Question

8. The shape of a park is a rectangle bounded by semi-circles at
the ends, each of radius 17.5 m, as shown in the adjoining
figure. Find the area and the perimeter of the park.
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Answer 1

Answer:

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Answer 2

★ AnswEr :-

• Perimeter = 190 m
• Area = 631.25 m^2

★ GiVen :-

• The shape of a park is a rectangle bounded by semi-circles at the ends, each of radius 17.5 m.
• The length and the breadth of the rectangle is 40 m and 35 m respectively.

★ To Find :-

• The area and the perimeter of the park.

★ Solution :-

We know, the formula of the perimeter of a circle is 2πr unit.

So, the perimeter of the semicircle would be,

$\sf \: (\pi \: r) = \dfrac{22}{7} \times 17.5$

$\sf \implies \: perimeter \: of \: circle = 55 \: m$

Therefore, the perimeter of the whole park would be,

$\sf \implies \: (55 + 40 + 55 + 40) \: m$

$\sf \implies perimeter = 190 \: m$

The area of the triangle is,

$\sf \implies \: 2(40 + 35) m^{2}$

$\sf \implies \: (2 \times 75) m^{2}$

$\sf \implies \: 150 m^{2}$

The area of two semicircle would be,

$\sf \: (\dfrac{1}{2} \pi \: r^{2} ) = \dfrac{1}{2} \times \dfrac{22}{7} \times ({17.5})^{2}$

$\sf \implies \dfrac{1}{2} \times \dfrac{22}{7} \times 306.25$

$\sf \implies \dfrac{3368.75}{7}$

$\sf \implies area = 481.25 \: m^{2}$

The whole area of the park is,(481.25 + 150) m^2 = 631.25 m^2.