The parallel sides of an isosceles trapezium are 25 cm and 11 cm, and the perpendicular distance between the parallel sides is 13 cm. Find the area of the trapezium.

150 cm²

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180 cm²

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234 cm²

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Solution

The correct option is C
234 cm²

When $h$ is the perpendicular distance between the parallel sides and the parallel sides are of length $a$ and $b$ .

Area of trapezium $= h \times$ $\frac{(a + b)}{2}$ sq. units

Here, $a = 25 c m$ , $b = 11 c m$ and $h = 13 c m$

Area of Trapezium $= \frac{1}{2} \times (25 + 11) \times 13 = 234 c m^{2}$

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The parallel sides of an isosceles trapezium are 25 cm and 11 cm, and the perpendicular distance between the parallel sides is 13 cm. Find the area of the trapezium.

The correct option is C 234 cm2 When h is the perpendicular distance between the parallel sides and the parallel sides are of length a and b. Area of trapezium =h× (a+b)2sq. units Here, a=25 cm, b=11 cm and h=13 cm Area of Trapezium=12×(25+11)×13=234 cm2