Question

Find the area of the following quadrilaterals (All measures are in cm)

• A. $296c{m}^2;174c{m}^2;297c{m}^2$
• B. $297c{m}^2;175c{m}^2;297c{m}^2$
• C. $297c{m}^2;174c{m}^2;299c{m}^2$
• D. $297c{m}^2;174c{m}^2;297c{m}^2$

Answer 1

The correct option is D $297c{m}^2;174c{m}^2;297c{m}^2$

(i)Area of triangle${}_1=\frac{\left(11\right)\left(22\right)}{2}=121c{m}^2$
Area of triangle${}_2=\frac{\left(16\right)\left(22\right)}{2}=176c{m}^2$
Area of quadrilateral$=121+176=297c{m}^2$
(ii)$9^2+{12}^2=81+144=225$
$\sqrt{225}=15$
Area of triangle$=\frac{\left(15\right)\left(16\right)}{2}=120c{m}^2$
Perimeter$s=\frac{9+12+15}{2}=18$
Area of triangle$=\sqrt{18(18-15)(18-9)(18-12)}=\sqrt{2916}=54c{m}^2$
Area of quadrilateral$=120+54=174c{m}^2$
(iii)Area of triangle${}_1=\frac{\left(27\right)\left(12\right)}{2}=162c{m}^2$
Area of triangle${}_2=\frac{\left(27\right)\left(10\right)}{2}=135c{m}^2$
Area of quadrilateral$=162+135=297c{m}^2$