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Ex.11.2 Q3 Mensuration Solution - NCERT Maths Class 8

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Question

Length of the fence of a trapezium shaped field \(ABCD\) is \(120 \,\rm{m.}\) If \(BC = 48\,\rm{ m},\; CD = 17\,\rm{m} \) and \(AD = 40\,\rm{ m},\) find the area of this field. Side \(AB\) is perpendicular to the parallel sides \(AD\) and \(BC.\)

 Video Solution
Mensuration
Ex 11.2 | Question 3

Text Solution

What is Known?

Length of the trapezium shaped field and side \(AB,\, DC,\) and \(BC\)

What is unknown?

Area of the field and side \(AB.\)

Reasoning:

Visually area of given figure (trapezium) is sum of the area of two triangles.

Steps:

Length of the hence of a trapezium shaped field \(ABCD\)

\[\begin{align}&\!=\! AB + BC + CD + AD\\120\,{\rm{m }} &\!= \!AB \!+\! 48\,{\rm{m}} \!+\! 17\,{\rm{m }} \!+\! 40\,{\rm{m}}\\120\,{\rm{m}}& \!= \!AB \!+\! 105\,{\rm{m}}\\AB{\rm{ }} &= \!{\rm{ }}120\,{\rm{m }}\!-\!105\,{\rm{m}}\\ &=\! {\rm{ }}15\,{\rm{m}}\end{align}\]

Area of the field \(ABCD\)

\[\begin{align}&= \! \frac{1}{2} \! \times\! \! \begin{bmatrix}\text{Length of the}\\\text{ parallel side}\end{bmatrix}\!\! \times \! \begin{bmatrix}\text{Distance between } \\\text{two parallel sides}\end{bmatrix}\\& = \! \frac{1}{2} \! \times \! (AD \! + \! BC) \! \times \! AB\\&= \! \frac{1}{2} \! \times \! (40 \! + \! 48) \! \times \! 15\,{\rm{m}}\\& = \! \frac{1}{2} \! \times \! (88\,{\rm{m}}) \! \times \! 15\,{\rm{m}}\\& = \! 44\,{\rm{m}} \! \times \! 15\,{\rm{m}}\\&= \! 660\,{{\rm{m}}^2}\end{align}\]

Thus, Area of the field \(ABCD\) is \(660\,{{\rm{m}}^2}\).

  
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