Question

A park in the shape of a quadrilateral ABCD has angle c=90 degree AB=9m BC=12m CD=5m and AD=8m. How much area does it occupy?

Answer 1

Clearly, BCD is a right triangle. So Area(BCD)=1/2(BC*CD)=1/2*12*5=30sq cm

For the other part use Heron’s Formula. BD is definitely 13cm from Pythagoras Theorem. Which gives BD=13cm, AB=9cm and AD=8cm. Now s=30cm.

So Area(ABD)=root(s/2(s-a)(s-b)(s-c)).

=root(15*2*6*7)=root(1295) sq cm

Hence, the total area would be 30root(1295)sq cm

Answer 2

area of quadrilateral can be written as area of triangle ABC +triangle ADC
now area of triangle ABC=
$x 1 (y2 - y3) + x2(y3 - y1) + x3(y1 - y2)$
sub in this formula you will get your answer




not sure about the answer