Question

A park, in the shape of a quadrilateral ABCD, has 2 C = 90°, AB = 9 m, BC = 12 m. CD= 5 m and AD=8 m. How much area does it occupy?​

Answer 1

Answer:

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 the Pythagoras theorem in order to find the length of BD

BD2 = BC2 + CD2 [Pythagoras theorem]

BD2 = 122 + 52

BD2 = 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 m2

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 m2 (approx.)

Area of ΔABD = 35.5 m2

Therefore,

Area of park ABCD = 30 m2 + 35.5 m2 = 65.5 m2

Thus, the park ABCD occupies an area of 65.5 m2.