**Quadrilaterals – Maharashtra Board Class 9 Solutions for Geometry**

AlgebraGeometryScience and TechnologyHindi

**Exercise – 5.1**

**Solution 1:**

Let measure of the fourth angle be x˚.

By using the angle sum property of quadrilaterals, we get,

130˚+ 82˚+ 40˚+ x˚= 360˚

∴ x˚= 360˚- 252˚

∴ x˚= 108˚

∴ The measure of the fourth angle is 108˚.

**Solution 2:**

The angles of the quadrilateral are in the ratio 2:3:5:8.

Let the measures of the angles of the quadrilateral be 2x, 3x, 5x and 8x.

Using the angle sum property of quadrilaterals, we get,

2x + 3x + 5x + 8x = 360˚

∴18x= 360˚

∴ x = 20˚

∴ 2x = 2 × 20˚ = 40˚

∴ 3x = 3 × 20˚ = 60˚

∴ 5x = 5 × 20˚ = 100˚

∴ 8x = 8 × 20˚ = 160˚

∴ The measures of the angles of the quadrilateral are 40˚, 60˚, 100˚ and 160˚.

**Solution 3:**

The measures of angles of a □PQRS are 3x, 4x, 5x and 6x respectively.

So ∠P = 3x, ∠Q = 4x, ∠R = 5x and ∠S = 6x.

Using the angle sum property of quadrilaterals, we get,

m∠P + m∠Q + m∠R + m∠S = 360˚

∴3x + 4x + 5x + 6x = 360˚

∴ 18x = 360˚

∴ x = 20˚

∴m∠P = 3x = 3 × 20˚ = 60˚

∴m∠Q = 4x = 4 × 20˚ = 80˚

∴m∠R = 5x = 5 × 20˚ = 100˚

∴m∠S = 6x = 6 × 20˚ = 120˚

∴ The measures of the angles of the quadrilateral are 60˚, 80˚, 100˚and 120˚.

**Solution 4:**

Let ∠B = ∠C = ∠D = x

Using the angle sum property of quadrilaterals we get,

m∠A + m∠B + m∠C + m∠D = 360˚

∴120˚ + x +x +x = 360˚

∴3x = 240˚

∴ x = 80˚

∴m∠B = x = 80˚

∴m∠C = x = 80˚

∴m∠D = x = 80˚

∴ The measures of angles ∠B, ∠C and ∠D are 80˚, 80˚and 80˚ respectively.

**Solution 5:**

Let the measures of all angles of a quadrilateral be x.

By using angle sum property of quadrilaterals, we get,

x + x + x + x = 360˚

∴ 4x = 360˚

∴ x = 90˚

∴ The measure of all the angles of the quadrilateral is 90˚.

**Solution 6:**

The measures of the angles of a quadrilateral are 120˚, 90˚, 72˚and x˚.

Using the angle sum property of quadrilaterals, we get,

120˚ + 90˚+ 72˚ + x = 360˚

∴ x = 360˚- 282˚

∴ x = 78˚

∴ The measure of x is78˚.

**Solution 7:**

In □KLMN, m∠K = 30˚, m∠M = 150˚ and m∠N = 110˚.

By using the angle sum property of quadrilaterals we get,

m∠K + m∠M + m∠N + m∠KLM = 360˚

∴30˚ + 150˚+ 110˚ + m∠KLM = 360˚

∴m∠KLM = 360˚- 290˚

∴m∠KLM = 70˚

m∠KLM + m∠MLP = 180˚ ….[Linear pair angles]

∴70˚ + m∠MLP = 180˚

∴m∠MLP = 110˚

∴ The measures of ∠KLM and ∠MLPare 70˚and 110˚ respectively.

**Solution 8:**

Let □KLMN be the quadrilateral withm∠K = 55˚,

m∠L = 55˚, and m∠M = 150˚.

Using the angle sum property of quadrilaterals, we get,

m∠K + m∠L + m∠M + m∠N = 360˚

∴55˚ + 55˚+ 150˚ + m∠N = 360˚

∴m∠N = 360˚- 260˚

∴m∠N = 100˚

∴ The measure of fourth angle is 100˚.

**Solution 9:**

**Solution 10:**

**Solution 11:**

**Exercise – 5.2**

**Solution 1:**

The opposite angles of a parallelogram are congruent.

∴∠A = ∠C = x˚

∴∠B = ∠D = 3x˚ + 20˚

In parallelogram ABCD

Using the angle sum property of quadrilaterals, we get,

m∠A + m∠B + m∠C + m∠D = 360˚

∴ x˚+ (3x˚ + 20˚) + x˚+ (3x˚ + 20˚) = 360˚

∴8x˚+ 40˚= 360˚

∴ 8x˚= 320˚

∴ x˚= 40˚

∴m∠C = x˚= 40˚

∴m∠D = 3x˚ + 20˚ = 120˚+ 20˚= 140˚

**Solution 2:**

The ratio of two sides of a parallelogram is 3:5.

Suppose the lengths of the sides of the parallelogram are 3x, 5x, 3x and 5x.

Perimeter of the parallelogram is 48cm.

∴3x + 5x + 3x + 5x = 48

∴16x= 48

∴ x = 3 cm

∴ 3x = 3 × 3 = 9 cm

∴5x = 5 × 3 = 15cm

∴The lengths of the sides of the parallelogram are 9cm, 15cm, 9cm and 15cm.

**Solution 3:**

Let the length of the first side of the parallelogram be x cm.

The other side of the parallelogram is greater than the first by 25cm.

So the length of other side is (25 + x) cm.

Perimeter of the parallelogram is 150cm.

∴ x + (25 + x) + x + (25 + x) = 150

∴ 4x + 50 = 150

∴ x = 25 cm

The length of the other side

= (25 + x) cm = (25 + 25) cm = 50 cm

∴The lengths of the sides of the parallelogram are 25cm, 50cm, 25cm and 50cm.

**Solution 4:**

Adjacent angles of a parallelogram are in the ratio 1:2.

Let the measures of the adjacent angles of the parallelogram be x, 2x.

We know that the opposite angles of a parallelogram are congruent.

So there are two angles having measure x and two angles having measure 2x.

By angle sum property of quadrilaterals we get,

x + 2x + x + 2x = 360˚

∴ 6x = 360˚

∴ x = 60˚

∴ 2x = 120˚

∴ The measures of all the angles of the parallelogram are 60˚, 120˚, 60˚and 120˚.

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**Exercise – 5.3**

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**Exercise – 5.4**

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