Question

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
T of this garden [Length of rectangle is
7m 20-(3.5 +3.5) metres].
20 m​

Answer 1

Answer:

• Perimeter of garden = 48 m
• Area of garden = 129.5 m²

Step-by-step explanation:

Diagram: Refers to attachment.

Given:

• Length of rectangle = [20 - (3.5 + 3.5) meters]

To find:

• Area
• Perimeter

⇒ Circumference of 1 semi - circle = πr

⇒ Circumference of 1 semi - circle = 3.14 × 3.5

⇒ Circumference of 1 semi - circle = 11 m (Approx)

Now, circumference of both semi - circle = 2 × 11 = 22 m

Now, we will calculate perimeter,

⇒ Perimeter of garden = AB + CD + circumference of both semi - circle

⇒ Perimeter of garden = 13 m + 13 m + 22 m

⇒ Perimeter of garden = 48 m

Now, we will calculate area of garden,

⇒ Area of garden = Area of rectangle + Area of both semi - circles.

⇒ Area of garden = [13 × 7] + [2 × 1/2 × 22/7 × 3.5 × 3.5]

⇒ Area of garden = [91 + 38.5]

⇒ Area of garden = 129.5 m²

Hence,

• Perimeter of garden = 48 m
• Area of garden = 129.5 m²

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#BAL

Answer 2

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Total area of the garden = Area of the rectangular portion + The sum of the areas of the pair of semi-circles

$l.b + 2 \times \frac{1}{2}\pi {r}^{2}$

$= (13 \times 7) {m}^{2} +$

$(2 \times \frac{1}{2} \times \frac{22}{7} \times 3.5 \times 3.5) {m}^{2}$

$= (91 + 38.5) {m}^{2} = 129.5 {m}^{2}$

Perimeter of the garden =2× length of rectangular portion + circumference of the circle

$= (2 \times 13 + 2 \times \frac{22}{7} \times 3.5)m$

$= (26 + 22)m = 48m$