An isosceles triangle has a perimeter of 36\text{ cm} and a base of length 16\text{ cm}. C… — JAMB UTME

An isosceles triangle has a perimeter of $36\text{ cm}$ and a base of length $16\text{ cm}$ . Calculate the area of the triangle in $\text{cm}^2$ .

Answer: 48 cm^2

The area of the isosceles triangle is

48\text{ cm}^2

The two equal sides of the isosceles triangle measure

\frac{36 - 16}{2} = 10\text{ cm}

each. An altitude dropped perpendicularly to the base bisects the

16\text{ cm}

base into two

8\text{ cm}

segments. By the Pythagorean theorem, the perpendicular height is

h = \sqrt{10^2 - 8^2} = 6\text{ cm}

. Therefore, the area is

\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2

Determine the length of the two equal sides

Each equal side is

10\text{ cm}

Subtract the base length from the total perimeter (

36 - 16 = 20\text{ cm}

) and divide by

2

Calculate the perpendicular height (altitude) to the base

Height

h = 6\text{ cm}

The perpendicular altitude bisects the base into two

8\text{ cm}

segments, creating right-angled triangles with hypotenuse

10\text{ cm}

. Use Pythagoras' theorem:

h = \sqrt{10^2 - 8^2} = 6\text{ cm}

Calculate the area of the triangle

Area =

48\text{ cm}^2

Multiply half the base by the perpendicular height:

\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2

Perimeter and Area of Isosceles Triangles using Pythagorean Theorem

Estimated Time:1m 30s

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