2019 WAEC Mathematics Theory In the diagram, PQRS is a quadrilateral, &lt; PQR = &lt; PRS = 90\(^o\), |PQ|...

In the diagram, PQRS is a quadrilateral, < PQR = < PRS = 90\(^o\), |PQ| =3cm, |QR| = 4cm and |PS| = 13 cm. Find the area of the quadrilateral.

In \(\triangle\)PQR

|PR|\(^2\) = |PQ|\(^2\) + |QR|\(^2\)

= 3\(^2\) + 4\(^2\)

|PR|\(^2\) = 9 + 16

|PR| = \(\sqrt{25}\) = 5cm

In \(\triangle\)PRS

|PS|\(^2\) = |PR|\(^2\) + |RS|\(^2\)

13\(^2\) = 5\(^2\) + |RS|\(^2\)

169 =25 + |RS|\(^2\)

|RS|\(^2\) = 169 - 25

= 144

|RS| = \(\sqrt{144}\) = 12cm

Area of quadrilateral PQRS = Area of \(\triangle\)PQR + Area of \(\triangle\)PQR + Area of \(\triangle\)PRS

= (\(\frac{1}{2} \times 3 \times 4\)) + (\(\frac{1}{2} \times 12 \times 5\))

= 6cm\(^2\) + 30cm\(^2\)

= 36cm\(^2\)

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