Question

Sides of a quadrilateral ABCD measure 9 m, 40 m, 28 m, and 15 m and the angle between the first two sides is a right angle. What is the area of quadilateral ABCD?

Answer 1

ABC is a right triangle; right angled at B,

AC = $\sqrt{A{B}^2+B{C}^2}$ ..... (Pythagoras Theorem)

=$\sqrt{9^2+(40{)}^2}$

=$\sqrt{81+1600}$

=$\sqrt{1681}=41m$ (9 -40 - 41 Pythagoras triplet)

Now, Area of $\Delta ACD$

$a=15m,b=28m,c=41m$

$s=\frac{15+28+41}{2}=\frac{84}{2}=42m$

$\therefore \text{Area}\Delta ACD=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{42\times 27\times 14\times 1}$

$=\sqrt{2\times 3\times 7\times 3\times 3\times 3\times 2\times 7}$

$=2\times 3\times 3\times 7=126sq.m$

Similarly, area of $\Delta ABC$

$=\frac{1}{2}\times 40\times 9=180sq.m$

$\therefore$ Area of quadrilateral ABCD

=Area of $\Delta ABC$ + Area of $\Delta ACD$

$=180+126=306sq.m$