# 9th Class Math Chapter 11 Parallelograms and Triangles Short Questions Answer

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## 9th Class Math Chapter 11 Parallelograms and Triangles Short Questions Answer

Solution:

n° ≅ 75° Opposite angles are congruent

n = 75

y° ≅ n° Alternate angles

y° ≅ n° ≅ 75°

y = 45

x° + y° = 180° Suppletory angles

x + y = 180

x + 75 = 180

x = 180 – 75 = 105

m° ≅ x° Opposite angles

m = x = 105

m° = 105°

Solution:

11x ≅ 55°

11x = 55

x = 5°

(5m + 10)° + 55° = 180°

Sum of interior angles of II lines

5m + 10 + 55 = 180

5m + 65 = 180

m = 23°

Solution:

As opposite sides of a parallelogram are congruent

8m – 4n = 8

Or 2m – n = 2 (i)

And 4m + n = 10 (ii)

Adding (i) and (ii)

6m = 12

Or m = 2

Putting m = 2 in (i)

We have

2(2) – n = 2

4 – n = 2

Or – n = – 2

Or n = 2

m = 2, n = 2

Solution:

Opposite angles of a parallelogram are congruent

∠ L ≅ ∠ N

But it is given that

m∠ L + m∠ N = 110

2(m∠ L) = 110 m∠ L = 55

m∠ L = m∠ N = 55°

m∠ L + m∠ P = 180°

Sum of interior angles between parallel lines

55 + m∠ P = 180°

m∠ P = 180° – 55° = 125°

Angles of the parallelogram are

55°, 125°, 55° and 125°

m∠ M = m∠ P = 125°