Question

The parallel sides of an isosceles trapezium are 10cm and 6cm respectively and one of the non-parallel sides is 8cm. If the angle between the non-parallel and parallel side is ${110}^{\circ }$, then the area of the trapezium is ___ $[sin{70}^{\circ }=0.93]$

• A. $55c{m}^2$
• B. $70c{m}^2$
• C. $50c{m}^2$
• D. $60.08c{m}^2$

Answer 1

The correct option is D

$60.08c{m}^2$

Let ABCD be the given trapezium having AB = 10 cm, CD = 6 cm, AD = 8 cm, $\angle D={110}^{\circ }$

Draw DE perpendicular to AB

$\angle A=(180-{110}^{\circ })={70}^{\circ },asDC||AB$

In right angled triangle ADE

$sin{70}^{\circ }=\frac{DE}{AD}$

$DE=AD.sin{70}^{\circ }=8\times 0.93=7.44cm$

Area of trapizium=$\frac{1}{2}\times (AB+CD)\times DE$

=$\frac{1}{2}\times (10+6)\times 7.44=59.52c{m}^2$