Question

Construct a trapezium ABCD in which $\overline{AB}$ is parallel to $\overline{DC}$ , AB = 7 cm, BC = 5 cm, CD = 4 cm and AD = 5 cm and calculate its area.

Answer 1

Given:
$\overline{AB}$ is parallel to $\overline{DC}$, BC = 5 cm, CD = 4 cm and AD = 5 cm.

Steps for construction:
Step 1 : Draw a rough diagram and mark the given measurements. Draw $\overline{CE}\parallel \overline{DA}$ . Now AECD is a parallelogram. $\therefore$ EC = 5 cm, AE = DC = 4 cm, EB = 3cm.

Step 2 : Draw a line segment AB = 7 cm.

Step 3 : Mark E on $\overline{AB}$ such that AE = 4 cm . [ $\because$ DC = 4 cm ]

Step 4 : With B and E as centres draw two arcs of radius 5 cm and let them cut at C.

Step 5 : With B as centre and radius 6 cm draw an arc cutting $\overline{BY}$ at C.

Step 6 : With C and A as centres and with 4 cm and 5 cm as radii draw two arcs. Let them cut at D.

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

Step 8 : From D draw $\overline{DF}\perp \overline{AB}$ and measure the length of DF. DF = h = 4.8 cm, AB = a = 7 cm, CD = b = 4 cm.

Calculation of area:In the trapezium ABCD, a = 7 cm, b = 4 cm and h = 4.8 cm.

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

=$\frac{1}{2}$ (4.8)(7+4)

=$\frac{1}{2}\times$ 4.8 $\times$ 11

= $\times$ 2.4 $\times$ 11
=26.4 $c{m}^2$