Question

Construct a trapezium ABCD in which $\overline{AB}$ is parallel to $\overline{DC}$ , AB = 7 cm,BC = 6 cm, $\angle$ BAD = ${80}^o$ and $\angle$ ABC = ${70}^o$ and calculate its area.

Answer 1

Given:
$\overline{AB}$ is parallel to $\overline{DC}$, BC = 6 cm, $\angle$ BAD = ${80}^o$ and $\angle$ ABC = ${70}^o$.
Steps for construction
Step 1 : Draw a rough diagram and mark the given measurements.
Step 2 : Draw a line segment AB = 7 cm.
Step 3 : On $\overline{AB}$ at A make $\angle$ BAX measuring ${80}^o$ .
Step 4 : On $\overline{AB}$ at B make $\angle$ ABY measuring ${70}^o$ .
Step 5 : With B as centre and radius 6 cm draw an arc cutting $\overline{BY}$ at C.
Step 6 : Draw $\overline{CZ}$ parallel to $\overline{AB}$ . This cuts $\overline{AX}$ at D. ABCD is the required trapezium.
Step 8 : From C draw $\overline{CE}\perp \overline{AB}$ and measure the length of CE. CE = h = 5.6 cm, CD = b = 4 cm Also AB = a = 7 cm.
Calculation of area:
In the trapezium PQRS, a = 7 cm, b = 4 cm and h = 5.6 cm.
Area of the trapezium ABCD = $\frac{1}{2}$ h(a+b)
=$\frac{1}{2}$ (5.6)(7+4)
=$\frac{1}{2}\times$ 5.6 $\times$ 11
=30.8 $c{m}^2$