##### Class 8^{th} Mathematics Term 1 Tamilnadu Board Solution

**Exercise 3.1**- Which of the following will be the angles of a triangle? Choose the correct…
- Which of the following statement is correct? Choose the correct answer:A.…
- The three exterior angles of a triangle are 130°, 140°, x° then x° is Choose…
- Which of the following set of measurements will form a triangle? Choose the…
- Which of the following will form a right-angled triangle, given that the two…
- The angles of a triangle are (x - 35) °, (x - 20) ° and (x + 40) °. Find the…
- In Δ ABC, the measure of ∠A is greater than the measure of ∠ B by 24° . If…
- The bisectors of ∠ B and ∠ C of a Δ ABC meet at O. Show that BOC = 90o + angle…
- Find the value of x° and y° from the following figures:
- Find the value of x° and y° from the following figures:
- Find the value of x° and y° from the following figures:
- Find the angles x°, y° and z° from the given figure.

**Exercise 3.2**- In the isosceles ∆XYZ, given XY = YZ then which of the following angles are…
- In Δ ABC and ∆DEF, ∠B = ∠E, AB = DE, BC = EF. The two triangles are congruent…
- Two plane figures are said to be congruent if they have Choose the correct…
- In a triangle ABC, ∠A = 40o and AB = AC, then ABC is _____ triangle. Choose the…
- In the triangle ABC, when ∠A = 90o the hypotenuse is ______ Choose the correct…
- In the Δ PQR the angle included by the sides PQ and PR is Choose the correct…
- In the figure, the value of x° is_______ Choose the correct answer:A. 80o B.…
- In the figure, ABC is a triangle in which AB = AC. Find x° and y°. delta…
- In the figure, Find x°. x
- In the figure ∆PQR and ∆SQR are isosceles triangles. Find x°.
- In the figure, it is given that BR = PC and ∠ACB= ∠QRP and AB ∥ PQ. Prove that…
- In the figure, AB = BC = CD, ∠A = xo. Prove that ∠DCF = 3∠A. a
- Find x°, y°, z° from the figure, where AB = BD, BC = DC and ∠DAC 30o. a…
- In the figure, ABCD is a parallelogram. AB is produced to E such that AB = BE.…
- In figure, BO bisects ∠ABC of ∆ABC. P is any point on BO. Prove that the…
- The Indian Navy flights fly in a formation that can be viewed as two…

**Exercise 3.1**

- Which of the following will be the angles of a triangle? Choose the correct…
- Which of the following statement is correct? Choose the correct answer:A.…
- The three exterior angles of a triangle are 130°, 140°, x° then x° is Choose…
- Which of the following set of measurements will form a triangle? Choose the…
- Which of the following will form a right-angled triangle, given that the two…
- The angles of a triangle are (x - 35) °, (x - 20) ° and (x + 40) °. Find the…
- In Δ ABC, the measure of ∠A is greater than the measure of ∠ B by 24° . If…
- The bisectors of ∠ B and ∠ C of a Δ ABC meet at O. Show that BOC = 90o + angle…
- Find the value of x° and y° from the following figures:
- Find the value of x° and y° from the following figures:
- Find the value of x° and y° from the following figures:
- Find the angles x°, y° and z° from the given figure.

**Exercise 3.2**

- In the isosceles ∆XYZ, given XY = YZ then which of the following angles are…
- In Δ ABC and ∆DEF, ∠B = ∠E, AB = DE, BC = EF. The two triangles are congruent…
- Two plane figures are said to be congruent if they have Choose the correct…
- In a triangle ABC, ∠A = 40o and AB = AC, then ABC is _____ triangle. Choose the…
- In the triangle ABC, when ∠A = 90o the hypotenuse is ______ Choose the correct…
- In the Δ PQR the angle included by the sides PQ and PR is Choose the correct…
- In the figure, the value of x° is_______ Choose the correct answer:A. 80o B.…
- In the figure, ABC is a triangle in which AB = AC. Find x° and y°. delta…
- In the figure, Find x°. x
- In the figure ∆PQR and ∆SQR are isosceles triangles. Find x°.
- In the figure, it is given that BR = PC and ∠ACB= ∠QRP and AB ∥ PQ. Prove that…
- In the figure, AB = BC = CD, ∠A = xo. Prove that ∠DCF = 3∠A. a
- Find x°, y°, z° from the figure, where AB = BD, BC = DC and ∠DAC 30o. a…
- In the figure, ABCD is a parallelogram. AB is produced to E such that AB = BE.…
- In figure, BO bisects ∠ABC of ∆ABC. P is any point on BO. Prove that the…
- The Indian Navy flights fly in a formation that can be viewed as two…

###### Exercise 3.1

**Question 1.**Choose the correct answer:

Which of the following will be the angles of a triangle?

A. 35°, 45°, 90°

B. 26°, 58°, 96°

C. 38°, 56°, 96°

D. 30°, 55°, 90°

**Answer:**Formula used: Angle Sum Property = ∠A + ∠B + ∠C = 180°

Option A: 35° + 45° + 90° = 170°. Hence, this cannot be correct because sum of all the angles of triangle.

Option B: 26° + 58° + 96° = 180°. Hence, this is correct because sum of all the angles of triangle.

Option C: 38° + 56° + 96° = 190°. Hence, this cannot be correct because sum of all the angles of triangle.

Option D: 35° + 45° + 90° = 175°. Hence, this cannot be correct because sum of all the angles of triangle.

**Question 2.**Choose the correct answer:

Which of the following statement is correct?

A. Equilateral triangle is equiangular.

B. Isosceles triangle is equiangular.

C. Equiangular triangle is not equilateral.

D. Scalene triangle is equiangular

**Answer:**Reason: All angles of an equilateral triangle are equal.

**Question 3.**Choose the correct answer:

The three exterior angles of a triangle are 130°, 140°, x° then x° is

A. 90°

B. 100°

C. 110°

D. 120°

**Answer:**Angle Sum Property = ∠A + ∠B + ∠C = 180°[T1]

∠ DAC = 130° and ∠ ABE = 140°

∠ DAC + ∠ DAB = 180° (sum of the angle on a straight line at point is 180°)

∴ ∠DAB = 180° - ∠ DAC

⇒ ∠ DAB = 180° - 130° = 50°

Similarly,

∠ ABE + ∠ ABD = 180° (sum of the angle on a straight line at point is 180°)

∴ ∠ABD = 180° - ∠ ABE

⇒ ∠ DAB = 180° - 140° = 60°

∠ BDB’ = ∠ DAB + ∠ DBA

(if a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior opposite angles)

∠ x° = 50° + 60° = 110°

Hence, option C is correct.

**Question 4.**Choose the correct answer:

Which of the following set of measurements will form a triangle?

A. 11 cm, 4 cm, 6 cm

B. 13 cm, 14 cm, 25 cm

C. 8 cm, 4 cm, 3 cm

D. 5 cm, 16 cm, 5 cm

**Answer:**Any Two sides of triangle together are greater than the third side.

Option A: AB + BC > AC = 11 + 4 > 6 = 15 > 6true

BC + AC > AB = 4 + 6 > 11 = 10 > 11not true

AC + AB > BC = 11 + 6 > 4 = 17 > 4 true.

∴, this cannot be true because only condition is satisfied.

Option B: AB + BC > AC = 13 + 14 > 25 = 27 > 25 true

BC + AC > AB = 14 + 25 > 13 = 39 > 13 true

AC + AB > BC = 25 + 13 > 14 = 38 > 13 true.

∴, this is true because all condition is satisfied.

Option C: AB + BC > AC = 8 + 4 > 3 = 12 > 3 true

BC + AC > AB = 4 + 3 > 8 = 7 > 8 not true

AC + AB > BC = 8 + 3 > 4 = 11 > 4 true.

∴, this cannot be true because only condition is satisfied.

Option D: AB + BC > AC = 5 + 16 > 5 = 21 > 5true

BC + AC > AB = 16 + 5 > 11 = 21 > 5 true

AC + AB > BC = 5 + 5 > 16 = 10 > 16 not true.

∴, this cannot be true because only condition is satisfied.

**Question 5.**Choose the correct answer:

Which of the following will form a right-angled triangle, given that the two angles are

A. 24°, 66°

B. 36°, 64°

C. 62°, 48°

D. 68°, 32°

**Answer:**For the triangle to be right-angled triangle one angle is 90° and sum of the two angles should be 90°

Option A: 24° + 66° = 90°. This option is correct.

Option B: 36° + 64° = 100°. This option is not correct.

Option C: 62° + 48° = 110°. This option is not correct.

Option D: 68° + 32° = 100°. This option is not correct.

**Question 6.**The angles of a triangle are (x – 35) °, (x – 20) ° and (x + 40) °.

Find the three angles.

**Answer:**Theorem 1: The sum of three angles is 180°.

i.e. ∠A + ∠B + ∠C = 180°

⇒ (x – 35)° + (x – 20)° + (x + 40)° = 180°

⇒ x – 35 + x – 20 + x + 40 = 180

⇒ 3x – 55 + 40 = 180

⇒ 3x – 15 = 180

⇒ 3x = 180 + 15

⇒ 3x = 195

⇒

⇒ x = 15

**Question 7.**In Δ ABC, the measure of ∠A is greater than the measure of ∠ B by 24° . If exterior angle ∠C is 108°. Find the angles of the Δ ABC.

**Answer:**Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

Let the ∠B be x.

∴ ∠ B = x

∴ ∠ A = ∠ B + 24°

Ext. [T2] ∠ C = ∠ B + ∠ A

⇒ 108° = x + x + 24°

⇒ 108° = 2x + 24°

⇒ 2x = 108° - 24°

⇒ 2x = 84°

⇒

⇒ x = 42°

**Question 8.**The bisectors of ∠ B and ∠ C of a Δ ABC meet at O.

Show that BOC = 90^{o} +

**Answer:**Theorem 1:

The sum of all three angles is 180°.

Let the ∠ B and ∠ C be 2x and 2y respectively.

In Δ ABC

∠A + ∠B + ∠C = 180° (Angle Sum Property)

⇒ ∠ A + 2x + 2y = 180°

⇒ 2x + 2y = 180° - ∠A

⇒ 2(x + y) = 180° - ∠ A

⇒

⇒ …(1)

In Δ OBC

∠OBC + ∠ OCB + ∠ BOC = 180° (Angle Sum Property)

⇒ x + y + ∠ BOC = 180°

⇒

⇒

⇒

Hence Proved.

**Question 9.**Find the value of x° and y° from the following figures:

**Answer:**Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two-interior angle.

Ext. ∠ B = ∠ A + ∠ C

⇒ y° = x° + 50° …(1)

Ext. ∠ B + 72° = 180°

⇒ Ext. ∠ B = 180° - 72°

⇒ Ext. ∠ B = 108°

⇒ y° = 108°

Put y = 108° in (1)

⇒ 108° = x° + 50°

⇒ x° = 108° - 50°

⇒ x° = 58°

**Question 10.**Find the value of x° and y° from the following figures:

**Answer:**theorem 1: Sum of all the angles of triangle is 180°

∠ ACD + ∠ ACB = 180°

⇒ 5y° + y° = 180°

⇒ 6y° = 180°

⇒

⇒ y° = 30°

In Δ ABC

∠ A + ∠ B + ∠ C = 180°

⇒ x° + 4y° + y° = 180°

⇒ x° + 4 × 30° + 30° = 180°

⇒ x° + 120° + 30° = 180°

⇒ x° + 150° = 180°

⇒ x° = 180° - 150°

⇒ x° = 30°

**Question 11.**Find the value of x° and y° from the following figures:

**Answer:**Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

∠ C = ∠ B (Alternate interior angles)

⇒ x° = 42°

In Δ CDE

Ext. ∠ E = ∠ C + ∠ D

⇒ 82° = 42° + y°

⇒ y° = 82° - 42°

⇒ y° = 40°

**Question 12.**Find the angles x°, y° and z° from the given figure.

**Answer:**Theorem 1: Sum of all the angles of triangles is 180°

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

In Δ ABC

∠ A + ∠ B + ∠ C = 180° (Angle Sum Property)

⇒ 101° + 26° + z° = 180°

⇒ 127° + z° = 180°

⇒ z° = 180° - 127°

⇒ z° = 53°

In Δ FBG

Ext. ∠ G = ∠ B + ∠ F

⇒ y° = 26° + 106°

⇒ y° = 132°

∠ AED + ∠ DEC = 180°

⇒ ∠ AED + 128° = 180°

⇒ ∠ AED = 180° - 128°

⇒ ∠ AED = 52°

In AED

Ext. ∠ D = ∠ A + ∠ C

⇒ x° = 101° + 52°

⇒ x° = 153°

**Question 1.**

Choose the correct answer:

Which of the following will be the angles of a triangle?

A. 35°, 45°, 90°

B. 26°, 58°, 96°

C. 38°, 56°, 96°

D. 30°, 55°, 90°

**Answer:**

Formula used: Angle Sum Property = ∠A + ∠B + ∠C = 180°

Option A: 35° + 45° + 90° = 170°. Hence, this cannot be correct because sum of all the angles of triangle.

Option B: 26° + 58° + 96° = 180°. Hence, this is correct because sum of all the angles of triangle.

Option C: 38° + 56° + 96° = 190°. Hence, this cannot be correct because sum of all the angles of triangle.

Option D: 35° + 45° + 90° = 175°. Hence, this cannot be correct because sum of all the angles of triangle.

**Question 2.**

Choose the correct answer:

Which of the following statement is correct?

A. Equilateral triangle is equiangular.

B. Isosceles triangle is equiangular.

C. Equiangular triangle is not equilateral.

D. Scalene triangle is equiangular

**Answer:**

Reason: All angles of an equilateral triangle are equal.

**Question 3.**

Choose the correct answer:

The three exterior angles of a triangle are 130°, 140°, x° then x° is

A. 90°

B. 100°

C. 110°

D. 120°

**Answer:**

Angle Sum Property = ∠A + ∠B + ∠C = 180°[T1]

∠ DAC = 130° and ∠ ABE = 140°

∠ DAC + ∠ DAB = 180° (sum of the angle on a straight line at point is 180°)

∴ ∠DAB = 180° - ∠ DAC

⇒ ∠ DAB = 180° - 130° = 50°

Similarly,

∠ ABE + ∠ ABD = 180° (sum of the angle on a straight line at point is 180°)

∴ ∠ABD = 180° - ∠ ABE

⇒ ∠ DAB = 180° - 140° = 60°

∠ BDB’ = ∠ DAB + ∠ DBA

(if a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior opposite angles)

∠ x° = 50° + 60° = 110°

Hence, option C is correct.

**Question 4.**

Choose the correct answer:

Which of the following set of measurements will form a triangle?

A. 11 cm, 4 cm, 6 cm

B. 13 cm, 14 cm, 25 cm

C. 8 cm, 4 cm, 3 cm

D. 5 cm, 16 cm, 5 cm

**Answer:**

Any Two sides of triangle together are greater than the third side.

Option A: AB + BC > AC = 11 + 4 > 6 = 15 > 6true

BC + AC > AB = 4 + 6 > 11 = 10 > 11not true

AC + AB > BC = 11 + 6 > 4 = 17 > 4 true.

∴, this cannot be true because only condition is satisfied.

Option B: AB + BC > AC = 13 + 14 > 25 = 27 > 25 true

BC + AC > AB = 14 + 25 > 13 = 39 > 13 true

AC + AB > BC = 25 + 13 > 14 = 38 > 13 true.

∴, this is true because all condition is satisfied.

Option C: AB + BC > AC = 8 + 4 > 3 = 12 > 3 true

BC + AC > AB = 4 + 3 > 8 = 7 > 8 not true

AC + AB > BC = 8 + 3 > 4 = 11 > 4 true.

∴, this cannot be true because only condition is satisfied.

Option D: AB + BC > AC = 5 + 16 > 5 = 21 > 5true

BC + AC > AB = 16 + 5 > 11 = 21 > 5 true

AC + AB > BC = 5 + 5 > 16 = 10 > 16 not true.

∴, this cannot be true because only condition is satisfied.

**Question 5.**

Choose the correct answer:

Which of the following will form a right-angled triangle, given that the two angles are

A. 24°, 66°

B. 36°, 64°

C. 62°, 48°

D. 68°, 32°

**Answer:**

For the triangle to be right-angled triangle one angle is 90° and sum of the two angles should be 90°

Option A: 24° + 66° = 90°. This option is correct.

Option B: 36° + 64° = 100°. This option is not correct.

Option C: 62° + 48° = 110°. This option is not correct.

Option D: 68° + 32° = 100°. This option is not correct.

**Question 6.**

The angles of a triangle are (x – 35) °, (x – 20) ° and (x + 40) °.

Find the three angles.

**Answer:**

Theorem 1: The sum of three angles is 180°.

i.e. ∠A + ∠B + ∠C = 180°

⇒ (x – 35)° + (x – 20)° + (x + 40)° = 180°

⇒ x – 35 + x – 20 + x + 40 = 180

⇒ 3x – 55 + 40 = 180

⇒ 3x – 15 = 180

⇒ 3x = 180 + 15

⇒ 3x = 195

⇒

⇒ x = 15

**Question 7.**

In Δ ABC, the measure of ∠A is greater than the measure of ∠ B by 24° . If exterior angle ∠C is 108°. Find the angles of the Δ ABC.

**Answer:**

Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

Let the ∠B be x.

∴ ∠ B = x

∴ ∠ A = ∠ B + 24°

Ext. [T2] ∠ C = ∠ B + ∠ A

⇒ 108° = x + x + 24°

⇒ 108° = 2x + 24°

⇒ 2x = 108° - 24°

⇒ 2x = 84°

⇒

⇒ x = 42°

**Question 8.**

The bisectors of ∠ B and ∠ C of a Δ ABC meet at O.

Show that BOC = 90^{o} +

**Answer:**

Theorem 1:

The sum of all three angles is 180°.

Let the ∠ B and ∠ C be 2x and 2y respectively.

In Δ ABC

∠A + ∠B + ∠C = 180° (Angle Sum Property)

⇒ ∠ A + 2x + 2y = 180°

⇒ 2x + 2y = 180° - ∠A

⇒ 2(x + y) = 180° - ∠ A

⇒

⇒ …(1)

In Δ OBC

∠OBC + ∠ OCB + ∠ BOC = 180° (Angle Sum Property)

⇒ x + y + ∠ BOC = 180°

⇒

⇒

⇒

Hence Proved.

**Question 9.**

Find the value of x° and y° from the following figures:

**Answer:**

Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two-interior angle.

Ext. ∠ B = ∠ A + ∠ C

⇒ y° = x° + 50° …(1)

Ext. ∠ B + 72° = 180°

⇒ Ext. ∠ B = 180° - 72°

⇒ Ext. ∠ B = 108°

⇒ y° = 108°

Put y = 108° in (1)

⇒ 108° = x° + 50°

⇒ x° = 108° - 50°

⇒ x° = 58°

**Question 10.**

Find the value of x° and y° from the following figures:

**Answer:**

theorem 1: Sum of all the angles of triangle is 180°

∠ ACD + ∠ ACB = 180°

⇒ 5y° + y° = 180°

⇒ 6y° = 180°

⇒

⇒ y° = 30°

In Δ ABC

∠ A + ∠ B + ∠ C = 180°

⇒ x° + 4y° + y° = 180°

⇒ x° + 4 × 30° + 30° = 180°

⇒ x° + 120° + 30° = 180°

⇒ x° + 150° = 180°

⇒ x° = 180° - 150°

⇒ x° = 30°

**Question 11.**

Find the value of x° and y° from the following figures:

**Answer:**

Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

∠ C = ∠ B (Alternate interior angles)

⇒ x° = 42°

In Δ CDE

Ext. ∠ E = ∠ C + ∠ D

⇒ 82° = 42° + y°

⇒ y° = 82° - 42°

⇒ y° = 40°

**Question 12.**

Find the angles x°, y° and z° from the given figure.

**Answer:**

Theorem 1: Sum of all the angles of triangles is 180°

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

In Δ ABC

∠ A + ∠ B + ∠ C = 180° (Angle Sum Property)

⇒ 101° + 26° + z° = 180°

⇒ 127° + z° = 180°

⇒ z° = 180° - 127°

⇒ z° = 53°

In Δ FBG

Ext. ∠ G = ∠ B + ∠ F

⇒ y° = 26° + 106°

⇒ y° = 132°

∠ AED + ∠ DEC = 180°

⇒ ∠ AED + 128° = 180°

⇒ ∠ AED = 180° - 128°

⇒ ∠ AED = 52°

In AED

Ext. ∠ D = ∠ A + ∠ C

⇒ x° = 101° + 52°

⇒ x° = 153°

###### Exercise 3.2

**Question 1.**Choose the correct answer:

In the isosceles ∆XYZ, given XY = YZ then which of the following angles are equal?

A. ∠X and ∠Y

B. ∠Y and ∠Z

C. ∠Z and ∠X

D. ∠X, ∠Y and ∠Z

**Answer:**Reason: Angles opposite XY and YZ are angle X and Z respectively.[T3]

**Question 2.**Choose the correct answer:

In Δ ABC and ∆DEF, ∠B = ∠E, AB = DE, BC = EF. The two triangles are congruent under _____ axiom

A. SSS

B. AAA

C. SAS

D. ASA

**Answer:**option C.

If any two sides and the included angle of a triangle are respectively equal to any two sides and the included angles of another triangle then the two triangles are congruent.

**Question 3.**Choose the correct answer:

Two plane figures are said to be congruent if they have

A. the same size

B. the same shape

C. the same size and the same shape

D. the same size but not same shape

**Answer:**Reason: If two geometrical figures are identical in shape and size then they are said to be congruent.

**Question 4.**Choose the correct answer:

In a triangle ABC, ∠A = 40^{o} and AB = AC, then ABC is _____ triangle.

A. a right angled

B. an equilateral

C. an isosceles

D. a scalene

**Answer:**In a triangle, When two sides are equal then triangle are said to be an isosceles triangle.

**Question 5.**Choose the correct answer:

In the triangle ABC, when ∠A = 90^{o} the hypotenuse is ______

A. AB

B. BC

C. CA

D. None of these

**Answer:**

**Question 6.**Choose the correct answer:

In the Δ PQR the angle included by the sides PQ and PR is

A. ∠P

B. ∠Q

C. ∠R

D. None of these

**Answer:**

**Question 7.**Choose the correct answer:

In the figure, the value of x° is_______

A. 80^{o}

B. 100^{o}

C. 120^{o}

D. 200^{o}

**Answer:**In Δ ABD and Δ CBD

AB = CB = 2cm

AD = CD = 3cm

BD = BD = common

∴ Δ ABD ≅ Δ CBD

∴ ∠ A = ∠ C

⇒ 100° = x°

Hence, option B is correct.

**Question 8.**In the figure, ABC is a triangle in which AB = AC. Find x° and y°.

**Answer:**Theorem 2:

If a side of a triangle is produced, the exterior angle so formed, is equal to the sum of the two interior angle.

∠ ACE + ∠ ACB = 180°

⇒ x + 48° + x° = 180°

⇒ 2x + 48° = 180°

⇒ 2x = 180° - 48°

⇒ 2x = 132°

⇒

⇒ x = 66°

Ext. ∠ A = ∠ B + ∠ C

⇒ y° = x° + x°

⇒ y° = 2x°

⇒ y° = 2 × 66°

⇒ y° = 132°

**Question 9.**In the figure, Find x°.

**Answer:**Theorem 1: Sum of all the angles of the triangles is 180°

In Δ COB

OD = DC

∴ ∠ COD = ∠DCO = 40°

In Δ AOB

OA = OB (Given)

∴ ∠ OAB = ∠ OBA = x°

∠ OAB + ∠ OBA + ∠ AOB = 180° (Angle Sum Property)

⇒ x° + x° + ∠ AOB = 180°

⇒ 2x° + ∠ AOB = 180°

⇒ ∠ AOB = 180° - 2x°

∠ AOB = ∠ COD (opposite angles are equal)

180 – 2x° = 40°

⇒ 180° = 40 + 2x°

⇒ 180° - 40° = 2x°

⇒ 140° = 2x°

⇒

⇒ x = 70°

**Question 10.**In the figure ∆PQR and ∆SQR are isosceles triangles. Find x°.

**Answer:**Theorem 1: Sum of all the angles of triangles is 180°

Let ∠ PQR be y

In Δ SQR,

SQ = SR

∴ ∠ SQR = ∠ SRQ

∠ SQR + ∠ SRQ + ∠ QSR = 180° (Angle Sum Property)

⇒ 2 ∠ SQR + 70° = 180°

⇒ 2 ∠ SQR = 180° - 70°

⇒ 2 ∠ SQR = 110°

⇒

⇒ ∠ SQR = 55° = ∠ SRQ

In Δ PQR,

∠ PQR + ∠ PRQ + ∠ RPQ = 180°

⇒ y + y + 40° = 180°

⇒ 2y + 40° = 180°

⇒ 2y = 180° - 40°

⇒ 2y = 140°

⇒

⇒ y = 70°

∠ PQR = ∠ PQS + ∠ SQR

⇒ 70° = x° + 55°

⇒ x° = 70° - 55°

⇒ x° = 15°

**Question 11.**In the figure, it is given that BR = PC and ∠ACB= ∠QRP and AB ∥ PQ. Prove that

AC = QR.

**Answer:**Given: BR = PC and ∠ ACB = ∠ QRP , AB || PQ

To Prove: AC = QR

Proof:

In Δ ABC, we have

BC = BR + RC

In Δ PQR

PR = PC + RC

But , BR = PC [Given]

So, BC = PC + RC and PR = BR + RC

⇒ BC = PR

So, in Δ ABC and Δ PQR, we have

∠ ACB = ∠ QRP [Given]

BC = PR [Proved Above]

∠ ABC = ∠ QPR [AB || PQ, alternate interior angles]

Thus, Δ ABC ≅ Δ PQR [Angle – Side – Angle]

∴ AC = QR [C. P. C. T]

**Question 12.**In the figure, AB = BC = CD, ∠A = x^{o}. Prove that ∠DCF = 3∠A.

**Answer:**Given: AB = BC = CD and ∠ A = x°

To Prove: ∠ DCF = 3 ∠ A

Proof:

In Δ ABC

AB = BC [Given]

∴ ∠ A = ∠ C = x°

Now,

∴ ext. ∠ B = ∠ A + ∠ C

⇒ Ext. ∠ B = x° + x°

⇒ Ext. ∠ B = 2x°

In Δ CBD

BC = CD [Given]

∴ ∠ B = ∠ D = 2x°

Now,

In Δ ADC,

Ext. ∠ DCF = ∠ CDA + ∠ CAD

⇒ ∠ DCF = 2x + x

⇒ ∠ DCF = 3x

⇒ ∠ DCF = 3 ∠ A [ ∠ A = x°, Given]

Hence Proved.

**Question 13.**Find x°, y°, z° from the figure, where AB = BD, BC = DC and ∠DAC 30^{o}.

**Answer:**Theorem 1: Sum of all the angles in the triangle is 180°.

In Δ ABD,

We know that, AB = BD

∴ ∠ A = ∠ D

⇒ 30° = x°

Hence, ∠ A + ∠ B + ∠ C = 180°

⇒ 30° + ∠ B + 30° = 180°

⇒ ∠ B + 60° = 180°

⇒ ∠ B = 180° - 60°

⇒ ∠ B = 120°

∠ DBA + ∠ DBC = 180° (Sum of adjacent angles is 180°)

⇒ 120° + y° = 180°

⇒ y° = 180° - 120°

⇒ y° = 60°

In Δ DBC

We know that, BC = DC

∴ ∠ B = ∠ D

⇒ y° = z°

⇒ 60° = z°

**Question 14.**In the figure, ABCD is a parallelogram. AB is produced to E such that AB = BE. AD produced to F such that AD = DF. Show that ∆FDC ≡ ∆CBE.

**Answer:**Given: Parallelogram ABCD and AB = BE and AD = FD

To prove: Δ FDC ≡ ΔCBE

Construction: Join DB

Proof:

We know that,

AB = DC [ opposite sides of parallelogram]

BE = DC [AB = BE, because B is the midpoint of AE]

Similarly,

AD = BC [ opposite sides of parallelogram]

DF = BC [ AD = DF, because B is the midpoint of AE]

Now, AD||BC and AB

∠ A = ∠ B [corresponding angles] …(1)

Now, AB||CD and AD

∠ A = ∠ D [corresponding angles] …(2)

∴ ∠ B = ∠ D (From 1 and 2)

In Δ FDC and Δ CBE

FD = CB [Proved Above]

DC = BE [Proved Above]

∠ D = ∠ B [Proved Above]

Thus, Δ FDC ≡ Δ CBE

Hence Proved.

**Question 15.**In figure, BO bisects ∠ABC of ∆ABC. P is any point on BO. Prove that the perpendicular drawn from P to BA and BC are equal.

**Answer:**Given: A Δ ABC in which BO is bisector of ∠ABC

Also, we have PD ⊥ AB and PE ⊥ BC

To Prove: PD = PE

Proof:

In Δ PBD and Δ PBE

PB = PB [common]

∠ PBD = ∠ PBE [ given]

∠ PDB = ∠ PEB = 90° [Given]

Thus, Δ PBD ≅ Δ PBE [Angle – Angle – Side]

∴ PD = PE

Hence Proved.

**Question 16.**The Indian Navy flights fly in a formation that can be viewed as two triangles with common side. Prove that ∆SRT ≡ ∆QRT, if T is the midpoint of SQ and SR = RQ.

**Answer:**Given: T is the mid-point of SQ and SR = RQ

To Prove: Δ SRT ≅ Δ QRT

Proof

In Δ SRT and QRT

RT = RT [common]

ST = QT [T is the mid-point of SQ]

SR = RQ [Given]

Thus, Δ SRT ≅ Δ QRT [Side – Side – Side]

Hence Proved.

**Question 1.**

Choose the correct answer:

In the isosceles ∆XYZ, given XY = YZ then which of the following angles are equal?

A. ∠X and ∠Y

B. ∠Y and ∠Z

C. ∠Z and ∠X

D. ∠X, ∠Y and ∠Z

**Answer:**

Reason: Angles opposite XY and YZ are angle X and Z respectively.[T3]

**Question 2.**

Choose the correct answer:

In Δ ABC and ∆DEF, ∠B = ∠E, AB = DE, BC = EF. The two triangles are congruent under _____ axiom

A. SSS

B. AAA

C. SAS

D. ASA

**Answer:**

option C.

If any two sides and the included angle of a triangle are respectively equal to any two sides and the included angles of another triangle then the two triangles are congruent.

**Question 3.**

Choose the correct answer:

Two plane figures are said to be congruent if they have

A. the same size

B. the same shape

C. the same size and the same shape

D. the same size but not same shape

**Answer:**

Reason: If two geometrical figures are identical in shape and size then they are said to be congruent.

**Question 4.**

Choose the correct answer:

In a triangle ABC, ∠A = 40^{o} and AB = AC, then ABC is _____ triangle.

A. a right angled

B. an equilateral

C. an isosceles

D. a scalene

**Answer:**

In a triangle, When two sides are equal then triangle are said to be an isosceles triangle.

**Question 5.**

Choose the correct answer:

In the triangle ABC, when ∠A = 90^{o} the hypotenuse is ______

A. AB

B. BC

C. CA

D. None of these

**Answer:**

**Question 6.**

Choose the correct answer:

In the Δ PQR the angle included by the sides PQ and PR is

A. ∠P

B. ∠Q

C. ∠R

D. None of these

**Answer:**

**Question 7.**

Choose the correct answer:

In the figure, the value of x° is_______

A. 80^{o}

B. 100^{o}

C. 120^{o}

D. 200^{o}

**Answer:**

In Δ ABD and Δ CBD

AB = CB = 2cm

AD = CD = 3cm

BD = BD = common

∴ Δ ABD ≅ Δ CBD

∴ ∠ A = ∠ C

⇒ 100° = x°

Hence, option B is correct.

**Question 8.**

In the figure, ABC is a triangle in which AB = AC. Find x° and y°.

**Answer:**

Theorem 2:

∠ ACE + ∠ ACB = 180°

⇒ x + 48° + x° = 180°

⇒ 2x + 48° = 180°

⇒ 2x = 180° - 48°

⇒ 2x = 132°

⇒

⇒ x = 66°

Ext. ∠ A = ∠ B + ∠ C

⇒ y° = x° + x°

⇒ y° = 2x°

⇒ y° = 2 × 66°

⇒ y° = 132°

**Question 9.**

In the figure, Find x°.

**Answer:**

Theorem 1: Sum of all the angles of the triangles is 180°

In Δ COB

OD = DC

∴ ∠ COD = ∠DCO = 40°

In Δ AOB

OA = OB (Given)

∴ ∠ OAB = ∠ OBA = x°

∠ OAB + ∠ OBA + ∠ AOB = 180° (Angle Sum Property)

⇒ x° + x° + ∠ AOB = 180°

⇒ 2x° + ∠ AOB = 180°

⇒ ∠ AOB = 180° - 2x°

∠ AOB = ∠ COD (opposite angles are equal)

180 – 2x° = 40°

⇒ 180° = 40 + 2x°

⇒ 180° - 40° = 2x°

⇒ 140° = 2x°

⇒

⇒ x = 70°

**Question 10.**

In the figure ∆PQR and ∆SQR are isosceles triangles. Find x°.

**Answer:**

Theorem 1: Sum of all the angles of triangles is 180°

Let ∠ PQR be y

In Δ SQR,

SQ = SR

∴ ∠ SQR = ∠ SRQ

∠ SQR + ∠ SRQ + ∠ QSR = 180° (Angle Sum Property)

⇒ 2 ∠ SQR + 70° = 180°

⇒ 2 ∠ SQR = 180° - 70°

⇒ 2 ∠ SQR = 110°

⇒

⇒ ∠ SQR = 55° = ∠ SRQ

In Δ PQR,

∠ PQR + ∠ PRQ + ∠ RPQ = 180°

⇒ y + y + 40° = 180°

⇒ 2y + 40° = 180°

⇒ 2y = 180° - 40°

⇒ 2y = 140°

⇒

⇒ y = 70°

∠ PQR = ∠ PQS + ∠ SQR

⇒ 70° = x° + 55°

⇒ x° = 70° - 55°

⇒ x° = 15°

**Question 11.**

In the figure, it is given that BR = PC and ∠ACB= ∠QRP and AB ∥ PQ. Prove that

AC = QR.

**Answer:**

Given: BR = PC and ∠ ACB = ∠ QRP , AB || PQ

To Prove: AC = QR

Proof:

In Δ ABC, we have

BC = BR + RC

In Δ PQR

PR = PC + RC

But , BR = PC [Given]

So, BC = PC + RC and PR = BR + RC

⇒ BC = PR

So, in Δ ABC and Δ PQR, we have

∠ ACB = ∠ QRP [Given]

BC = PR [Proved Above]

∠ ABC = ∠ QPR [AB || PQ, alternate interior angles]

Thus, Δ ABC ≅ Δ PQR [Angle – Side – Angle]

∴ AC = QR [C. P. C. T]

**Question 12.**

In the figure, AB = BC = CD, ∠A = x^{o}. Prove that ∠DCF = 3∠A.

**Answer:**

Given: AB = BC = CD and ∠ A = x°

To Prove: ∠ DCF = 3 ∠ A

Proof:

In Δ ABC

AB = BC [Given]

∴ ∠ A = ∠ C = x°

Now,

∴ ext. ∠ B = ∠ A + ∠ C

⇒ Ext. ∠ B = x° + x°

⇒ Ext. ∠ B = 2x°

In Δ CBD

BC = CD [Given]

∴ ∠ B = ∠ D = 2x°

Now,

In Δ ADC,

Ext. ∠ DCF = ∠ CDA + ∠ CAD

⇒ ∠ DCF = 2x + x

⇒ ∠ DCF = 3x

⇒ ∠ DCF = 3 ∠ A [ ∠ A = x°, Given]

Hence Proved.

**Question 13.**

Find x°, y°, z° from the figure, where AB = BD, BC = DC and ∠DAC 30^{o}.

**Answer:**

Theorem 1: Sum of all the angles in the triangle is 180°.

In Δ ABD,

We know that, AB = BD

∴ ∠ A = ∠ D

⇒ 30° = x°

Hence, ∠ A + ∠ B + ∠ C = 180°

⇒ 30° + ∠ B + 30° = 180°

⇒ ∠ B + 60° = 180°

⇒ ∠ B = 180° - 60°

⇒ ∠ B = 120°

∠ DBA + ∠ DBC = 180° (Sum of adjacent angles is 180°)

⇒ 120° + y° = 180°

⇒ y° = 180° - 120°

⇒ y° = 60°

In Δ DBC

We know that, BC = DC

∴ ∠ B = ∠ D

⇒ y° = z°

⇒ 60° = z°

**Question 14.**

In the figure, ABCD is a parallelogram. AB is produced to E such that AB = BE. AD produced to F such that AD = DF. Show that ∆FDC ≡ ∆CBE.

**Answer:**

Given: Parallelogram ABCD and AB = BE and AD = FD

To prove: Δ FDC ≡ ΔCBE

Construction: Join DB

Proof:

We know that,

AB = DC [ opposite sides of parallelogram]

BE = DC [AB = BE, because B is the midpoint of AE]

Similarly,

AD = BC [ opposite sides of parallelogram]

DF = BC [ AD = DF, because B is the midpoint of AE]

Now, AD||BC and AB

∠ A = ∠ B [corresponding angles] …(1)

Now, AB||CD and AD

∠ A = ∠ D [corresponding angles] …(2)

∴ ∠ B = ∠ D (From 1 and 2)

In Δ FDC and Δ CBE

FD = CB [Proved Above]

DC = BE [Proved Above]

∠ D = ∠ B [Proved Above]

Thus, Δ FDC ≡ Δ CBE

Hence Proved.

**Question 15.**

In figure, BO bisects ∠ABC of ∆ABC. P is any point on BO. Prove that the perpendicular drawn from P to BA and BC are equal.

**Answer:**

Given: A Δ ABC in which BO is bisector of ∠ABC

Also, we have PD ⊥ AB and PE ⊥ BC

To Prove: PD = PE

Proof:

In Δ PBD and Δ PBE

PB = PB [common]

∠ PBD = ∠ PBE [ given]

∠ PDB = ∠ PEB = 90° [Given]

Thus, Δ PBD ≅ Δ PBE [Angle – Angle – Side]

∴ PD = PE

Hence Proved.

**Question 16.**

The Indian Navy flights fly in a formation that can be viewed as two triangles with common side. Prove that ∆SRT ≡ ∆QRT, if T is the midpoint of SQ and SR = RQ.

**Answer:**

Given: T is the mid-point of SQ and SR = RQ

To Prove: Δ SRT ≅ Δ QRT

Proof

In Δ SRT and QRT

RT = RT [common]

ST = QT [T is the mid-point of SQ]

SR = RQ [Given]

Thus, Δ SRT ≅ Δ QRT [Side – Side – Side]

Hence Proved.