Students can Download Maths Chapter 15 Quadrilaterals Ex 15.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

## Karnataka Board Class 8 Maths Chapter 15 Quadrilaterals Ex 15.1

Question 1.

Two angles of a quadrilateral are 70° and 130° and the other two angles are equal. Find the measure of these two angles.

Answer:

Let the equal angles be x

70 + 130 + x + x = 360°

[Sum of the angles of a quadrilateral 360°]

200 + 2x = 360°

2x = 160

x =

x = 80°

Each of the other two angles is 80°

Question 2.

In the figure. Suppose ∠P and ∠Q are supplementary angles and ∠R = 125 °. Find the measure of∠S.

Answer:

∠P + ∠Q +∠R + ∠s = 360°

[Sum of the angles of a quadrilateral]

180 + 125° + ∠S = 360°

[∠P and ∠Q are supplementary]

305 + ∠S = 360°

∠ S = 360 – 305

∠S = 55°

Question 3.

Three angles of a quadrilateral are in the ratio 2:3:5 and the fourth angles are 90°. Find the measure of the other three angles.

Answer:

Three angles are in the ratio 2 : 3 : 5 Let the angles be 2x, 3x and 5x

[Sum of the angles of a quadrilateral]

2x + 3x + 5x + 90° = 360°

10x + 90 = 360

10x = 360 – 90

10x = 270°

x = = 27°

∴ 2x = 2 × 27 = 54°

3x = 3 × 27 = 81°

5x = 5 × 27 = 135°

Three angle are 54°, 81° & 135°

Question 4.

In the adjoining figure, ABCD is a quadrilateral such that ∠D + ∠C = 100°. The bisectors of ∠A and ∠B meet at P.

Determine ∠APB.

Answer:

∠A + ∠B + ∠C + ∠D = 360°

[Sum of the angles of a quadrilateral]

∠A + ∠B = 360 – 100

∠A + ∠B = 260

∠A + ∠B = × 260°

[Multiplying by ]

∠PAB + ∠PBA = 130°

[AP and BP are bisectors of ∠A and ∠B]

In ∆ APB. ∠APB +∠PAB +∠PB A = 180°

∠PAB + ∠PBA + ∠APB = 180°

[Sum of the angles of a triangle]

∠APB + 130 = 180°

∠APB = 180 – 130

∠ APB = 50°