Question

the
The shape of a garden is rectangular in the middle and semi circular
at the ends as shown in the diagram. Find the area and the perimeter
Т.
of this garden [Length of rectangle is
7m 20-(3.5 +3.59 metres).
1
DI
20 m​

Answer 1

Answer:

$Area = 129.5m^2$

$Perimeter = 48m$

Step-by-step explanation:

Given

See attachment

Required

Determine the area and the perimeter of the garden

Calculating Area

First, we calculate the $area\ of\ the\ rectangle$

$A_1 = L * B$

Where:

$L = 20-(3.5 +3.5)$

$B = 7$

So:

$A_1 = (20 - (3.5 + 3.5)) * 7$

$A_1 = (20 - 7) * 7$

$A_1 = 13 * 7$

$A_1 = 91$

Next, we calculate the area of the two semi-circles.

Two semi-circles = One Circle

So:

$A_2 = \pi r^2$

Where

$r = \frac{7}{2}$

$A_2 = \frac{22}{7} * (\frac{7}{2})^2$

$A_2 = \frac{22}{7} * \frac{49}{4}$

$A_2 = \frac{22}{1} * \frac{7}{4}$

$A_2 = \frac{22*7}{4}$

$A_2 = \frac{154}{4}$

$A_2 = 38.5$

Area of the garden is

$Area = A_1 + A_2$

$Area = 91 + 38.5$

$Area = 129.5m^2$

Calculating Perimeter

First, we calculate the perimeter of the rectangle

But in this case, we only consider the length because the widths have been covered by the semicircles

$P_1 = 2 * L$

Where:

$L = 20-(3.5 +3.5)$

So:

$P_1 =2 * (20-(3.5 +3.5))$

$P_1 =2 * (20-7)$

$P_1 =2 * 13$

$P_1 =26$

Next, we calculate the perimeter of the two semi-circles.

Two semi-circles = One Circle

So:

$P_2 = 2\pi r$

Where

$r = \frac{7}{2}$

$P_2 = 2 * \frac{22}{7} * \frac{7}{2}$

$P_2 = \frac{2 * 22 * 7}{7 * 2}$

$P_2 = \frac{308}{14}$

$P_2 = 22$

Perimeter of the garden is

$Perimeter = P_1 + P_2$

$Perimeter = 26 + 22$

$Perimeter = 48m$