A sports field consists of a central rectangular section of length 100\text{ m} bounded on… — JAMB UTME

A sports field consists of a central rectangular section of length $100\text{ m}$ bounded on two opposite ends by semicircular regions, each having a radius of $35\text{ m}$ . What is the total area of the sports field? (Take $\pi = \frac{22}{7}$ )

A
$7,350\text{ m}^2$
B
$8,925\text{ m}^2$
$10,850\text{ m}^2$ Answer
D
$14,700\text{ m}^2$

Answer

The total area of the sports field is

10,850\text{ m}^2

The sports field is composed of a central rectangle of dimensions

100\text{ m}

70\text{ m}

(since width is equal to the diameter

2 \times 35\text{ m} = 70\text{ m}

) and two semicircular ends of radius

35\text{ m}

. The rectangular area is

100 \times 70 = 7,000\text{ m}^2

. The two semicircles join to make one full circle with an area of

\frac{22}{7} \times 35^2 = 3,850\text{ m}^2

. Adding both parts gives

7,000 + 3,850 = 10,850\text{ m}^2

Step-by-Step Solution

Calculate the width of the rectangular region.

Width =

2 \times r = 2 \times 35\text{ m} = 70\text{ m}

The diameter of the semicircular ends forms the width of the central rectangle.

Calculate the area of the rectangular region.

Area of rectangle =

\text{length} \times \text{width} = 100\text{ m} \times 70\text{ m} = 7,000\text{ m}^2

Formula for the area of a rectangle is

\text{length} \times \text{width}

Calculate the combined area of the two semicircular ends.

Combined area =

\pi r^2 = \frac{22}{7} \times 35 \times 35 = 22 \times 5 \times 35 = 3,850\text{ m}^2

Two identical semicircles of radius

r

combine to form one full circle of radius

r

Add the rectangular area and the combined circular area.

Total Area =

7,000\text{ m}^2 + 3,850\text{ m}^2 = 10,850\text{ m}^2

The total area is the sum of the composite plane shapes.

Key Concept

Area of Composite Plane Figures

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