Ex.11.1 Q5 Mensuration Solutions - NCERT Maths Class 8

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Question

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.

Text Solution

What is Known?

Shape and dimensions of various food pieces.

What is unknown?

Perimeter of different shaped food pieces.

Reasoning:

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

Steps:

(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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