CTET 2 Math and Science 3 Jan 2022 Paper

Area of trapezium = 1 / 2 × (sum of parallel sides) × height
ABCD is a trapezium AD II BC, AB = 5 cm, BC = 11 cm, AD = 7 cm, DA is produced to a point F such that AF = 3 cm and BF perpendicular to DF.In triangle BFA,
${\mathrm{BF}}^2={\mathrm{AB}}^2-{\mathrm{AF}}^2$
${\mathrm{BF}}^2=5^2-3^2$
${\mathrm{BF}}^2=16\mathrm{cm}$
$\mathrm{BF}=4\mathrm{cm}$
Area of trapezium $=1∕2\times ($ sum of parallel sides $)\times$ height
$\Rightarrow 1∕2\times \left(7+11\right)\times 4$
$\Rightarrow 1∕2\times 18\times 4$
$\Rightarrow 36{\mathrm{cm}}^2$