In quadrilateral ABCD, we have AD = 9 cm, BC = 8 cm and CD = 17 cm, AD ⊥ AB and diagonal BD ⊥ BC. Find the area of the quadrilateral ABCD.

Correct option is D

Given:

In quadrilateral ABCD, AD = 9cm, BC =8cm and CD = 17cm

AD ⊥ AB and diagonal BD ⊥ BC

Formula Used:

Area of triangle =$$\frac{1}{2}bh$$​​

Solution:

Area of quadrilateral ABCD = Area of triangle ABD + Area of triangle BCD

In the figure we need BD to find the area of triangle BCD and AB to find the area of triangle ABD

For BD we use Pythagoras’ Theorem

​$$CD^2 = BC^2 + BD^2$$​​

​$$(17)^2 = (8)^2 + BD^2$$​​

​$$BD^2 = 289 - 64 = 225$$​​

BD =$$\sqrt{225}$$​ = 15cm

For area of triangle ABD we need the measure of AB using Pythagoras’ Theorem

​$$(BD)^2 = (AB)^2 + (AD)^2$$​​

​$$(15)^2 = (AB)^2 + (9)^2$$​​

​$$AB^2$$​ = 225-81 = 144

AB =$$\sqrt{144}$$​ = 12cm

Area of triangle ABD =$$\frac{1}{2} \times 12 \times 9 = 54cm^2$$​​

Area of triangle BCD =$$\frac{1}{2} \times 15 \times 8 =60cm^2$$​​

Area of quadrilateral ABCD = 60+54 = 114cm$$^2$$​​