Question

ncert solutions for class 8 maths chapter 11 try these pg 175​

Answer 1

Answer:

this is the answer of your question

Answer 2

(1) Given,

A quadrilateral ABCD with diagonal AC = 6 cm

Heights = 3 cm and 6 cm

To find,

Area of the quadrilateral.

Solution,

The diagonal AC is dividing the quadrilateral into two triangles.

Area of triangle = 1/2 × base × height

Area of quadrilateral = area of ∆ ABC + area of ∆ ACD

Area = 1/2 × base × ( Height of ∆ ABC + height of ∆ ACD)

Area = 1/2 × 6 × ( 3+5)

Area = 1/2 × 6 × 8

Area = 24 cm²

(2) Given,

The quadrilateral is a rhombus.

diagonal 1 = 7 cm

diagonal 2 = 6 cm

To find,

Area of the rhombus.

Solution,

We know,

Area of rhombus = 1/2 × diagonal 1 × diagonal 2

Area = 1/2 × 7 × 6

Area = 21 cm².

(3) Given,

The quadrilateral is a parallelogram.

Diagonal is dividing it into two triangles of the same bases and heights.

Base = 8 cm

Height = 2 cm

To find,

Area of the parallelogram.

Solution,

Area of parallelogram = sum of the area of two triangles

Area = 2 × 1/2 × base × height ( The base and height are the same of two triangles. So, the formula is multiplied by 2)

Area = base × height

Area = 8 × 2

Area = 16 cm².