Ex.11.1 Q5 Mensuration Solutions - NCERT Maths Class 8

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An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression \(c = 2{\rm{\pi }}r\) , where \(r\) is the radius of the circle.

 Video Solution
Ex 11.1 | Question 5

Text Solution

What is Known?

Shape and dimensions of various food pieces.

What is unknown?

Perimeter of different shaped food pieces.


Hence, we will calculate perimeter of different shaped food piece. Larger the perimeter, longer will be the path to take round.


(a) Radius of the semicircle part 

\[\begin{align} = \left( {\frac{{2.8}}{2}} \right)\,{\rm{cm}} = 1.4\,{\rm{cm}}\end{align}\]

Perimeter of the circle \( = 2{{ \pi }}r\)

\(\therefore \) Perimeter of the semicircle = \(\pi r\)

The perimeter of the food piece

\[\begin{align}&= 2.8\,{\rm{cm}} + \pi r\\ &= 2.8\,{\rm{cm}} + \left( {\frac{{22}}{{\not7}} \times{{\not\!\!\!\!{1.4}}^{0.2}}} \right)\,{\rm{cm}}\\ &= {\rm{ }}2.8\,{\rm{cm }} + {\rm{ }}4.4\,{\rm{cm}}\\ &= 7.2\,{\rm{cm}}\end{align}\]

(b). Radius of semicircle part

\[\begin{align} = \frac{{2.8}}{2} = 1.4\,\,{\rm{cm}}\end{align}\]

Perimeter of the food piece

\[\begin{align}&= 1.5\,{\rm{cm}} + 2.8\,{\rm{cm}} + 1.5\,{\rm{cm}} + \pi r\\&= 5.8\,{\rm{cm}} + \left( {\frac{{22}}{{\not7}} \times \not\!\!\!\!{{1.4}^{0.2}}} \right){\rm{cm}}\\&= 5.8\,\,{\rm{cm}} + 4.4\,\,c{\rm{m}}\\& = 10.2\,\,{\rm{cm}}\end{align}\]

(c). Radius of the food piece

\[\begin{align} = \frac{{2.8}}{2} = 1.4\,\,{\rm{cm}}\end{align}\]

Perimeter of the food piece

\[\begin{align}&= 2{\rm{cm}} + \pi r + 2{\rm{cm}}\\& = 4\,{\rm{cm}} + \left( {\frac{{22}}{{\not7}} \times \not\!\!\!\!{{1.4}^{0.2}}} \right){\rm{cm}}\\& = 4\,\,{\rm{cm}} + 4.4\,\,c{\rm{m}}\\& = {\rm{8}}{\rm{.4}}\,\,{\rm{cm}}\end{align}\]

Thus, the ant will have to take a longer round for food piece (b) because the perimeter of the figure given in (b) is the greatest among all.

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