Question

In fig. 9.11, $\square$ABCD is trapezium seg AD || seg BC
AB = 10, BP = 6, AD = 7, DC = 17.
Find BC and area of $\square$ABCD.

Answer 1

In $△$ABP, we have:

$$\begin{gathered} \mathrm{AB}{}^2={\mathrm{AP}}^2+{\mathrm{BP}}^2(\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {10}^2={\mathrm{AP}}^2+6^2 \\ \Rightarrow {\mathrm{AP}}^2=100-36=64 \\ \Rightarrow \mathrm{AP}=8\mathrm{units} \end{gathered}$$

DQ = AP = 8 units (Perpendicular distance between two parallel lines is constant)

Now, in $△$ DQC, we have:
$$\begin{gathered} {\mathrm{DC}}^2={\mathrm{DQ}}^2+{\mathrm{QC}}^2(\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {17}^2=8^2+{\mathrm{QC}}^2 \\ \Rightarrow {\mathrm{QC}}^2=289-64=225 \\ \Rightarrow \mathrm{QC}=15\mathrm{units} \end{gathered}$$

Also, PQ = AD = 7 units

Now, BC = BP + PQ + QC
⇒ BC = (6 + 7 + 15) units = 28 units
∴ Area of $\square$ABCD = $\frac{1}{2}\times 8\times (7+28)$ = $\frac{1}{2}\times 35\times 8=140\mathrm{sq}\mathrm{units}$