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

$=\frac{1}{2}(16+8) \times 15 \mathrm{~cm}^{2}$

$=\frac{1}{2} \times 24 \times 15$

$=180 \mathrm{~cm}^{2}$

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

35 Views

Kickstart Your Career

Get certified by completing the course

Get Started