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 = = 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

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:

∠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.

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:

∠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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