Sign in
Please select an account to continue using cracku.in
↓ →
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

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, right angled at B.
Area of ABD = $$\dfrac{1}{2}\times\ 24\times\ 18$$ = 216
Area of BCD = $$\dfrac{1}{2}\times\ 30\times\ 40$$= 600
Area of ABCD = 816
Create a FREE account and get:
Book Free CAT Mentorship
Get personalized CAT strategy from a 99%iler
500+ students mentored
OTP Verification
Enter the 6-digit code sent to your phone
Booking Summary
Enter OTP
Didn't receive the OTP?
Start your IIM journey with the right preparation and crack CAT 2026.
Educational materials for CAT preparation