Ex.11.2 Q1 Mensuration Solution - NCERT Maths Class 8

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The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are \(1\,\rm{ m}\) and \(1.2\,\rm{ m}\) and perpendicular distance between them is \(0.8\rm{ m}.\)

 Video Solution
Ex 11.2 | Question 1

Text Solution

What is Known?

Shape of the table is trapezium and dimensions of the table.

What is unknown?

Area of the table.


The table is rectangular in the middle and triangle at the end. Usually the area of the problem is sum of areas of two right angle triangle and rectangle.


Base of the triangles

\[\begin{align}(CE+FD) &= (CD-EF)\\&= 1.2\;\rm{cm} – 1\,\rm{m} \\&= 0.2\,\rm{m}\end{align}\]

Base of one triangle \(\begin{align} = {\rm{CE}} = {\rm{FD}} = \frac{{0.2}}{2} = 0.1{\rm{m}} \end{align}\)

Height of the triangle \(= AE = BF = 0.3\,\rm{m}\)

Area of the triangle \(ACE\) \(=\) Area of the triangle \(BDF\)
\[\begin{align}  &= \frac{1}{2} \times \rm{base} \times \rm{height}\\& = \frac{1}{2} \times 0.1{\rm{m}} \times 0.8{\rm{m}} = 0.04{{\rm{m}}^2}\end{align}\]

Area of the rectangle \(ABEF\)
\[\begin{align} &= 1{\rm{m}} \times 0.8{\rm{m}}\\ &= 0.8\,\rm{m^2}\end{align}\]

Area of the table \(=\) Area of the triangle \(ACE\) \(+\) Area of the rectangle \( ABEF\) \(+\) Area of the triangle \( BFD\)

\[\begin{align}&= 0.04\,{{\rm{m}}^2} \times 0.8\,{{\rm{m}}^2} + 0.04\,{{\rm{m}}^2}\\&= 0.88\,{{\rm{m}}^2}\end{align}\]

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