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JAMB Mathematics 2024 Paper

Question : 21

Total: 50

Solution:

Concept:

The perimeter of a triangle is the sum of the distances between its three pairs of vertices.

Explanation:

Let the vertices be

A(3,2)

B(4,5)

, and

C(6,2)

Use the distance formula:

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Distance

AC

from

(3,2)

(6,2)

AC = \sqrt{(6-3)^2 + (2-2)^2} = \sqrt{3^2 + 0^2} = 3

Distance

BC

from

(4,5)

(6,2)

BC = \sqrt{(6-4)^2 + (2-5)^2} = \sqrt{2^2 + (-3)^2} = \sqrt{4+9} = \sqrt{13}

Distance

AB

from

(3,2)

(4,5)

AB = \sqrt{(4-3)^2 + (5-2)^2} = \sqrt{1^2 + 3^2} = \sqrt{1+9} = \sqrt{10}

Perimeter

= AC + BC + AB = 3 + \sqrt{13} + \sqrt{10}

Answer:

3 + \sqrt{10} + \sqrt{13}

Correct option: C.

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