These Class 7 Maths Extra Questions Chapter 6 The Triangle and its Properties will help students prepare well for the exams.

## Class 7 Maths Chapter 6 Extra Questions The Triangle and its Properties

### Class 7 Maths The Triangle and its Properties Extra Questions – 2 Marks

Question 1.

The ratio of angles of a triangle are 1 : 2 : 3. Find the largest angle.

Solution:

Let the ratio of angles of a triangle = 1 : 2 : 3

Let the first angle = x°

Second angle = 2x°; Third angle = 3x°

By angle sum property of a triangle

Sum of the angles in a triangle = 180°

x° + 2x° + 3x° = 180°

6x° = 180°

x° = \(\frac{180^{\circ}}{6}\) ⇒ x° = 30°

Largest angle = 3x° = 3 × 30° = 90°

Question 2.

Find x in the figure.

Solution:

By the exterior angle property of a triangle

In △ABC, x = 80° + 30° ⇒ x = 110°

Question 3.

State True (or) False.

i) The sum of all angles in a triangle is 180°.

ii) In a right angle triangle the largest angle is 98°.

Solution:

i) True

ii) False.

Question 4.

Two angles of a triangle are 60° and 70°, then find the third angle.

Solution:

Given the two angles of a triangle are 60° and 70°

Let the third angle = x°

According to Angle sum property in triangles

60° + 70° + x° = 180°

130° + x° = 180°

x° = 180° – 130° ⇒ x = 50°

∴ The third angle = 50°

Question 5.

Find the angles in an isosceles right angled triangle?

Solution:

△PQR is an isosceles right angled triangle

∠Q = 90°

PQ = QR, then ∠P =∠R

Let ∠P = ∠R = x°

By angle sum property

∠P + ∠Q +∠R = 180°

x° + 90° + x° = 180°

2x° + 90° = 180°

2x° = 180° – 90°

2x° = 90° ⇒ x° = \(\frac{90^{\circ}}{2}\) ⇒ x° = 45°

∴ The angles of an isosceles right angled triangle are 45°, 45° and 90°.

Question 6.

Find x° in the figure.

Solution:

In △MNR, ∠M = 60°, ∠N = x°

∠MRT = 120°

By exterior angle property in triangles

60° + x° = 120°

x° = 120° – 60° ⇒ x° = 60°

Question 7.

In the figure find ‘b’.

Solution:

By exterior angle property

b = 65° + 45° ⇒ b = 110°

Question 8.

State True (or) False.

i) Each angle in an equilateral triangle is 60°.

ii) There can be two obtuse angles in a right angled triangle.

Solution:

i) True

ii) False

Question 9.

Find ‘a’ in the figure.

Solution:

By angle sum property

a° + a° + a° = 180°

3a° = 180°

a° = \(\frac{180^{\circ}}{3}\) ⇒ a° = 60°

∴ Each angle is an equilateral triangle = 60°

Question 10.

State True (Or) false.

i) A triangle can have 60°, 80°, 100° as its measurements.

ii) The largest side in a right angled triangle is called Hypotenuse.

Solution:

i) False

ii) True

Question 11.

Find the exterior angle of an equilateral triangle?

Solution:

Each angle in an equilateral triangle = 60°

Let x° be the exterior angle.

By exterior angle property

x° = 60° + 60°

x° = 120°

∴ Exterior angle of an equilateral triangle = 120°

Question 12.

Two angles of a triangle are 110° and 35°. Find the measure of third angle.

Solution:

Given two angles are 110° and 35°

Let the third angle = x°

By angle sum property is triangles

110° + 35° + x° = 180°

145° + x° = 180°

x° = 180° – 145° ⇒ x° = 35°

∴ The measure of third angle = 35°

Question 13.

State True or False.

i) A triangle has 4 sides

ii) A triangle has 3 angles

Solution:

i) False

ii) True

Question 14.

Write Yes or No for the following statements.

i) A triangle has no diagonals

ii) A triangle has 3 vertices.

Solution:

i) Yes,

ii) Yes

Question 15.

In the figure find x.

Solution:

In △ABC, ∠B = 90°

By Pythagoras property

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

x^{2} = 3^{2} + 4^{2}

x^{2} = 9 + 16

x^{2} = 25 ⇒ x^{2} = 5^{2} ⇒ x = 5 cm

Question 16.

One of the acute angles of a right angled triangle is 54°. Find other angle.

Solution:

Let in △ABC ∠B = 90°, ∠C = 54°

By angle sum property

∠A + ∠B + ∠C = 180°

∠A + 90° + 54° = 180°

∠A + 144° = 180°

∠A = 180° – 144° ⇒ ∠A = 36°

∴ Other angle = 36°

Question 17.

In the figure find x.

Solution:

In △PQR, ∠Q = 90°,

PR = 13 cm, PQ = x cm, RQ = 12 cm

By Pythagoras theorum

PR^{2} = PQ^{2} + QR^{2}

13^{2} = x^{2} + 12^{2}

169 = x^{2} + 144

x^{2} = 169 – 144

x^{2} = 25 ⇒ x^{2} = 5^{2} ⇒ x = 5cm

Question 18.

In a triangle the sum of two angles is equal to third angle. Find the greatest angle of triangle.

Solution:

Let △ABC is a triangle

Let ∠A = ∠B + ∠C

By angle sum property

∠A + ∠B +∠C = 180°

A + ∠A = 180°

2∠A = 180° ⇒ ∠A = \(\frac{180^{\circ}}{2}\) = 90°

∴ The greatest angle = 90°

Question 19.

Is it possible to have the angles 45°, 62°, 73° in a triangle.

Solution:

Sum of given angles = 45° + 62° + 73° = 180°

∴ It is possible as angle sum property is satisfied.

Question 20.

Is it possible to have the angles 30°, 70°, 60° in a triangle

Solution:

Sum of given angles = 30° + 70° + 60°

= 160° ≠ 180°

Angle sum property of triangle is not satisfied.

∴ The triangle will not have given angles

### The Triangle and its Properties Extra Questions Class 7 – 3 Marks

Question 1.

The ratio of angles of a triangles are 2 : 3 : 4. Find the angles.

Solution:

Given ratio = 2 : 3 : 4

Let the first angle = 2x°

Second angle = 3x°;

Third angle = 4x°

By angle sum property

2x° + 3x° + 4x° = 180°

9x° = 180° ⇒ x° = \(\frac{180^{\circ}}{9}\) ⇒ x° = 20°

First angle = 2 × 20° = 40°

Second angle = 3 × 20° = 60°

Third angle = 4 × 20° = 80°

Question 2.

State True (Or) False.

i) 1 cm, 2 cm, 3 cm are the measurements of a right triangle.

ii) A triangle will have 2 vertices

iii) Each angle in an equilateral triangle is equal and equal to 60°

Solution:

i) False

ii) False

iii) True

Question 3.

The angles of a triangle are x°, x – 20°, x -40°, Find the angles.

Solution:

Let the angles of triangle are

x° + (x-20)°, (x-40)°

By angle sum property

x° + (x – 20)° + (x – 40)° = 180°

3x° – 60° = 180°

3x° = 180° + 60°

3x° = 240° ⇒ x° = \(\frac { 240 }{ 3 }\) ⇒ x° = 80°

∴ First angle = 80°

Second angle = x – 20° = 80° – 20° = 60°

Third angle = x – 40° = 80° – 40° = 40°

Question 4.

Verify 15 cm, 12 cm and 17 cm will form a right angled triangle (or) not.

Solution:

15^{2} = 225; 12^{2} = 144 ; 17^{2} = 289

here 289 ≠ 225 + 144

289 ≠ 369

∴ They will not form a right angled triangle.

Question 5.

In the figure find a and b.

Solution:

By exterior angle property

a = 30° + 40° ⇒ a = 70°

a + b = 180° (Linear Pair)

70° + b = 180°

b = 180° – 70° ⇒ b = 110°

Question 6.

From the below figure find the points i) inside ii) outside iii) on the triangle.

Solution:

i) Inside-A, B

ii) Outside – H, I, J, K, L

iii) On the triangle-D, E, F.

Question 7.

If one angle of a triangle is 50° and other two angles are in the ratio 2 : 3. Find the other angles.

Solution:

One angle = 50°

Let Second angle = 2x°

Third angle = 3x°

By angle sum property

50° + 2x° + 3x° = 180°

50° + 5x° = 180°

5x° = 180° – 50°

x° = \(\frac{130^{\circ}}{5}\) ⇒ x° = 26°

2x° = 2 × 26° = 52°

3x° = 3 × 26° = 78°

∴ The other angles are 52° and 78°

∴ The other angles are 52° and 78°

Question 8.

In △XYZ, ∠X = 120°, XA bisects ∠X and XA ⊥ YZ and find ∠Y.

Solution:

In △XYZ, ∠X = 120°

XA bisects ∠X

XA ⊥ YZ

In △XYA, ∠A = 90°; ∠X = 60°

By angle same property

∠X + ∠A + ∠Y = 180°

60° + 90° + ∠Y = 180°

150° + ∠Y = 180°

∠Y = 180° – 150° ⇒ ∠Y = 30°

Question 9.

From the figure, find a and b

Solution:

b + 100° = 180° (Linear Pair)

b = 180° – 100° ⇒ b = 80°

By exterior angle property

a = 40° + b

a = 40° + 80°

a = 120°

Question 10.

In the figure given find all the angles of △ABC.

Solution:

Let us draw the figure

∠1 = 40° (Vertically Opposite angles)

∠3 + 120° = 180° (Linear Pair)

3 = 180° – 120° ⇒∠3 = 60°

By angle sum property

∠A + ∠B +∠C = 180°

∠1 + ∠2 + ∠3 = 180°

40° + ∠2 + 60° = 180°

∠2 + 100° = 180°

2 = 180° – 100° ⇒ ∠2 = 80°

Question 11.

Find a and b in the figure.

Solution:

b = 45° +50° (Exterior angle property)

b = 95°

a = 40 + b (Exterior angle property)

a = 40° + 95°

a = 135°

Question 12.

Find x in the figure.

Solution:

Let us draw the figure

∠1 + 110° = 180° (Linear Pair)

∠1 = 180° – 110° = 70°

∠2 = 120° = 180° (Linear Pair)

∠2 = 180° – 120°

∠2 = 60°

120° = ∠1 + x° (Exterior angle property)

120° = 70° + x°

x° = 120° – 70° ⇒ x = 50°

Question 13.

Verify (5, 8, 17) will form a pythagoras triplet (or) not.

Solution:

(5,8,17)

5^{2} = 25; 8^{2} = 64

17^{2} = 289

289 = 25 + 64

289 = 89 (False)

∴ (5,8,17) will not form a pythagoras triplet.

Question 14.

Verify (8,15,17) is a pythagoras triplet (or) not.

Solution:

8^{2} = 64 ; 15^{2} = 225

17^{2} = 289

289 = 225 + 64

289 = 289 (True)

∴ (8,15,17) will form a pythagoras triplet.

Question 15.

The angles of a triangle are in the ratio 3 : 4 : 5. Is a right angled triangle ? Justify.

Solution:

Let the first angle = 3x

Second angle = 4x

Third angle = 5x

By angle sum property

3x°+ 4x° + 5x° = 180°

12x° = 180° ⇒ x° = \(\frac{180^{\circ}}{12}\) ⇒ x° = 15°

First angle = 3 × 15° = 45°

Second angle = 4 × 15° = 60°

Third angle = 5 × 15°= 75°

∴ They will not form a right angled triangle.

### Extra Questions of The Triangle and its Properties Class 7 – 5 Marks

Question 1.

The lengths of diagonals of a Rhombus are 42 cm and 40 cm . Find the perimeter of the Rhombus.

Solution:

In a rhombus, the diagonals intersect at 90°.

Let the diagonals intersect at ‘ O ‘, then four right angled triangles formed.

∴ In △BOC, ∠BOC = 90°

By pythagoras property

x^{2} = OC^{2} + OB^{2}

Also diagonals bisect each other

∴ OD = OB = \(\frac { 40 }{ 2 }\) = 20 cm

∴ OC = OA = \(\frac { 42 }{ 2 }\) = 21 cm

x^{2} = 21^{2} + 20^{2}

x^{2} = 441 + 400

x^{2} = 841 ⇒ x^{2} = 29^{2} ⇒ x = 29 cm

Perimeter of rhombus

= 4 × 29 cm = 116 cm

Question 37.

In the figure find x.

Solution:

In △PQR, ∠Q = 90°

By pythagoras property

PR^{2} = PQ^{2} + QR^{2}

(x – 2)^{2} = 5^{2} + (x – 3)^{2}

x^{2} – 4x + 4 = 25 + x^{2} – 6x + 9

[∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

-4x + 4 = 34 – 6x

-4x + 6x = 34 – 4

2x = 30 ⇒ x = \(\frac { 30 }{ 2 }\) ⇒ x = 15

Question 38.

Find x in the fugure.

Solution:

In △ABC, ∠B = 90°

By pythagoras property

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

17^{2} = x^{2} + 15^{2}

289 = x^{2} + 225

x^{2} = 289 – 225

x^{2} = 64 ⇒ x^{2} = 8^{2} ⇒ x = 8 cm

Question 39.

Observe the following figure and answer the following questions.

In △ABC,

i) which is the median?

ii) which is the altitude ?

iii) what is ∠ADB?

iv) write the relation between BE and CE.

v) If BC = 12 cm what is BE ?

Solution:

i) AE

ii) AD

iii) 90°

iv) BE = CE

v) BE = \(\frac { 12 }{ 2 }\) = 6 cm

Question 40.

In the figure, find ∠P + ∠Q + ∠R +∠A + ∠B + ∠C.

Solution:

In △PQR, by angle sum property

∠P + ∠Q +∠R = 180°

In △ABC by angle sum property

∠A + ∠B + ∠C = 180°

(1) + (2) ⇒

∠P + ∠Q + ∠R + ∠A + ∠B + ∠C

= 180° + 180° = 360°

Question 41.

Prove that the sum of form angles in a quadrilateral is 360°by using angle sum property in triangles.

Solution:

Let us consider PQRS is a quadrilaterial. Join QS diagonal

In △PQS

∠P + a + d = 180° …(1) (By angle sum property)

in △RSQ

∠R + ∠b + ∠c = 180° …(2) (By angle sum property)

(1) + (2) ⇒

∠P + a + d + ∠R + ∠b +∠c = 180° + 180° = 360°

∠P + ∠c + ∠d + ∠b + ∠a + ∠R

∠P + ∠Q + ∠R + ∠S = 360°

Question 42.

In the figure, DE || BC find a, b and c.

Solution:

Given in △ABC, DE || BC

∠B = 60°, ∠C = 40°

By angle sum property

∠A + ∠B + ∠C = 180°

C + 60° + 40° = 180°

C = 180° – 100°

C = 80°

DE || BC

∠a = 60° (Corresponding angle)

∠b = 40° (Corresponding angle)

Question 43.

Find ∠p and ∠q and also ∠r from the below figure.

Solution:

In △ABD, By angle sum property

40° + 60° + ∠p = 180°

100° + ∠p = 180°

p = 180° – 100° ⇒ ∠p = 80°

∠p + ∠q = 180° (Linear Pair)

80° + ∠q = 180°

∠q = 180° – 80° ⇒ ∠q = 100°

In △AOC by angle sum property

∠A + q + r = 180°

35° + 100° + r = 180°

135° + ∠r = 180°

∠r = 180° – 135° ⇒ ∠r = 45°

Question 44.

The angle of a triangle are (a – 40)°, (a – 20)° and \(\left(\frac{1}{2} a-10\right)^{\circ}\) Find ‘a’.

Solution:

By angle sum property of traingle

a – 40° + a – 20° + \(\frac { 1 }{ 2 }\) a – 10° = 180°

2a + \(\frac { a }{ 2 }\) – 70° = 180°

\(\frac { 5a }{ 2 }\) = 180° – 70°

\(\frac { 5a }{ 2 }\) = 110°

a = 110 × \(\frac { 2 }{ 5 }\) = 22 × 2 ⇒ a = 44°

Question 45.

In △PQR, ∠P = 90°, D lies on QP produced and DE ⊥ QR intersecting QR at E and PR at F . If △PFE = 120°, find i)∠QDE ii) ∠QRP

Solution:

In △PQR, ∠P = 90°

DE ⊥ QR, ∠PFE = 120°

∠PFD = 180° – 120° = 60° (Linear Pair)

∠QDE = 180° – 90° – 60°

= 180° – 150° = 30° (By angle sum property)

∠EFR = 60° (Vertically opposite angle)

∠QRP = 180° – 90° – 60°

= 180° – 150° = 30°

Question 46.

In △ABC, AB = AC, ∠A = 90°, then prove that BC^{2} = 2AC^{2}.

Solution:

In △ABC, ∠A = 90°

AB = AC

By pythagoras property

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

BC^{2} = AC^{2} + AC^{2} (∵AB = AC)

BC^{2} = 2 AC^{2}

Question 47.

A ladder is placed against a wall of 20 m from a point 15 m on the ground from the foot of wall. Find the length of ladder.

Solution:

In right angle triangle, ABC ,

∠A = 90°, AB = 20 m, AC = 15 m

BC = xm

By pythagoras property

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

x^{2} = 20° + 15^{2}

x^{2} = 400 + 225

x^{2} = 625 ⇒ x^{2} = 25° ⇒ x = 25 m

∴ Length of ladder = 25 m

Question 48.

A cat walks towards each 3 km and then to North 4 km. How far is the shortest-distance from standing point it is?

Solution:

In ABO right angled triangle ∠B = 90°

AB = 4m; OB = 3m

By pythagoras property

AO^{2} = AB^{2} + BO^{2}

AO^{2} = 4^{2} + 3^{2} = 16 + 9 = 25

AO^{2} = 5^{2}

AO = 5 cm

Shortest Distance = 5m.

Question 49.

The hypotenuse of a right angled triangle is 2.5 m. If one of its side is 1.5 m. Find the length of other side.

Solution:

In △ABC, ∠B = 90°

By pythagoras property

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

(2.5)^{2} = AB^{2} + (1.5)^{2}

6.25 = AB^{2} + 2.25

6.25 – 2.25 = AB^{2}

AB^{2} = 4

AB^{2} = 2^{2}

AB = 2 m

∴ Length of other side = 2 m.

Question 50.

Two sides of a right angled triangle are equal and the square of its hypotenuse is 50 cm. Find the length of equal side.

Solution:

In △PQR, ∠Q = 90°

By pythagoras property

PR^{2} = PQ^{2} + QR^{2} (∵ PQ = QR = x)

50 = x^{2} + x^{2}

2x^{2} = 50; x^{2} = 25

x = √ 25 = x = 5

∴ Length of equal side = 5 cm