KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex 15.3

In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 15 Quadrilaterals Ex 15.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 15 Quadrilaterals Ex 15.3 pdf, free KSEEB SSLC Class 8 Maths Chapter 15 Quadrilaterals Ex 15.3 pdf download. Now you will get step by step solution to each question.

Karnataka State Syllabus Class 8 Maths Chapter 15 Quadrilaterals Ex 15.3

Question 1.
The adjacent angles of a parallelogram are in the ratio 2 : 1. Find the measures of all the angles.
Answer:
The adjacent angles are in the ratio 2 : 1.
Let the angles be 2x and x
2x + x = 180°
[adjacent angles of a parallelogram are supplementary]
3x = 180° .
x = \frac { 180 }{ 3 } = 60°
2x = 2 × 60° = 120°
∴The angles of the parallelogram are 60°, 120°. 60° and 120°

Question 2.
A field is in the form of a parallelogram whose perimeter is 450m and one of its sides is larger than the other by 75m. Find the lengths of all the sides.
Answer:
Let one side be x, then the other side is x + 75
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.3 1
Perimeter = 450
x + (x + 75) + x + (x + 75) = 450
4x + 150 = 450
4x = 450 – 150
4x = 300
x = 3004 = 75
x = 75m
x + 75 = 75 + 75 = 150 m
∴ The four sides are 75m, 150m, 75m, 150m.

Question 3.
In the figure ABCD is a parallelogram. The diagonals AC and BD intersect at 0; and ∠DAC = 40°, ∠CAB = 35° and ∠DOC = 110° calculate the measure ∠ABO, ∠ADC, ∠ACB and ∠CBD
Answer:
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.3 2
∠AOB = ∠DOC = 110°
[Vertically opposite angles]
In ΔAOB,
∠AOB + ∠OAB + ∠OBA = 180°
110 + 55 +∠OBA = 180°
∠OBA = 180 – 165
∠ABO = 15°
∠ADC + ∠DAB = 180°
[adjacent angles of a parallelogram , are supplementary]
∠ADC + 95 = 180°
∠ADC = 180 – 95
∠ ADC = 85°
∠ACB = ∠DAC = 40°
[AD 11BC alternate angle]
∠ACB = 40°
∠ABC = ∠ ADC = 85°
[Opposite angles pf a parallelogram]
[∠ABC = ∠ABO + ∠CBO]
15 + ∠CBO = 85°
∠CBO = 85 – 15
∠CBO = 70
∴ ∠CBO = ∠CBD = 70°
∴ ∠CBD = 70°

Question 4.
In a parallelogram ABCD, the side DC is produced to E and ∠BCE = 105° calculate ∠A, ∠E, ∠C, and∠D.
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.3 3
Answer:
∠DCB + ∠DCE = 180° [Linear pair] ∠DCB + 1050 = 180°
∠DCB = 180 – 105
∠DCB = 75°
∠A = ∠DCB = 75°
∠A = ∠C = 75°
[Opposite angles of a parallelogram]
∠A +∠B = 180°
[adjacent angles of a parallelogram]
75° + ∠B = 180°
∠B = 180 + 75
∠B = 105°
∠B = ∠D = 105°
[Opposite angles of a parallelogram]

Question 5.
In a parallelogram KLMN, ∠K = 60 °. Find the measures of all the angles.
Answer:
KSEEB Solutions for Class 8 Maths Chapter 15 Quadrilaterals Ex. 15.3 4
∠K = ∠M = 60° [Opposite angles of parallelogram]
∠K + ∠N = 180° [adjacent angles of a parallelogram]
60 + ∠N = 180°
∠N = 180 – 60
∠N = 120°
∠N = ∠L = 120°
[Opposite angles of a parallelograom]
∴ Angles of a paralelogram are 60°, 120°, 60° & 120°

All Chapter KSEEB Solutions For Class 8 maths

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All Subject KSEEB Solutions For Class 8

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