Question

1. In Fig. 14.74, compute the area of quadrilateral ABCD. ​

Answer 1

Answer: 114 cm^2

Step-by-step explanation:

Let AB=x

BD must be 15 and angle DBC must be 90 as BC^2 +BD^2=DC^2,  it forms a pythagorean triplet.

So,

$81+x^{2} =DB^{2} \\and \\DB^2=225\\hence,\\81+x^{2} =225\\x^2=144\\x=12$

a(ABD)=(9*12)/2=54

a(DBC)=(15*8)/2=60

a(ABCD=a(DBC)+a(ABD)=114