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:
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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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