Question

Find the area of the given trapezium.

• A. 148 $m^2$
• B. 250 $m^2$
• C. 152 $m^2$
• D. 130 $m^2$

Answer 1

The correct option is C 152 $m^2$

Given :
Trapezium COLD in which OL = 25 cm, CD = 13 cm, CO = DL = 10 cm.
Construction: Through D, draw DR ∥ CO, meeting OL at R.

Also, draw DG ⊥ OL.
Now, RL = (OL - OR) = OL - CD

= (25 - 13) m
= 12 m

DR = CO = 10 m; OR = CD = 13 m.

Now, in ∆DRL, we have DR = DL = 10 m.

$\therefore$ ∆DRL is an isosceles triangle.

Also, DG ⊥ RL

So, G is the midpoint of RL.

$\therefore$ RG = $\frac{1}{2}\times RL$

=$\frac{1}{2}\times 12=6m.$

Thus, in right-angled ∆DGR, we have DR = 10 m, RG = 6 m.

By Pythagoras’ theorem, we have

DG = $\sqrt{D{R}^2-R{G}^2}$

= $\sqrt{{10}^2-6^2}$

= $\sqrt{64}$

= $\sqrt{(8\times 8)}$

= 8 m.

Thus, the distance between the parallel sides is 8 cm.

Area of trapezium COLD = $\frac{1}{2}\times$ (sum of parallel sides) (height)

= $\frac{1}{2}\times$ (25 + 13) × 8 m

= $\frac{1}{2}\times 38m\times 8m$

= 152 m²