3.The shape of a garden is rectangular in the middle and semicircular at the ends as shown in the diagram. Find the area and the perimeter of this garden. [Length of rectangle is 20 – (3.5 + 3.5) metres]

We have,Total area of the garden = Area of the rectangular portion + The sum of the areas of the pair of semi-circles
We have, Length of the rectangle = \(20 – (3.5 + 3.5) = 20 – 7 = 13 m\)
Therefore, area of the rectangle =\( l \times b = 13 \times 7 = 91 m^2\)

Area of two circular ends = \(2\times \frac { 1 }{ 2 }\pi r^2\)

\(= \pi r^2\)

\(= \frac { 22 }{ 7 } \times \frac { 7 }{ 2 } \times \frac { 7 }{ 2 }\)

\(=\frac { 77 }{ 2 } m^2\)

\(= 38.5 m^2\)

Total area = Area of the rectangle + Area of two ends = \(91 m^2 + 38.5 m^2 = 129.5 m^2\)

Total perimeter = Perimeter of the rectangular portion + Perimeter of two circular ends

\(= 2 (l + b) + 2 \times (\pi r) – 2(2r)\)

\(= 2 (13 + 7) + 2(\frac { 22 }{ 7 } \times \frac { 7 }{ 2 }) – 4 \times \frac { 7 }{ 2 }\)

\(= 2 \times 20 + 22 – 14\)

\(= 40 + 22 – 14\)

\(= 48 m\)

So the total perimeter of the figure =48m