# A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

**Solution:**

By dividing the quadrilateral into two triangles and applying Heron’s formula, we can calculate the area of triangles.

Heron's formula for the area of a triangle is: Area = √s(s - a)(s - b)(s - c)

Where a, b, and c are the sides of the triangle, and s = Semi-perimeter = Half the Perimeter of the triangle

Now, ABCD is the park shown in the figure below

We have ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m.

Let's connect B and D, such that BCD is a right-angled triangle.

In ∆BDC, apply Pythagoras theorem in order to find the length of BD

BD^{2} = BC^{2} + CD^{2} [Pythagoras theorem]

BD^{2} = 12^{2} + 5^{2}

BD^{2} = 144 + 25

BD = √169

BD = 13 m

Area of quadrilateral ABCD = area of ∆BCD + area of ∆ABD

Now, Area of ∆BCD = 1/2 × base × height

= 1/2 × 12 m × 5 m

= 30 m^{2}

Now, in ∆ABD, AB = a = 9 m, AD = b = 8 m, BD = c = 13 m

Semi Perimeter of ΔABD

s = (a + b + c)/2

= (9 + 8 + 13)/2

= 30/2

= 15 m

By using Heron’s formula,

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

= √15(15 - 9)(15 - 8)(15 - 13)

= √15 × 6 × 7 × 2

= 6√35

= 35.5 m^{2} (approx.)

Area of ΔABD = 35.5 m^{2}

Therefore,

Area of park ABCD = 30 m^{2} + 35.5 m^{2} = 65.5 m^{2}

Thus, the park ABCD occupies an area of 65.5 m^{2}.

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

**Video Solution:**

## A park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?

Class 9 Maths NCERT Solutions Chapter 12 Exercise 12.2 Question 1

**Summary:**

It is given that there is a park, in the shape of a quadrilateral ABCD, has ∠C = 90°, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. We have found that it occupies an area of 65.5 m^{2}.

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