Question

Construct a trapezium PQRS in which $\overline{PQ}$ is parallel to $\overline{SR}$ , PQ = 8 cm $\angle$ PQR = ${70}^o$, QR = 6 cm and PS = 6 cm. Calculate its area.

Answer 1

Given:
$\overline{PQ}$ is parallel to $\overline{SR}$, PQ = 8 cm, $\angle$ PQR = ${70}^o$ QR = 6 cm and PS = 6 cm.

Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.

Step 2 : Draw a line segment PQ = 8 cm.

Step 3 : At Q on $\overline{PQ}$ make $\angle$ PQX whose measure is ${70}^o$ .

Step 4 : With Q as centre and 6 cm as radius draw an arc. This cuts $\overline{QX}$ at R.

Step 5 : Draw $\overline{RY}$ parallel to $\overline{OP}$ .

Step 6 : With Q as centre and radius 6 cm draw an arc cutting $\overline{RY}$ at S.

Step 7 : Join $\overline{PS}$ . PQRS is the required trapezium.

Step 8 : From S draw $\overline{ST}\perp \overline{PQ}$ and measure the length of ST. ST = h = 5.6 cm, RS = b = 6 cm, PQ = a = 8 cm.

Calculation of area:
In the trapezium PQRS, a = 8 cm, b = 3.9 cm and h = 5.6 cm.

Area of the trapezium ABCD = $\frac{1}{2}$ h(a+b)

=$\frac{1}{2}\left(5.6\right)(8+3.9)$

=$\frac{1}{2}\times$ $5.6\times 11.9$

=$33.32c{m}^2$