## ML Aggarwal Class 7 Solutions for ICSE Maths Chapter 10 Lines and Angles Ex 10.2

Question 1.

Identity each of the given pair of angles as alternate interior angles, co-interior angles or corresponding angles or none of these in the given figure:

(i) ∠2, ∠6

(ii) ∠1, ∠6

(iii) ∠3, ∠5

(iv) ∠2, ∠7

(v) ∠3, ∠6

(vi) ∠4, ∠8

Solution:

In the given figure,

(i) ∠2 and ∠6 – are corresponding angles.

(ii) ∠1 and ∠6 – none

(iii) ∠3 and ∠5 – alternate interior angles

(iv) ∠2 and ∠7 – none

(v) ∠3 and ∠6 – co-interior angles

(vi) ∠4 and ∠8 – corresponding angles

Question 2.

State the property that is used in each of the following statements:

(i) If a || b, then ∠1 = ∠5.

(ii) If ∠4 = ∠6, then a || b.

(iii) If ∠4 + ∠5 = 180°, then a || b.

Solution:

In the given figure,

(i) If a || b, then

∠1 = ∠5 (Property of the corresponding pair of angles)

(ii) If ∠4 = ∠6

Property of interior alternate angles are equal

a || b

(iii) If ∠4 + ∠5 = 180°

Property of co-interior angles are supplementary, then a || b.

Question 3.

In each of the following figures, a pair of parallel lines is cut by a transversal. Find the value of x:

Solution:

(i) In the given figure,

x = 100 (Corresponding angles)

(ii) x + 110° = 180°

(Cointerior Angles are Supplementary)

x = 180° – 110° = 70°

(iii) Let ∠1 opposite to ∠110°

∠1 = 110° (Vertically opposite angles)

and ∠1 + x = 180°

(Cointerior Angles are Supplementary)

⇒ 110° + x = 180°

⇒ x = 180° – 110°

⇒ x = 70°

Question 4.

In the following figures, a pair of parallel lines are cut by a transversal. Find the value of x in each figure.

Solution:

(i) AB || CD and EF its transversal

⇒ ∠AGH + ∠GHC = 180°

⇒ 2x + 60 + 3x + 54° = 180°

(Sum of co-interior angles)

⇒ 5x + 60° = 180°

⇒ 5x = 180° – 60° = 120°

⇒ x = 24°

x = 24°

(ii) AB || CD and EF is its transversal

Which intersects them at G and H.

∠HGB = ∠FHD (Corresponding angles)

∠HGB = 3x + 30°

But ∠AGH + ∠HGB = 180° (Linear pair)

2x+ 15° + 3x + 30° = 180°

⇒ 5x + 45° = 180°

⇒ 5x = 180° – 45° = 135°

⇒ 5x = 135°

⇒ x = 27°

x = 27°

Question 5.

In the following figures (i) to (vi), a pair of parallel lines are cut by a transversal. Find the size of each lettered angle.

Solution:

In the given figure,

(i) Given angle is 60°

x = 60° (Vertically opposite angles)

But ∠y = ∠x (corresponding angles)

∠y = 60°

Hence, ∠x = 60°, ∠y = 60°

(ii) In the given figure,

Given angle is 135°

∠q = 135° (Vertically opposite angles)

But ∠p + ∠q = 180° (Sum of cointerior angles)

135° + ∠q = 180°

∠q = 180° – 135° = 45°

∠p = 45,°, ∠q = 135°

(iii) In the given figure,

Given angle is 70°

a = 70° (Alternate angles)

But ∠a + ∠b = 180° (Linear pair)

70° + ∠b = 180°

∠b = 180° – 70° = 110°

∠a = 70°, ∠b = 110°

(iv) In the given figure,

Given angle is 128°

∠x + 128° = 180° (Linear pair)

∠x = 180° – 128° = 52°

But ∠x = ∠y (Corresponding angles)

∠y = 52°

Also, y + z = 180° (Linear pair)

52° + ∠z = 180°

∠z = 180° – 52°

∠z = 128°

Hence, x = 52°, y = 52°, z = 128°

(v) In the given figure,

Given angle is 75°

∠a + 75° = 180° (Linear pair)

∠a = 180° – 75° = 105°

∠b = 75° (Vertically opposite angles)

∠c = 75° (Corresponding angles)

But ∠c + ∠d = 180° (Linear pair)

75° + ∠d = 180°

∠d = 180° – 75° = 105°

∠a = 105°, ∠b = 75°, ∠c = 75°, ∠d = 105°

(vi) Given angle = 62°

62 ° + ∠q = 180° (Linear pair ∠s)

∠q = 180° – 62° = 118°

∠p = 62° (Vertically opposite ∠s)

∠s = p (Corresponding ∠s)

∠s = 62°

∠s + ∠r = 180° (Linear pair ∠s)

62° + ∠r = 180°

∠r = 180° – 62° = 118°

Question 6.

In the given diagram, lines AB, CD and EF are parallel. Calculate the values of x and y. Hence, find the reflex angle ECA.

Solution:

In the given figure,

AB || CD || EF

∠E = 120°, ∠A = 140°

y + 140° = 180° (Sum of co-interior angles)

y = 180° – 140° = 40°

Similarly, x + 120° = 180°

x = 180° – 120° = 60°

x = 60°, y = 40°

Now, ∠ECA = x + y = 60° + 40° = 100°

and reflex ∠ECA = 360° – 100° = 260°

Question 7.

In the given figure, l || m. Find the values of x, y and z.

Solution:

In the given figure, x = 45° (Alternate angles)

Similarly, z = 55° (Alternate angles)

But x + y + z = 180°

(Angles on the one side of a straight line)

⇒ 45° + y + 55° = 180°

⇒ y + 100° = 180°

⇒ y = 180° – 100° = 80°

x = 45°, y = 80°, z = 55°

Question 8.

Calculate the measure of each lettered angle in the following figure (parallel lines, segment or rays are denoted by thick matching arrows):

Solution:

(i) In the given figure,

∠x + 143° = 180°

(Sum of co-interior angles)

∠x = 180° – 143° = 37°

Similarly, ∠z + 60° = 180°

∠y = 180° – 60° = 120°

But ∠x + ∠y + ∠z = 180°

(Angles on one side of a straight line)

37° + 120° + ∠y = 180°

157° + ∠y = 180°

∠y = 180° – 157° = 23°

∠x = 37°, ∠y = 23°, ∠z = 120°

(ii) In the given figure,

∠c = 72° (Alternate angles)

∠a = 55° (Corresponding angles)

But ∠b + ∠c + 55° = 180°

(Angles on one side of a straight line)

∠b + 72° + 55° = 180°

∠b + 127° = 180°

∠b = 180° – 127° = 53°

Here, ∠a = 55°, ∠b = 53°, ∠c = 72°

(iii) In the given figure,

∠a + 75° = 180° (Linear pair)

∠a = 180° – 75° = 105°

∠b = 75° (Alternate angles)

∠d = b (Corresponding angles)

∠d = 75°

75° = ∠c (Corresponding angles)

∠c = 75°

∠a = 105°, ∠b = 15°, ∠c = 75°, ∠d = 75°

Question 9.

In the figure given below, state whether the lines l and m are parallel or not.

Solution:

(i) In the given figure,

∠106° and ∠64° are co-interior angles

106° + 64° = 170°

Sum of co-interior angles is not equal to 180°

l and m are not parallel

(ii) ∠1 = 75° (Vertically opposite angles)

But ∠1 and 75° are co-interior angles

and ∠1 + 75° = 75° + 75° = 150°

It the sum is not equal to 180°

l and m are not parallel

(ii) ∠1 = 57° (Vertically opposite angles)

∠1 + 123° = 57° + 123° = 180° (Sum of co-interior)

∠1 + 123° = 180°

l || m