Question:

The parallel sides of a trapezium are in the ratio 3 : 4 and the perpendicular distance between them is 12 cm. If the area of the trapezium is 630 cm2, then its shorter of the parallel sides is

(a) 45 cm

(b) 42 cm

(c) 60 cm

(d) 36 cm

Solution:

(a) 45 cm

Let the length of the parallel sides be $3 \mathrm{x} \mathrm{cm}$ and $4 \mathrm{x} \mathrm{cm}$, respectively.

Then, area of the trapezium $=\left\{\frac{1}{2} \times(3 x+4 x) \times 12\right\} \mathrm{cm}^{2}$

$=\left(\frac{1}{2} \times 7 x \times 12\right) \mathrm{cm}^{2}$

$=42 x \mathrm{~cm}^{2}$

But it is given that the area of the trapezium is $630 \mathrm{~cm}^{2}$.

$\therefore 42 x=630$

$\Rightarrow x=\frac{630}{42}$

$\Rightarrow x=15 \mathrm{~cm}$

$L$ ength of the parallel sides $=(3 \times 15) \mathrm{cm}=45 \mathrm{~cm}$

$(4 \times 15) \mathrm{cm}=60 \mathrm{~cm}$

Hence, the shorter of the parallel sides is $45 \mathrm{~cm}$.