Name: Area of a Quadrilateral and Rhombus
Uploaded: 2019-07-19
Duration: 8 min 51 s

Question

A trapezium with its parallel sides in the ratio

11 : 3

is cut off from a rectangle whose sides measure

98 cm

and

12 cm

. The area of the trapezium is

\frac{3}{7}

of the area of the rectangle. Find the lengths of the parallel sides of the trapezium, when its height is equal to the smaller side of the rectangle.

Accepted Answer

Given, in a trapezium, the parallel sides are in the ratio $11 : 3$ .

Let the parallel sides of the trapezium be $11 x cm, 3 x cm$ .

Also, height of the trapezium is equal to the smaller side of the rectangle.

i.e. height of the trapezium $= 12 cm$ (Given)

Let us consider the figure:

Also given, area of a trapezium $= \frac{3}{7} \times$ Area of a rectangle

And measurement of rectangle are $98 cm$ and $12 cm$ .

We need to find the length of the parallel sides.

From given conditions:

$\Rightarrow \frac{1}{2} \times$ (Sum of the parallel sides) $\times$ (Perpendicular distance between the parallel sides) $= \frac{3}{7} \times$ Area of a rectangle

$\Rightarrow \frac{1}{2} \times (11 x + 3 x) \times 12 = \frac{3}{7} \times (98 \times 12)$

$\Rightarrow \frac{1}{2} \times 14 x \times 12 = \frac{3}{7} \times 98 \times 12$

$\Rightarrow x = 6$

Then, length of parallel sides is

$3 x = 3 \times 6 = 18$

And $11 x = 11 \times 6 = 66$

Hence, lengths of the parallel sides of a trapezium are $18 cm$ and $66 cm$ .

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