The area of trapezium is 64 $c m^{2}$ and height is 2 cm. The parallel sides are in the ratio 3:5. Find the length of the bases.

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Solution

Let the parallel sides be $3 x$ and $5 x$
Area of trapezium $= \frac{1}{2} \times h (b a s e_{1} + b a s e_{2})$
$64 = \frac{1}{2} \times 2 \times (3 x + 5 x)$
$64 = 8 x$
$x = 8$
The parallel sides are $3 \times 8$ and $5 \times 8$
Therefore, the parallel sides are $24$ cm and $40$ cm.

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The area of trapezium is 64 cm2 and height is 2 cm. The parallel sides are in the ratio 3:5. Find the length of the bases.

Let the parallel sides be 3x and 5xArea of trapezium =12×h(base1+base2)64=12×2×(3x+5x) 64=8xx=8The parallel sides are 3×8 and 5×8Therefore, the parallel sides are 24 cm and 40 cm.

The area of trapezium is 64 $c m^{2}$ and height is 2 cm. The parallel sides are in the ratio 3:5. Find the length of the bases.

Let the parallel sides be $3 x$ and $5 x$
Area of trapezium $= \frac{1}{2} \times h (b a s e_{1} + b a s e_{2})$
$64 = \frac{1}{2} \times 2 \times (3 x + 5 x)$
$64 = 8 x$
$x = 8$
The parallel sides are $3 \times 8$ and $5 \times 8$
Therefore, the parallel sides are $24$ cm and $40$ cm.