x** *= 112.5° and 3*x** *= 67.5°

Thus, the measures of these angles are 112.5°, 67.5° and 22.5°.

#### Page No 171:

#### Question 108:

In the given figure, find the value of *x*.

#### Answer:

The sum of all three angles in a triangle is equal to 180°.

$\angle \mathrm{ADC}+\angle \mathrm{ADB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+\angle \mathrm{ADB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ADB}=90\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3\mathrm{ADB},\mathrm{by}\mathrm{Angle}\mathrm{Sum}\mathrm{Property}\mathrm{of}\mathrm{Triangle}\phantom{\rule{0ex}{0ex}}\angle \mathrm{ADB}+\angle \mathrm{DBA}+\angle \mathrm{BAD}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+55\xb0+x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=180\xb0-145\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=35\xb0\phantom{\rule{0ex}{0ex}}$

Thus, the value of *x *is 35°*.*

#### Page No 172:

#### Question 109:

In the given figure (i) and (ii), find the values of *a*, *b* and *c*.

#### Answer:

$\mathrm{In}\u25b3\mathrm{ADC}\phantom{\rule{0ex}{0ex}}60\xb0+70\xb0+c=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 130\xb0+c=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow c=50\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}c+b=180\xb0\left(\mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 50\xb0+b=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow b=130\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3\mathrm{ADB}\phantom{\rule{0ex}{0ex}}30\xb0+a+b=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+a+130\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow a=20\xb0$

$\mathrm{In}\u25b3\mathrm{PQS}\phantom{\rule{0ex}{0ex}}60\xb0+55\xb0+a=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 115\xb0+a=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow a=65\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}a+b=180\xb0\left(\mathrm{Linear}\mathrm{pair}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 65\xb0+b=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow b=115\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3\mathrm{PSR}\phantom{\rule{0ex}{0ex}}40\xb0+c+b=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 40\xb0+c+115\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow c=25\xb0$

#### Page No 172:

#### Question 110:

In triangle XYZ, the measure of angle X is 30° greater than the measure of angle Y and angle Z is a right angle. Find the measure of ∠Y.

#### Answer:

Given: In triangle XYZ, the measure of angle X is 30° greater than the measure of angle Y and angle Z is a right angle.

$\angle \mathrm{Z}=90\xb0\phantom{\rule{0ex}{0ex}}\angle \mathrm{X}=30\xb0+\angle \mathrm{Y}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3\mathrm{XYZ},\mathrm{by}\mathrm{Angle}\mathrm{Sum}\mathrm{Property}\mathrm{of}\mathrm{Triangle}\phantom{\rule{0ex}{0ex}}\angle \mathrm{X}+\angle \mathrm{Y}+\angle \mathrm{Z}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+\angle \mathrm{Y}+\angle \mathrm{Y}+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\angle \mathrm{Y}=60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{Y}=30\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 172:

#### Question 111:

In a triangle ABC, the measure of angle A is 40° less than the measure of angle B and 50° less than that of angle C. Find the measure of ∠ A.

#### Answer:

Given: In a triangle, ABC, the measure of angle A is 40° less than the measure of angle B and 50° less than that of angle C.

$\angle \mathrm{A}=\angle \mathrm{B}-40\xb0.....\left(1\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}=\angle \mathrm{C}-50\xb0.....\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)\phantom{\rule{0ex}{0ex}}\angle \mathrm{B}-40\xb0=\angle \mathrm{C}-50\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{B}=\angle \mathrm{C}-10\xb0.....\left(3\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3\mathrm{ABC},\mathrm{by}\mathrm{Angle}\mathrm{Sum}\mathrm{Property}\mathrm{of}\mathrm{triangle}\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(\angle \mathrm{C}-50\xb0\right)+\left(\angle \mathrm{C}-10\xb0\right)+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 3\angle \mathrm{C}-60\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 3\angle \mathrm{C}=180\xb0+60\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 3\angle \mathrm{C}=240\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{C}=80\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{A}=80\xb0-50\xb0=30\xb0\phantom{\rule{0ex}{0ex}}\mathrm{And}\angle \mathrm{B}=80\xb0-10\xb0=70\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 172:

#### Question 112:

I have three sides. One of my angle measures 15°. Another has a measure of 60°. What kind of a polygon am I? If I am a triangle, then what kind of triangle am I?

#### Answer:

The polygon with three sides is called a triangle.

$\mathrm{Consider}\u25b3\mathrm{ABC}\mathrm{with}\angle \mathrm{A}=15\xb0\mathrm{and}\angle \mathrm{B}=60\xb0.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{In}\u25b3\mathrm{ABC},\phantom{\rule{0ex}{0ex}}\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 15\xb0+60\xb0+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{C}=105\xb0\phantom{\rule{0ex}{0ex}}$

Since one angle of the triangle is greater than 90°, it is an obtuse-angled triangle.

#### Page No 172:

#### Question 113:

Jiya walks 6 km due east and then 8 km due north. How far is she from her starting place?

#### Answer:

Given: Jiya walks 6 km due east and then 8 km due north.

Distance from the starting point to the final position is the hypotenuse of the right triangle $\u25b3\mathrm{ABC}$.

${\mathrm{AC}}^{2}={\mathrm{CB}}^{2}+{\mathrm{AB}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={\left(6\right)}^{2}+{\left(8\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=36+64\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=100\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=10\mathrm{cm}$

Thus, she is 10 km from her starting place.

#### Page No 172:

#### Question 114:

Jayanti takes shortest route to her home by walking diagonally across a rectangular park. The park measures 60 metres × 80 metres. How

much shorter is the route across the park than the route around its edges?

#### Answer:

Given: The park measures 60 metres × 80 metres.

$\mathrm{In}\u25b3\mathrm{ABC}\phantom{\rule{0ex}{0ex}}{\mathrm{AC}}^{2}={\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}={\left(60\right)}^{2}+{\left(80\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=3600+6400\phantom{\rule{0ex}{0ex}}\Rightarrow {\mathrm{AC}}^{2}=10000\phantom{\rule{0ex}{0ex}}\Rightarrow \mathrm{AC}=100\mathrm{m}$

If she walks through the sides of the rectangle = AB + BC = 80 m + 60 m = 140 m

If she walks through the diagonal of the rectangle = AC = 100 m

Difference between the two paths = 140 m − 100 m = 40 m

#### Page No 173:

#### Question 115:

In âˆ†PQR of in the given figure, PQ = PR. Find the measures of ∠Q and ∠R.

#### Answer:

Given: In âˆ†PQR PQ = PR, $\angle \mathrm{QPR}=30\xb0$

Since PQ = PR

⇒ ∠PQR = ∠PRQ .....(1)

In âˆ†PQR, by Angle Sum Property of triangle

∠PQR + ∠PRQ + ∠QPR = 180°

⇒ ∠PQR + ∠PQR + 30° = 180°

⇒ 2∠PQR = 150°

⇒ ∠PQR = 75°

Thus, ∠PQR = ∠PRQ = 75°

#### Page No 173:

#### Question 116:

In the given figure, find the measures of ∠*x* and ∠*y*.

#### Answer:

By Angle Sum Property of triangle.

$45\xb0+60\xb0+x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=180\xb0-105\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=75\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}y+45\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y=135\xb0$

#### Page No 173:

#### Question 117:

In the given figure, find the measures of ∠PON and ∠NPO.

#### Answer:

In $\u25b3\mathrm{LMO}$, by Angle Sum Property of triangle

$\angle \mathrm{LMO}+\angle \mathrm{MOL}+\angle \mathrm{OLM}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 20\xb0+\angle \mathrm{MOL}+70\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{MOL}=90\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\angle \mathrm{PON}=\angle \mathrm{MOL}\left(\because \mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right)$

In $\u25b3\mathrm{PON}$, by Angle Sum Property of triangle

$\angle \mathrm{PON}+\angle \mathrm{NPO}+\angle \mathrm{ONP}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+\angle \mathrm{NPO}+70\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{NPO}=20\xb0$

#### Page No 173:

#### Question 118:

In the given figure, QP || RT. Find the values of *x* and *y*.

#### Answer:

Given: QP|| RT, where PR is a transversal line.

So, ∠*x =* ∠TRP = 70° (Alternate interior angles)

In $\u25b3\mathrm{PQR}$, by Angle Sum Property of triangle

$30\xb0+x+y=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+70\xb0+y=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y=80\xb0$

Thus, ∠*x =* 70° and ∠*y =* 80°.

#### Page No 174:

#### Question 119:

Find the measure of ∠A in the given figure.

#### Answer:

In $\u25b3\mathrm{ABC}$, by Exterior angle property

$\angle \mathrm{ABC}+\angle \mathrm{BAC}=115\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 65\xb0+\angle \mathrm{BAC}=115\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{BAC}=50\xb0\phantom{\rule{0ex}{0ex}}$

#### Page No 174:

#### Question 120:

In a right-angled triangle if an angle measures 35°, then find the measure of the third angle.

#### Answer:

Consider $\u25b3\mathrm{ABC}\mathrm{with}\angle \mathrm{A}=90\xb0\mathrm{and}\angle \mathrm{B}=35\xb0$

In $\u25b3\mathrm{ABC}$, by Angle Sum Property of triangle

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 90\xb0+35\xb0+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{C}=55\xb0$

Thus, the measure of the third angle is 55°.

#### Page No 174:

#### Question 121:

Each of the two equal angles of an isosceles triangle is four times the third angle. Find the angles of the triangle.

#### Answer:

Let the third angle be *x*.

Then, the other two angles are 4*x* and 4*x*, respectively.

By Angle Sum Property of the triangle

$x+4x+4x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 9x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=20\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore 4x=80\xb0$

Thus, the angles of the triangle are 20°, 80°, and 80°.

#### Page No 174:

#### Question 122:

The angles of a triangle are in the ratio 2 : 3 : 5. Find the angles.

#### Answer:

Let measures of the given angles of a triangle be 2*x*, 3*x*, and 5*x*.

By Angle Sum Property of the triangle

$2x+3x+5x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 10x=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=18\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore 2x=36\xb0,3x=54\xb0\mathrm{and}5x=90\xb0$

#### Page No 174:

#### Question 123:

If the sides of a triangle are produced in an order, show that the sum of the exterior angles so formed is 360°.

#### Answer:

In âˆ†ABC, by exterior angle property,

Exterior ∠1 = Interior ∠A + Interior ∠B .....(1)

Exterior ∠2 = Interior ∠B + Interior ∠C .....(2)

Exterior ∠3 = Interior ∠A + Interior ∠C .....(3)

Adding (1), (2) and (3), we get

∠1 + ∠2 + ∠3 = 2(∠A + ∠B + ∠C)

By angle sum property of a triangle, ∠A + ∠B + ∠C = 180°

∴ ∠1 + ∠2 + ∠3 = 2 ×180°

∠1 + ∠2 + ∠3 = 360°

Hence, the sum of exterior angles is 360°.

#### Page No 174:

#### Question 124:

In âˆ†ABC, if ∠A = ∠C, and exterior angle ABX = 140°, then find the angles of the triangle.

#### Answer:

Given: In âˆ†ABC, if ∠A = ∠C, and exterior angle ABX = 140°.

By Exterior angle property of triangle

$\angle \mathrm{ABX}=\angle \mathrm{B}+\angle \mathrm{C}\phantom{\rule{0ex}{0ex}}\Rightarrow 40\xb0=\angle \mathrm{B}+\angle \mathrm{B}\phantom{\rule{0ex}{0ex}}\Rightarrow 40\xb0=2\angle \mathrm{B}\phantom{\rule{0ex}{0ex}}\Rightarrow 20\xb0=\angle \mathrm{B}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \angle \mathrm{C}=20\xb0$

By Angle sum property of triangle

$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}+20\xb0+20\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{A}=140\xb0$

Thus, the angles of the triangles are 20°, 20° and 140°.

#### Page No 174:

#### Question 125:

Find the values of *x* and *y* in the given figure.

#### Answer:

In $\u25b3\mathrm{TQR}$, by Exterior angle property of triangle

$\angle \mathrm{Q}+\angle \mathrm{R}=\angle \mathrm{RTP}\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+45\xb0=y\phantom{\rule{0ex}{0ex}}\Rightarrow 75\xb0=y$

$\angle \mathrm{QTS}+\angle \mathrm{STR}+\angle \mathrm{RTP}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 50\xb0+\angle 2+75\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle 2+125\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle 2=55\xb0$

In $\u25b3\mathrm{STR},$ by Angle sum property of triangle

$\angle 2+x+\angle \mathrm{R}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 55\xb0+x+45\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=80\xb0$

Thus, $x=80\xb0\mathrm{and}y=75\xb0$.

#### Page No 175:

#### Question 126:

Find the value of *x* in the given figure.

#### Answer:

In $\u25b3\mathrm{ABC},$ by Angle sum property of triangle

$\angle \mathrm{BAC}+\angle \mathrm{ABC}+\angle \mathrm{ACB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 80\xb0+30\xb0+\angle \mathrm{ACB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{ACB}=70\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\angle \mathrm{ACB}+\angle \mathrm{ACE}+\angle \mathrm{ECD}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 70\xb0+x+90\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=20\xb0$

#### Page No 175:

#### Question 127:

The angles of a triangle are arranged in descending order of their magnitudes. If the difference between two consecutive angles is 10°, find the three angles.

#### Answer:

Let one of the angles of a triangle be *x*.

If angles are arranged in descending order. Then, angles will be *x*, (*x* -10°) and (*x* – 20°).

By Angle sum property of triangle

$x+\left(x-10\xb0\right)+\left(x-20\xb0\right)=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x-30\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 3x=210\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=70\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\therefore \left(x-10\xb0\right)=70\xb0-10\xb0=60\xb0\phantom{\rule{0ex}{0ex}}\left(x-20\xb0\right)=70\xb0-20\xb0=50\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Thus, the three angles are $70\xb0,60\xb0\mathrm{and}50\xb0$.

#### Page No 175:

#### Question 128:

In âˆ† ABC, DE || BC . Find the values of *x*, *y* and *z*.

#### Answer:

In âˆ† ABC

DE || BC

∠ADE = ∠ABC (Corresponding angles)

⇒ *x* = 30°

∠AED = ∠ACB (Corresponding angles)

⇒ *y* = 40°

In $\u25b3\mathrm{ADE}$, by Angle sum property of triangle

$x+y+z=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+40\xb0+z=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow z=110\xb0\phantom{\rule{0ex}{0ex}}$$x+y+z=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 30\xb0+40\xb0+z=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow z=110\xb0$

#### Page No 175:

#### Question 129:

In the given figure, find the values of *x*, *y* and *z*.

#### Answer:

In ABD, by Angle sum property of triangle

$\angle \mathrm{ABD}+\angle \mathrm{ADB}+\angle \mathrm{DAB}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0+y+60\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y=60\xb0$

$x+y=180\xb0\left(\mathrm{Straight}\mathrm{line}\right)\phantom{\rule{0ex}{0ex}}\Rightarrow 60\xb0+y=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y=120\xb0$

In ADC, by Angle sum property of triangle

$z+y+\angle \mathrm{DAC}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow z+120\xb0+30\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow z=30\xb0$

#### Page No 175:

#### Question 130:

If one angle of a triangle is 60° and the other two angles are in the ratio 1 : 2, find the angles.

#### Answer:

Given: One angle of a triangle is 60°.

Let the other two angles be *x* and 2*x*.

Since the sum of all angles in a triangle is equal to 180°.

So,*x *+ 2*x* + 60° = 180°

3*x* = 180° − 60°

3*x* = 120°*x *= 40°

So, the other two angles will be *x* = 40° and 2*x* = 2 × 40° = 80°.

#### Page No 175:

#### Question 131:

In âˆ†PQR, if 3∠P = 4∠Q = 6∠R, calculate the angles of the triangle.

#### Answer:

Given: In âˆ†PQR, if 3∠P = 4∠Q = 6∠R.

Let 3∠P = 4∠Q = 6∠R =* x*

$3\angle \mathrm{P}=x,4\angle \mathrm{Q}=x,6\angle \mathrm{R}=x\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{P}=\frac{x}{3},\angle \mathrm{Q}=\frac{x}{4},\angle \mathrm{R}=\frac{x}{6}$

In âˆ†PQR

$\angle \mathrm{P}+\angle \mathrm{Q}+\angle \mathrm{R}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{x}{3}+\frac{x}{4}+\frac{x}{6}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{4x+3x+2x}{12}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \frac{9x}{12}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow x=20\xb0\times 12\phantom{\rule{0ex}{0ex}}\Rightarrow x=240\xb0\phantom{\rule{0ex}{0ex}}$

$\therefore \angle \mathrm{P}=\frac{240\xb0}{3},\angle \mathrm{Q}=\frac{240\xb0}{4},\angle \mathrm{R}=\frac{240\xb0}{6}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{P}=80\xb0,\angle \mathrm{Q}=60\xb0,\angle \mathrm{R}=40\xb0$

#### Page No 175:

#### Question 132:

In âˆ†DEF, ∠D = 60°, ∠E = 70° and the bisectors of ∠E and ∠F meet at O. Find (i) ∠F (ii) ∠EOF.

#### Answer:

Given: In âˆ†DEF, ∠D = 60°, ∠E = 70° and the bisectors of ∠E and ∠F meet at O.

(i)

$\angle \mathrm{E}+\angle \mathrm{D}+\angle \mathrm{F}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \left(35\xb0+35\xb0\right)+60\xb0+\angle \mathrm{F}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{F}=50\xb0$

(ii) OF is the bisector of ∠F.

$\text{\u2234\u2220OFE=}\frac{1}{2}\angle \mathrm{F}\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{OFE}=\frac{1}{2}\times 50\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{OFE}=25\xb0$

$\mathrm{In}\u25b3\mathrm{EOF},\phantom{\rule{0ex}{0ex}}\angle \mathrm{EOF}+\angle \mathrm{OFE}+\angle \mathrm{OEF}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{EOF}+25\xb0+35\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow \angle \mathrm{EOF}=120\xb0$

#### Page No 175:

#### Question 133:

In the given figure, âˆ†PQR is right-angled at P. U and T are the points on line QRF. If QP || ST and US || RP, find ∠S.

#### Answer:

If QP || ST and QT is a transversal, then ∠PQR = ∠STU (Alternate interior angles)

and if DS || RP and QT is a transversal, then ∠PRQ = ∠SUT (Alternate interior angles)

Hence, ∠S must be equal to ∠Pi.e. 90°.

#### Page No 176:

#### Question 134:

In each of the given pairs of triangles of in the given figure, applying only ASA congruence criterion, determine which triangles are congruent. Also, write the congruent triangles in symbolic form.

(c)

(e)

(f)

#### Answer:

(a) Not possible, because the side is not included between two angles.

(b) âˆ†ABD ≅ âˆ†CDB

(c) âˆ†XYZ ≅ âˆ†LMN

(d) Not possible, because there is not any included side equal.

(e) âˆ†MNO ≅ âˆ†PON

(f) âˆ†AOD ≅ âˆ†BOC

#### Page No 177:

#### Question 135:

In each of the given pairs of triangles of in the given figure, using only RHS congruence criterion, determine which pairs of triangles are congruent. In case of congruence, write the result in symbolic form:

#### Answer:

(a) In âˆ†ABD and âˆ†ACD,

AB = AC (Given)

AD = AD (Common)

∠ADB = ∠ADC = 90° (Given)

By RHS congruence criterion, âˆ†ABD ≅ âˆ†ACD.

(b) In âˆ†XYZ and âˆ†UZY,

∠XYZ = ∠UZY = 90° (Given)

XZ = YU (Given)

ZY = ZY (Common)

By RHS congruence criterion, âˆ†XYZ ≅ âˆ†UZY.

(c) In âˆ†AEC and âˆ†BED,

CE = DE (Given)

AE = BE (Given)

∠ACE = ∠BDE = 90° (Given)

By RHS congruence criterion, âˆ†AEC ≅ âˆ†BED.

(d) Here, CD = BD − BC = 14 − 8 = 6 cm

In right-angled âˆ†ABC,

AC = $\sqrt{{\mathrm{AB}}^{2}+{\mathrm{BC}}^{2}}=\sqrt{{6}^{2}+{8}^{2}}=\sqrt{100}=10\mathrm{cm}$

In right-angled âˆ†CDE,

CD = 6 cm

DE = $\sqrt{{\mathrm{CE}}^{2}-{\mathrm{CD}}^{2}}=\sqrt{{10}^{2}-{6}^{2}}=\sqrt{100-36}=\sqrt{64}=8\mathrm{cm}$ (By Pythagoras theorem)

In âˆ†ABC and âˆ†CDE,

AC = CE = 10 cm

BC = DE = 8 cm

∠ABC = ∠CDE = 90°

By RHS congruence criterion, âˆ†ABC ≅ âˆ†CDE.

(e) Not possible, because there is not any right angle.

(f) In âˆ†LOM and âˆ†LON,

LM = LN = 8cm

LO = LO (Common)

∠LOM = ∠LON = 90°

By RHS congruence criterion, âˆ†LOM ≅ Aâˆ†LON

#### Page No 178:

#### Question 136:

In the given figure, if RP = RQ, find the value of *x*.

#### Answer:

Given: RP = RQ

$\angle \mathrm{RPQ}=50\xb0\left(\mathrm{Vertically}\mathrm{oppposite}\mathrm{angles}\right)$

In $\u25b3\mathrm{PQR}$,

RP = RQ

$\angle \mathrm{RQP}=\angle \mathrm{RPQ}\phantom{\rule{0ex}{0ex}}\Rightarrow x=50\xb0$

#### Page No 178:

#### Question 137:

In the given figure, if ST = SU, then find the values of *x* and *y*.

#### Answer:

Given: ST = SU

$\angle \mathrm{TSU}=78\xb0\left(\mathrm{Vertically}\mathrm{oppposite}\mathrm{angles}\right)$

In $\u25b3\mathrm{STU}$,

ST = SU

$\angle \mathrm{STU}=\angle \mathrm{SUT}\phantom{\rule{0ex}{0ex}}$ (By isosceles triangle property)

In $\u25b3\mathrm{STU}$, by Pythagoras Theorem

$\angle \mathrm{STU}+\angle \mathrm{SUT}+\angle \mathrm{TSU}=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y+y+78\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2y+78\xb0=180\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow 2\mathrm{y}=102\xb0\phantom{\rule{0ex}{0ex}}\Rightarrow y=51\xb0\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

#### Page No 178:

#### Question 138:

Check whether the following measures (in cm) can be the sides of a right-angled triangle or not.

1.5, 3.6, 3.9

#### Answer:

For a right angled triangle, the sum of square of two shorter sides must be equal to the square of the third side.

Now, 1.5^{2} + 36^{2} = 2.25 + 12.96 = 15.21

Also, (3.9)^{2 }= 15.21

(1.5)^{2} + (3.6)^{2} = (3.9)^{2}

Hence, the given sides form a right-angled triangle.

#### Page No 178:

#### Question 139:

Height of a pole is 8 m. Find the length of rope tied with its top from a point on the ground at a distance of 6 m from its bottom.

#### Answer:

Given: Height of a pole is 8 m.

Distance between the bottom of the pole and a point on the ground is 6 m.

Let the length of the rope be *x* m.

AB = Height of the pole

BC = Distance between the bottom of the pole and a point on ground, where rope was tied.

In right-angled MBC, by â€‹Pythagoras theorem

(AC)^{2 }= (AB)^{2} + (BC)^{2}

⇒ (*x*)^{2} = (8)^{2} + (6)^{2}

⇒ (*x*)^{2} = 64 + 36

⇒ (*x*)^{2} = 100

⇒ (*x*) = 10 m

Hence, the length of the rope is 10 m.

#### Page No 179:

#### Question 140:

In the given figure, if *y* is five times *x*, find the value of *z*.

#### Answer:

Given: *y* is five times *x*

According to the question,*y* = 5*x* ......(1)

According to the angle sum property of a triangle,

600 + *x *+ *y *= 180°

⇒ 60° + *x *+ 5*x *= 180° [From (1)]

⇒ 60° + 6*x* = 180°

⇒ 6*x* = 180° − 60°

⇒ 6*x* = 120°

⇒ *x* = 20°

âˆµ* y* = 5*x*

⇒ *y *= 5 × 20° =100°

According to the exterior angle property,*z* = 60° + *y* = 60° + 100° =160°

#### Page No 179:

#### Question 141:

The lengths of two sides of an isosceles triangle are 9 cm and 20 cm. What is the perimeter of the triangle? Give reason.

#### Answer:

Third side must be 20 cm, because sum of two sides should be greater than the third side.

∴ Perimeter of the triangle = Sum of all sides

= (9 + 20 + 20) cm

= 49 cm

#### Page No 179:

#### Question 142:

Without drawing the triangles write all six pairs of equal measures in each of the following pairs of congruent triangles.

(a) âˆ†STU ≅ âˆ†DEF (b) âˆ†ABC ≅ âˆ†LMN

(c) âˆ†YZX ≅ âˆ†PQR (d) âˆ†XYZ ≅ âˆ†MLN

#### Answer:

Since corresponding parts of congruent triangles are equal.

(a) ΔSTU ≅ ΔDEF

∠S = ∠D,

∠T = ∠E,

∠U = ∠F,

ST = DE,

And TU = EF,

SU = DF

(b) ΔABC ≅ ΔLMN

∠A = ∠L,

∠B = ∠M,

∠C = ∠N,

AB = LM,

BC = MN

And AC = LN

(c) ΔYZX ≅ APQR

∠T = ∠P,

∠Z = ∠Q,

∠X = ∠R

YZ = PQ,

ZX = QR

And YX = PR

(d) ΔXYZ ≅ ΔMLN

∠X = ∠M,

∠Y = ∠L,

∠Z = ∠N,

XY = ML,

YZ = LN

And XZ = MN

#### Page No 179:

#### Question 143:

In the following pairs of triangles of in the given figure, the lengths of the sides are indicated along the sides. By applying SSS congruence criterion, determine which triangles are congruent. If congruent, write the results in symbolic form.

#### Answer:

(a) ΔABC ≅ ΔNLM

(b) ΔLMN ≅ ΔGHI

(c) ΔLMN ≅ ΔLON

(d) ΔZYX ≅ ΔWXY

(e) ΔOAB ≅ ΔDOE

(f) ΔSTU ≅ ΔSVU

(g) ΔPSR ≅ ARQP

(h) ΔSTU ≅ ΔPQR

#### Page No 180:

#### Question 144:

ABC is an isosceles triangle with AB = AC and D is the mid-point of base BC

(a) State three pairs of equal parts in the triangles ABD and ACD.

(b) Is âˆ†ABD ≅ âˆ†ACD. If so why?

#### Answer:

Given, AB = AC and BD = CD

(a) In âˆ†ABD and âˆ†ACD.

AB = AC (given)

BD =CD (given)

AD = AD (common side)

(b) Yes, by SSS congruence criterion, âˆ†ABD ≅ âˆ†ACD.

#### Page No 181:

#### Question 145:

In the given figure, it is given that LM = ON and NL = MO

(a) State the three pairs of equal parts in the triangles NOM and MLN.

(b) Is âˆ†NOM ≅ âˆ†MLN. Give reason?

#### Answer:

(a) In âˆ†MLN and âˆ†NOM,

LM = ON (Given)

MN = MN (Common)

LN = OM (Given)

(b) Yes, by SSS congruence criterion, âˆ†MLN ≅ âˆ†NOM.

#### Page No 181:

#### Question 146:

Triangles DEF and LMN are both isosceles with DE = DF and LM = LN, respectively. If DE = LM and EF = MN, then, are the two triangles congruent? Which condition do you use? If ∠ E = 40°, what is the measure of ∠ N?

#### Answer:

Given: DE = DF, LM = LN, and DE = LM

DE = OF = LM = LN ......(1)

In âˆ†DEF and âˆ†MNL,

DE = LM (Given)

EF = MN (Given)

DF = LN (From 1)

By SSS congruence criterion, âˆ†DEF = âˆ†LMN

∠E = ∠M (by CPCT)

∠M = 40°

Also, ∠M = ∠N, and angles opposite to equal sides are equal.

∴ ∠N = 40°

#### Page No 181:

#### Question 147:

If âˆ†PQR and âˆ†SQR are both isosceles triangle on a common base QR such that P and S lie on the same side of QR. Are triangles PSQ and PSR congruent? Which condition do you use?

#### Answer:

In âˆ†PQR and âˆ†SQR

PQ = SR (Given)

SQ = SR (Given)

QR = QR (Given)

By SSS Congruency criterion, âˆ†PQR ≅ âˆ†SQR.

#### Page No 181:

#### Question 148:

In the given figure, which pairs of triangles are congruent by SAS congruence criterion (condition)? If congruent, write the congruence of the two triangles in symbolic form.

#### Answer:

(i) In âˆ†PQR and âˆ†TUS,

PQ = TU = 3cm

OR = US = 5cm

∠PQR = ∠SUT = 40°

By SAS congruence criterion, âˆ†PQR ≅ âˆ†TUS.

(ii) Not congruent, because the angle is not included between two sides.

(iii) In âˆ†BCD and âˆ†BAE,

AB = CB = 5.2 cm

DC = EA = 5 cm

∠EAS = ∠DCB = 50°

By SAS congruence criterion, âˆ†BCD ≅ âˆ†BAE.

(iv) In âˆ†STU and âˆ†XZY,

TU = ZY = 4cm

TS = ZX = 3cm

∠STU = ∠XZY = 30°

By SAS congruence criterion, âˆ†STU ≅ âˆ†XZY.

(v) In âˆ†DOF and âˆ†HOC,

DO = HO (Given)

CO = FO (Given)

∠DOF = ∠HOC (Vertically opposite angles)

By SAS congruence criterion, âˆ†DOF ≅ âˆ†HOC.

(vi) Not congruent, because the angle is not included between two sides.

(vii) In âˆ†PSQ and âˆ†RQS,

PS = RQ = 4cm

SQ = SQ (Common side)

∠PSQ = ∠RQS = 40°

By SAS congruence criterion, âˆ†PSQ ≅ âˆ†RQS.

(viii) In âˆ†LMN and âˆ†OMN,

LM = OM Given)

MN = MN (Common side)

∠LMN = ∠OMN = 40°

By SAS congruence criterion, âˆ†LMN ≅ âˆ†OMN.

#### Page No 183:

#### Question 149:

State which of the following pairs of triangles are congruent. If yes, write them in symbolic form (you may draw a rough figure).

(a) âˆ† PQR : PQ = 3.5 cm, QR = 4.0 cm, ∠ Q = 60°

âˆ† STU : ST = 3.5 cm, TU = 4 cm, ∠ T = 60°

(b) âˆ†ABC : AB = 4.8 cm, ∠ A = 90°, AC = 6.8 cm

âˆ†XYZ : YZ = 6.8 cm, ∠ X = 90° , ZX = 4.8 cm

#### Answer:

(a)

Both the triangles are congruent.

$\u25b3\mathrm{PQR}\cong \u25b3\mathrm{STU}$ (By SAS Congruency)

(b)

Both triangles are not congruent.

#### Page No 183:

#### Question 150:

In the given figure, PQ = PS and ∠ 1 = ∠ 2.

(i) Is âˆ†PQR ≅ âˆ†PSR? Give reasons.

(ii) Is QR = SR? Give reasons.

#### Answer:

(i) Yes,

In âˆ†PQR anC âˆ†PSR,

PQ = PS (Given)

∠1 = ∠2 (Given)

PR = PR (Common side)

By SAS congruence criterion, âˆ†PQR ≅ âˆ†PSR.

(ii) Yes, QR = SR (by CPCT)

#### Page No 183:

#### Question 151:

In the given figure, DE = IH, EG = FI and ∠ E = ∠ I. Is âˆ†DEF ≅ âˆ†HIG? If yes, by which congruence criterion?

#### Answer:

Given: EG = FI

EG + GF = Fl + OF (Adding GF on both sides)

EF = IG .....(1)

In âˆ†DEF and âˆ†HIG,

DE = IH (Given)

EF = IG [From (1)]

∠E = ∠I (Given)

By SAS congruence criterion, âˆ†DEF ≅ âˆ†HIG.

#### Page No 184:

#### Question 152:

In the given figure, ∠1 = ∠ 2 and ∠ 3 = ∠ 4.

(i) Is âˆ†ADC ≅ âˆ† ABC? Why ?

(ii) Show that AD = AB and CD = CB.

#### Answer:

(i) In âˆ†ADC and âˆ†ABC,

∠1 = ∠2 (Given)

AC = AC (Common side)

∠3 = ∠4 (Given)

By ASA congruence criterion, âˆ†ADC ≅ âˆ†ABC

(ii) AD = AB (by CPCT)

CD = CB (by CPCT)

#### Page No 184:

#### Question 153:

Observe in the given figure and state the three pairs of equal parts in triangles ABC and DBC.

(i) Is âˆ†ABC ≅ âˆ†DCB? Why?

(ii) Is AB = DC? Why?

(iii) Is AC = DB? Why?

#### Answer:

(i) In âˆ†ABC and âˆ†DCB,

BC = BC (Common side)

∠ABC = ∠DCB = 70°

∠ACB = ∠DBC = 30°

By ASA congruence criterion, âˆ†ABC ≅ âˆ†DCB

(ii) AB = DC (by CPCT)

(iii) AC = DB (by CPCT)

#### Page No 184:

#### Question 154:

In the given figure, QS ⊥ PR, RT ⊥ PQ and QS = RT.

(i) Is âˆ† QSR ≅ âˆ† RTQ? Give reasons.

(ii) Is ∠ PQR = ∠ PRQ? Give reasons.

#### Answer:

(i) In âˆ†QSR and âˆ†RTQ

QS = RT (Given)

∠OSR = ∠QTR = 90°

OR = OR (Common side)

By RHS congruence criterion, âˆ†QSR ≅ âˆ†RTQ

(ii) Yes, ∠POR = ∠PRO (by CPCT)

#### Page No 184:

#### Question 155:

Points A and B are on the opposite edges of a pond as shown in the given figure. To find the distance between the two points, the surveyor makes a right-angled triangle as shown. Find the distance AB.

#### Answer:

Since âˆ†ACD is a right-angled triangle.

In right-angled âˆ†ADC, by Pythagoras theorem,

(AC)^{2} = (AD)^{2} + (CD)^{2}

(AC)^{2} = (30)^{2} + (40)^{2}

(AC)^{2} = 900 + 1600

(AC)^{2} = 2500

AC= 50 m

Now, AB = AC − BC = 50 − 12 = 38 m

Hence, the distance AB is 38 m.

#### Page No 185:

#### Question 156:

Two poles of 10 m and 15 m stand upright on a plane ground. If the distance between the tops is 13 m, find the distance between their feet.

#### Answer:

Let BC =* x* m

In right-angled âˆ†ACB,

AB^{2} = AC^{2} + BC^{2} (by Pythagoras theorem)

⇒ (13)^{2} = (5)^{2} +* x*^{2}

⇒ 169 − 25 = *x*^{2}

⇒ 144 = *x*^{2}

⇒ *x* =12 m

Hence, the distance between the feet of the two poles is 12 m.

#### Page No 185:

#### Question 157:

The foot of a ladder is 6 m away from its wall and its top reaches a window 8 m above the ground, (a) Find the length of the ladder. (b) If the ladder is shifted in such a way that its foot is 8 m away from the wall, to what height does its top reach?

#### Answer:

(a) Let the length of the ladder be *x* m.

In right-angled âˆ†ABC, by Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}

⇒ (*x*)^{2} = (8)^{2} + (6)^{2}

⇒ (*x*)^{2 }= 64 + 36

⇒ *x* = 10 m

Hence, the length of the ladder is 10 m.

(b) Let the height of the top be *x* m.

In right angled MCB,

AC^{2} = AB^{2} + BC^{2} [by Pythagoras theorem]

⇒ *x*^{2 }= (10)^{2} − (8)^{2 }

⇒ *x*^{2 }= 100 − 64

⇒ *x*^{2 }= 36 m

⇒ *x* = 6 m

Hence, the height of the top is 6 m.

#### Page No 185:

#### Question 158:

In the given figure, state the three pairs of equal parts in âˆ†ABC and âˆ†EOD. Is âˆ†ABC ≅ âˆ† EOD? Why?

#### Answer:

In âˆ†ABC and âˆ†EOD,

AB = EO (Given)

AC = ED (Given)

∠ABC = ∠EOD = 90°

By RHS congruence criterion, âˆ†ABC ≅ âˆ†EOD.

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