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The shape of a garden is rectangular in the middle and semicircular at each end as shown in the figure. Find the area and perimeter of the garden.

The shape of a garden is rectangular in the middle and semicircular at each end as shown in the figure. Find the area and perimeter of the garden. Area of a Trapezium and a Polygon, Concise Mathematics Solutions ICSE Class 8.

Area Trapezium Polygon

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Answer

Given:

Total length of garden = 20 m

Radius of the circle = 72\dfrac{7}{2} m

Length of rectangle = total length - 2 x radius of the circle

= 20 - 2 x 72\dfrac{7}{2} m

= 20 - 7 m

= 13 m

Perimeter of the figure = l + l + 2 x 12\dfrac{1}{2} x circumference of semicircle

As we know that circumference of the circle = 2πr

= 13 + 13 + 2 x 12\dfrac{1}{2} x 2πr

= 26 + 2 x 227×72\dfrac{22}{7} \times \dfrac{7}{2}

= 26 + 2×227×72\cancel{2} \times \dfrac{22}{\cancel{7}} \times \dfrac{\cancel{7}}{\cancel{2}}

= 26 + 22

= 48 m

Area of the figure = Area of rectangle + 2 x Area of semicircle

As we know that area of rectangle = length x breadth

And, area of the circle = πr2

So, area

=13×7+227×(72)2=91+227×494=91+22×497×4=91+107828=91+38.5=129.5 m2= 13 \times 7 + \dfrac{22}{7} \times \big( \dfrac{7}{2} \big)^2 \\[1em] = 91 + \dfrac{22}{7} \times \dfrac{49}{4} \\[1em] = 91 + \dfrac{22 \times 49}{7 \times 4} \\[1em] = 91 + \dfrac{1078}{28} \\[1em] = 91 + 38.5 \\[1em] = 129.5 \text{ m}^2

Hence, the area of the garden is 129.5 m2 and the perimeter is 48 m.

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