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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