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Compute the area of trapezium $PQRS$ in the figure.
"
Given:
Trapezium $PQRS$
To do:
We have to find the area of trapezium $PQRS$.
Solution:
In $\triangle TQR$,
$\angle RTQ = 90^o$
This implies,
$QR^2 = TQ^2 + RT^2$
$(17)^2 = 8^2 + RT^2$
$289 = 64 + RT^2$
$RT^2 = 289 - 64$
$= 225$
$= 15^2$
$\Rightarrow RT = 15\ cm$
$PQ = 8 + 8$
$= 16\ cm$
Area of trapezium $PQRS=\frac{1}{2}$ Sum of parallel sides $\times$ Height
$=\frac{1}{2}(16+8) \times 15 \mathrm{~cm}^{2}$
$=\frac{1}{2} \times 24 \times 15$
$=180 \mathrm{~cm}^{2}$
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