Question

In fig. 9.28 Seg AD || seg DC and Seg DA $\perp$Seg AB. Further DC = 7cm, CB = 10cm and AB= 13cm. FInd the area of $\square$ABCD.

Answer 1

Given:
Seg AB $\parallel$ seg DC and seg DA $\perp$ seg AB

Construction:
Draw a perpendicular from CE to AB.

AE = DC = 7 cm

$\therefore$ EB = AB – AE = (13 – 7) cm = 6 cm

In $△$BEC, we have:
$$\begin{gathered} {\mathrm{BC}}^2={\mathrm{CE}}^2+{\mathrm{EB}}^2(\mathrm{By}\mathrm{pythagoras}\mathrm{theorem}) \\ \Rightarrow {10}^2={\mathrm{CE}}^2+6^2 \\ \Rightarrow {\mathrm{CE}}^2=100-36 \\ \Rightarrow {\mathrm{CE}}^2=64 \\ \Rightarrow \mathrm{CE}=8\mathrm{cm} \end{gathered}$$

$\therefore$ Area of the trapezium = $\frac{1}{2}$× (sum of parallel sides) × height
= $\frac{1}{2}$× (7 + 13) × 8 cm²
= 80 cm²