Question

Construct an isosceles trapezium ABCD in which $\overline{AB}$ is parallel to $\overline{DC}$ ,AB = 11 cm, DC = 7 cm, AD = BC = 6 cm and calculate its area.

Answer 1

Given :
$\overline{AB}$ is parallel to $\overline{DC}$ ,AB = 11 cm, DC = 7 cm, AD = BC = 6 cm

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

Step 2 : Draw a line segment AB = 11 cm.

Step 3 : Mark E on $\overline{AB}$ such that $\overline{AE}$ = 7 cm ( since DC = 7 cm)

Step 4 : With E and B as centres and (AD = EC = 6 cm) radius 6 cm draw two arcs. Let them cut at C.

Step 5 : Join $\overline{BC}$ and $\overline{EC}$ .

Step 6 : With C and A as centres draw two arcs of radii 7 cm and 6 cm respectively and let them cut at D.

Step 7 : Join $\overline{AD}$ and $\overline{CD}$ . ABCD is the required isosceles trapezium.

Step 8 : From D draw $\overline{DF}\perp \overline{AB}$ and measure the length of DF. DF = h = 5.6 cm. AB = a = 11 cm and CD = b = 7 cm.

Calculation of area:
In the isosceles trapezium ABCD, a = 11 cm, b = 7 cm and h = 5.6 cm.

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

=$\frac{1}{2}\left(5.6\right)(11+7)$

=$\frac{1}{2}\times 5.6\times 18$

= 50.4 $c{m}^2$