Selina Concise Class 7 Math Chapter 14 Lines And Angles Exercise 14B Solution

Selina Concise Class 7 Math Chapter 14 Lines And Angles Exercise 14B Solution

EXERCISE 14B

In questions 1 and 2, given below, identify the given pairs of angles as corresponding angles, interior alternate angles, exterior alternate angles, adjacent angles, vertically opposite angles or allied angles:

(1) (i) Interior alternate angles

(ii) Adjacent angles

(iii) Corresponding angles

(iv) Exterior alternate angles

(v) Co-interior (allied) angles

(vi) Exterior alternate angles

(vii) Corresponding angles

(viii) Vertically opposite angles

(ix) Adjacent angles

(2) (i) Vertically opposite angles

(ii) Interior alternate angles

(iii) Vertically opposite angles

(iv) Corresponding angles

(v) Vertically opposite angles

(vi) Co-interior angles

(vii) Vertically opposite angles

(viii) Adjacent angles

(ix) Co-interior angles

(x) Adjacent angles

(xi) Co-interior angles

(3) In the following figures, the arrows indicate parallel lines. State which angles are equal. Give reasons.

(i) ∠a = ∠b (corresponding angles)

∠b = ∠c (Vertically opposite angles)

∠a = ∠c (alternate angles)

Then, ∠a = ∠b = ∠c

(ii) x = y (Vertically opposite angles)

y = l (Alternate angles)

x = l (corresponding angles)

l = n (vertically opposite angles)

n = r (Corresponding angles)

Then, x = y = l = n = r

k = m (Vertically opposite angles)

m = q (Alternate angles)

Then, k = m = q

(4) In the given figure, find the measure of the unknown angles:

Ans: f = 110° (vertically opposite angles)

f = d = 110° (alternate angles)

d = a = 110° (vertically opposite angles)

a + b = 180° (linear pair)

⇒ 110° + b = 180°

⇒ b = 180° – 110°

⇒ b = 70°

b = c = 70° (Vertically opposite angles)

c = g = 70° (alternate angles)

g = e = 70° (vertically opposite angles)

Hence, a = 110°, b = 70°, c = 70°, d = 110°, e = 70°, f = 110°, g = 70°

(5) Which pair of the dotted line segments, in the following figures, are parallel. Give reason:

Ans: (ii), (iv) and (v)

(6) In the given figure, the directed lines are parallel to each other. Find the unknown angles.

(i) b = 60° (vertically opposite angles)

b = a = 60° (alternate angles)

a = c = 60° (vertically opposite angles)

(ii) y = 55° (vertically opposite angles)

z + 55° = 180° (linear pair)

⇒ z = 180° – 55°

⇒ z = 125°

z = x = 125° (alternate angles)

(iii) 120° = c (alternate angles)

c + a = 180° (linear pair)

⇒ a + 120° = 180°

⇒ a = 180° – 120° = 60°

a = b = 60° (vertically opposite angles)

(iv) x= 50° (corresponding angles)

y + 120° = 180° (linear pair)

⇒ y = 180° – 120°

⇒ y = 60°

x + y + z = 360°

⇒ 50° + 60° + z = 360°

⇒ z + 110° = 360°

⇒ z = 360° – 110°

⇒ z = 250°

(v) x = 90° (linear pair)

We know sum of three angles of a triangle 180°

Then, 30° + 90° + z = 180°

⇒ z + 120° = 180°

⇒ z = 180° – 120°

⇒ z = 60°

z + y = 180° (linear pair)

⇒ 60° + y = 180°

⇒ y = 180° – 60°

⇒ y = 120°

Again, y + k = 180° (linear pair)

⇒ 120° + k = 180°

⇒ k = 180° – 120°

⇒ k = 60°

(vi) x = 110°(vertically opposite angles)

x + p = 180° (linear pair)

⇒ 110° + p = 180°

⇒ p = 180° – 110°

⇒ p = 70°

p = 60° (interior angles)

p + q = 180° (linear pair)

⇒ q = 180° – 60°

⇒ q = 120°

r + 60° = 180° (linear pair)

⇒ r = 180° – 60° = 120°

r = t = 120° (vertically opposite angles)

s = y = 70° (alternate angles)

(vii) y = 110° (vertically opposite angles)

x + 120° = 180° (linear pair)

⇒ x = 180° – 120°

⇒ x = 60°

x = p = 60° (parallel lines)

p + q = 180° (linear pair)

⇒ 60° + q = 180°

⇒ q = 180° – 60°

⇒ q = 120°

z + y = 180° (linear pair)

⇒ z = 180° -110°

⇒ z = 70°

(viii) x + 112° = 180° (linear pair)

⇒ x = 180° -112°

⇒ x = 68°

x + 75° + z = 180° (linear pair)

⇒ z + 75° + 68° = 180°

⇒ z = 180° – 143°

⇒ z= 37°

y = 75° (alternate angles)

(ix) a + 115° = 180° (linear pair)

⇒ a = 180° – 115° = 65°

c + 120° = 180° (linear pair)

⇒ c = 180° – 120° = 60°

a + b + c = 180° (linear pair)

⇒ 65° + b + 60° = 180°

⇒ b + 125° = 180°

⇒ b = 180° – 125° = 55°

(x) x + 110° = 180° (co-interior angles)

⇒ x = 180° – 110° = 70°

x + y = 180° (co-interior angles)

⇒ y = 180° – 70° = 110°

Y = z = 110° (corresponding angles)

(xi) y = 160° + 130° (alternate angles)

⇒ y = 290°

x + y = 360° (angles at a point)

⇒ x = 360° – 290° = 70°

(xii) b = 50° + 40° (alternate angles)

⇒ b = 90°

a + b = 360° (angles at a point)

⇒ a = 360° – 90° = 270°

(7) Find x, y and p in the given figures:

(i) x + 40° + 270° = 360° (angle at a point)

⇒ x + 310° = 360°

⇒ x = 360° – 310° = 50°

x = z = 50° (corresponding angles)

y = 40° (corresponding angles)

z + p = 180° (linear pair)

⇒ p = 180° – 50°

⇒ p = 130°

(ii) p + 25° + 110° = 180° (linear pair)

⇒ p = 180° – 135° = 45°

y = 110° (corresponding angles)

x + y + 25° = 180° (sum of three angles of a triangle)

⇒ x + 110° + 25° = 180°

⇒ x = 180° – 135°

⇒ x = 45°

(8) Find x in the following cases:

(i) x + 2x = 180° (co-interior angles)

⇒ 3x = 180°

⇒ x = 180°/3 = 60°

(ii) 4x + 5x = 180° (vertically opposite angles)

⇒ 9x = 180°

⇒ x = 180°/9 = 20°

(iii) x + 4x = 180° (linear pair)

⇒ 5x = 180°

⇒ x = 180°/5 = 36°

(iv) 2x + 5° + 3x + 55° = 180° (co-interior angles)

⇒ 5x + 60° = 180°

⇒ 5x = 180° – 60° = 120°

⇒ x = 120°/5 = 24°

(v) 3x + 25° + 2x + 20° = 180° (linear pair)

⇒ 5x + 45° = 180°

⇒ 5x = 180° – 45° = 135°

⇒ x = 135°/5 = 27°

(vi) 4x + 6x = corresponding angles = 130°

⇒ 10x = 130°

⇒ x = 130°/10

⇒ x = 13°

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