Question

Find area of shaded region in figure

Answer 1

in ∆ABD (right angled ∆)
AB²=AD²+BD² (pythagoras theorem)
=(12)²+(16)²
=144+25
AB²=400
AB=√400
AB=20 cm

Area of ∆ABD
$\frac{1}{2} \times base \times height$
1/2×12×16
=96cm²
perimeter of ∆ABC = sum of all the sides
perimeter=52+48+20
=120cm
semiperimeter (s) =120/2
=60cm
area of ∆ABC
$s \sqrt{(s - a)(s - b)(s - c)}$
60√(60-52)(60-48)(60-20)
=60√8×12×40
=60×4×4√15
=960×3.87298
=3718.06401cm²
Area of shaded region = area of ∆ABC - area of ∆ABD
=3718.06401-96
Area of shaded region =3622.06401cm² (ANSWER)

Answer 2

the answer is 3622.06