Question 37

In the quadrilateral $$ABCD$$ below, $$\angle DAB$$ = 90° and $$AB = 24$$cm , $$BC = 40$$cm, $$CD = 50$$cm and $$AD = 18$$cm(The diagram is not drawn to scale) Find the area of the quadrilateral

Solution

Join BD

Area of ABCD = area of ABD + area of BCD

In ABD $$AB^2\ +\ AD^2\ =BD^2$$

Therefore $$24^{2\ }+\ 18^{2\ }=\ BD^2$$ , on solving BD= 30

also $$30^{2\ }+\ 40^2\ =50^2$$ which implies BCD is also a right-angled triangle

Area of ABD = $$\frac{1}{2}\times\ 24\times\ 18$$ = 216

Area of BCD = $$\frac{1}{2}\times\ 30\times\ 40$$= 600

Area of ABCD = 816

Share on Whatsapp Share on Whatsapp

Create a FREE account and get: