# The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

**Solution:**

Given, ABCD is a quadrilateral

The sides of the quadrilateral are 6 cm, 8 cm, 12 cm and 14 cm (taken in order)

The angle between the first two sides is a right angle.

We have to find the area of the quadrilateral.

Given, ABC is a right triangle with B at right angle.

By using Pythagorean theorem,

AC² = AB² + BC²

AC² = (6)² + (8)²

AC² = 36 + 64

AC² = 100

Taking square root,

AC = 10 cm

Area of quadrilateral ABCD = area of triangle ABC + area of triangle ACD

Area of triangle = 1/2 × base × height

Area of triangle ABC = 1/2 × BC × AB

= 1/2 × 8 × 6

= 4 × 6

Area of triangle ABC = 24 cm²

Considering triangle ACD,

a = 10 cm

b = 12 cm

c = 14 cm

By Heron’s formula,

Area of triangle = √s(s - a)(s - b)(s - c)

Where s = semiperimeter

s = (a + b + c)/2

So, s = (10 + 12 + 14)/2

= 36/2

s = 18 cm

Area of triangle ACD = √18(18 - 10)(18 - 12)(18 - 14)

= √18(8)(6)(4)

= √9 × 2 × 4 × 2 × 3 × 2 × 4

= (3 × 2 × 4)√3 × 2

Area of triangle ACD = 24√6 cm²

Area of ABCD = 24 + 24√6

= 24(1 + √6) cm²

Therefore, the area of quadrilateral ABCD is24(1 + √6) cm²

**✦ Try This: **A field in the form of a parallelogram has sides 70 m and 50 m and one of its diagonals is 90 m long. Find the area of the parallelogram.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 12

**NCERT Exemplar Class 9 Maths Exercise 12.3 Problem 8**

## The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. Find its area.

**Summary:**

The sides of a quadrilateral ABCD are 6 cm, 8 cm, 12 cm and 14 cm (taken in order) respectively, and the angle between the first two sides is a right angle. The area is 24(1 + √6) cm²

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