The Triangle and its Properties Class 7 Extra Questions with Answers

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
AC² = AB² + BC²
x² = 3² + 4²
x² = 9 + 16
x² = 25 ⇒ x² = 5² ⇒ 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
PR² = PQ² + QR²
13² = x² + 12²
169 = x² + 144
x² = 169 – 144
x² = 25 ⇒ x² = 5² ⇒ 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:
15² = 225; 12² = 144 ; 17² = 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)
5² = 25; 8² = 64
17² = 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:
8² = 64 ; 15² = 225
17² = 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
x² = OC² + OB²
Also diagonals bisect each other
∴ OD = OB = \(\frac { 40 }{ 2 }\) = 20 cm
∴ OC = OA = \(\frac { 42 }{ 2 }\) = 21 cm
x² = 21² + 20²
x² = 441 + 400
x² = 841 ⇒ x² = 29² ⇒ 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
PR² = PQ² + QR²
(x – 2)² = 5² + (x – 3)²
x² – 4x + 4 = 25 + x² – 6x + 9
[∵ (a – b)² = a² – 2ab + b²]
-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
AC² = AB² + BC²
17² = x² + 15²
289 = x² + 225
x² = 289 – 225
x² = 64 ⇒ x² = 8² ⇒ 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 BC² = 2AC².
Solution:

In △ABC, ∠A = 90°
AB = AC
By pythagoras property
BC² = AB² + AC²
BC² = AC² + AC² (∵AB = AC)
BC² = 2 AC²

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
BC² = AB² + AC²
x² = 20° + 15²
x² = 400 + 225
x² = 625 ⇒ x² = 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
AO² = AB² + BO²
AO² = 4² + 3² = 16 + 9 = 25
AO² = 5²
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
AC² = AB² + BC²
(2.5)² = AB² + (1.5)²
6.25 = AB² + 2.25
6.25 – 2.25 = AB²
AB² = 4
AB² = 2²
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
PR² = PQ² + QR² (∵ PQ = QR = x)
50 = x² + x²
2x² = 50; x² = 25
x = √ 25 = x = 5
∴ Length of equal side = 5 cm