A metal plate is initially in the form of a rectangle ABCD measuring 14\text{ cm} by 10\te… — JAMB UTME

A metal plate is initially in the form of a rectangle $ABCD$ measuring $14\text{ cm}$ by $10\text{ cm}$ , where $AB = 14\text{ cm}$ . A semicircular piece with diameter $AB$ is cut out from side $AB$ . On the opposite side $CD$ , an isosceles triangular plate with base $CD$ and height $24\text{ cm}$ is attached externally. Taking $\pi = \frac{22}{7}$ , what is the total area of the resulting plate in $\text{cm}^2$ ?

Answer: 231 cm^2

Answer

The total area of the resulting plate is

231\text{ cm}^2

The net area of the plate is found by starting with the area of the rectangle (

140\text{ cm}^2

), subtracting the area of the semicircular cutout (

77\text{ cm}^2

), and adding the area of the attached isosceles triangle (

168\text{ cm}^2

), resulting in

140 - 77 + 168 = 231\text{ cm}^2

Step-by-Step Solution

Calculate the area of the original rectangle ABCD

140 cm²

The area of a rectangle is calculated as

\text{length} \times \text{width} = 14\text{ cm} \times 10\text{ cm} = 140\text{ cm}^2

Calculate the area of the removed semicircular section

77 cm²

The radius of the semicircle is

r = \frac{14}{2} = 7\text{ cm}

. The area of a semicircle is

\frac{1}{2}\pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2

Calculate the area of the attached triangular section

168 cm²

The area of a triangle is

\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14\text{ cm} \times 24\text{ cm} = 168\text{ cm}^2

Combine the area components to determine the final net area

231 cm²

Subtract the removed semicircular area from the rectangle's area and add the attached triangle's area:

140 - 77 + 168 = 231\text{ cm}^2

Key Concept

Perimeter and Area of Composite Plane Figures

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