ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3

Question 1.
Find the value of x in each of the following figures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q1.1
Solution:
(i) In ∆ABC,
AB = AC
∠B = ∠C (Angles opposite to equal sides)
x = 50°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q1.2
(ii) In ∆ABC,
AB = AC
∠B = ∠C (Angles opposite to equal sides)
∠B = ∠C = x
and Ext. ∠BAD = ∠B + ∠C = x + x
110° = ∠B + ∠C
110° = x + x = 2x
x = 55°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q1.3
(iii) In the given figure,
CA = CB
∠A = ∠ABC = x
∠EBD = 36°
∠ABC = ∠EBD = 36° (Vertically opposite angles)
x = 36°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q1.4

Question 2.
Find the value of x in each of the following figures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q2.1
Solution:
(i) In ∆ABC,
AC = BC, ∠C = 40°
∠A = ∠B (Angles opposite to equal sides)
∠A = ∠B = x
But A + B + C = 180° (Angles of a triangle)
⇒ x + x + 40° = 180°
⇒ 2x = 180° – 40° = 140°
⇒ x = 70°
x = 70°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q2.2
(ii) In ∆ABC,
∠B = 45°
AC = BC
∠A = ∠B = 45°
But ∠A + ∠B + ∠C = 180°
⇒ 45° + 45° + x° = 180°
⇒ x + 90°= 180°
⇒ x = 180° – 90° = 90°
x = 90°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q2.3
(iii) In the given figure, EF || BC
In ∆ABC, Ext. ∠DAF = 50°
AB = BC
∠A = ∠C = x
EF || BC
∠DAF = ∠ABC = 50°
Now in ∆ABC
∠BAC + ∠ABC + ∠BCA = 180°
⇒ x + 50° + x = 180°
⇒ 2x = 180° – 50° = 130°
⇒ x = 65°
x = 65°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q2.4

Question 3.
Find the values of x and y in each of the following figures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q3.1
Solution:
(i) In the given figure of ∆ABC
AB = AC
∠ABC = ∠ACB = y
But, Ext. ∠ACD + ∠ACB = 120° (Linear pair)
∠ACB = 180° – 120° = 60°
y = 60°
Now in ∆ABC,
∠A + ∠B + ∠ACB = 180° (Angles of a triangle)
⇒ x + y + y = 180°
⇒ x + 60° + 60° = 180°
⇒ x + 120° = 180°
⇒ x = 180° – 120° = 60°
Here, x = 60°, y = 60°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q3.2
(ii) Here, ∠A = ∠ACB
∠ACB = x (∵ angles opposite to equal sides)
⇒ x + 115° = 180°
⇒ x = 180° – 115° = 65
∠ACB = ∠A = 65°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q3.3
Now, ∠A + ∠B + ∠ACB = 180° (Sum of ∠s of a ∆)
65° + 65° + ∠y = 180°
130° + ∠y = 180°
∠y = 180° – 130° = 50°
(iii) In ∆ABC,
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q3.4
∠ABC = ∠ACB (∠s opposite to equal sides)
∠ABC = x = ∠ACB
Now, In ∆ABC
48° + x + x = 180° (Sum of ∠s of a ∆)
⇒ 2x = 180° – 48°
⇒ 2x = 132°
⇒ x = 66°
In ∆ACD,
∠CAD = ∠CDA (∠s opposite to equal sides)
∠CAD = y = ∠CDA
Now, x + ∠ACD = 180° (Linear pair ∠s)
66° + ∠ACD = 180°
∠ACD = 180° – 66° = 114°
Now, in ∆ACD
y + y + 114°= 180° (Sum of ∠s of a ∆)
⇒ 2y + 114° = 180°
⇒ 2y = 180° – 114°
⇒ 2y = 66°
⇒ y = 33°
Hence, x = 66° and y = 33°

Question 4.
Calculate the size of each lettered angle in the following figures:
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q4.1
Solution:
(i) In ∆ABC,
AC = BC (Given)
∠ABC = ∠BAC (∠s opposite to equal sides)
∠ABC = 32°
⇒ x = 32°
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q4.2
Now, y = 32° + 32° = 64°
(∵ Exterior angle = Sum of two opposite interior ∠s)
In ∆BCD,
BC = BD
∠BDC = ∠BCD (∠s opposite to equal sides)
∠BDC = y = ∠BCD
∠BDC = 64° = ∠BCD
Now, z + 64° + 64° = 180° (Sum of ∠s of a ∆)
⇒ z + 128°= 180°
⇒ z = 180° – 128° = 52°
Hence, x = 32°, y = 64°, z = 52°
(iii) In ∆ABC,
AC = BC (Given)
∠ABC = ∠BAC (∠s opposite to equal sides)
∠ABC = 35° = ∠BAC
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q4.3
Now, ∠ACD = ∠ABC + ∠BAC (Exterior angle = Sum)
= 35° + 35° = 70°.
In ∆ACD,
AC = AD (Given)
∠ADC = ∠ACD (∠s opposite to equal sides)
∠ADC = 70°
Now, ∠x = ∠ABD + ∠ADB
(Exterior angle = Sum of two opposite interior ∠s)
= 35° + 70° = 105°
Hence, ∠x = 105°

Question 5.
If the angles of a triangle are in the ratio 1 : 2 : 1, find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Ratio in the angles of a triangle are 1 : 2 : 1
Sum of angles of a triangle = 180°
Let first angle = x
Then second = 2x
and third angle x
x + 2x + x = 180°
⇒ 4x = 180°
⇒ x = 45°
Angles are 45°, 45° × 2 = 90° and 45°
Two angles are equal
Their opposite sides are also equal
It is an isosceles triangle
It’s one angle is 90°
It is a right-angled triangle.

Question 6.
In an isosceles triangle, a base angle is four times its vertical angle. Find all the angles of the triangle.
Solution:
In an isosceles triangle ABC, AB = AC
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q6.1
Base angles are equal
Let ∠B = ∠C = x
ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 11 Triangles and its Properties Ex 11.3 Q6.2
Angles of the triangle will be 80°, 80°, 20°

ML Aggarwal Class 7 Solutions for ICSE Maths

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