# Ex.11.2 Q9 Mensuration Solution - NCERT Maths Class 8

## Question

Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

## Text Solution

**What is Known?**

Top surface of a raised platform is octagonal shaped.

**What is unknown?**

Area of the octagonal surface

**Reasoning:**

Visually, the area of the octagonal surface will be the sum of the area of two trapezium and area of rectangular**.**

**Steps:**

Area of octagon \( ABCDEFGH =\) Area of trapezium \(ABCH\, +\) Area of rectangular \(HCDG \;+\) Area of trapezium \(EFGD\)

Side of the regular octagon \(= 5\,\rm{cm}\)

Area of trapezium \(ABCH\, =\,\) Area of trapezium \(EFGD\)

Area of trapezium \(ABCH\)

\[\begin{align} &= \frac{1}{2} \times (AB + CH) \times AI\\&= \frac{1}{2} \times (5\,\rm{m} + 11\,\rm{m}) \times 4\,\rm{m}\\&= \frac{1}{2} \times 16\,\rm{m} \times 4\,\rm{m}\\&= 32\,\rm{m^2}\end{align}\]

\(\therefore \) Area of trapezium \(ABCH\,=\,\) Area of trapezium \( EFGD= 32 \rm{m^2}\)

Area of rectangle \(HCDG\)

\[\begin{align} &= HC \times CD \\&= 11{\rm{m}} \times 5{\rm{m}}\\& = 55{{\rm{m}}^2}\end{align}\]

Area of \(ABCDEFGH\) \(=\) Area of trapezium \(ABCH\) \(+\) Area of rectangle \(HCDG\) \(+\) Area of trapezium \(EFGD\)

\[\begin{align}& = 32\,{{\rm{m}}^2} + 55\,{{\rm{m}}^2} + 32\,{{\rm{m}}^2}\\& = 119\,\rm{m^2}\end{align}\]

Thus, the area of the octagonal surface is \(119{{\rm{m}}^2}\)

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