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Find the perimeter of ABC with vertices A (1,1), B (7,1), and C (1,9)

asked
User Znik
by
7.9k points

1 Answer

2 votes

Answer:

24 unit

Explanation:

Given,

The vertices of the triangle ABC are,

A (1,1), B (7,1), and C (1,9),

By the distance formula,


AB=√((7-1)^2+(1-1)^2)=√(6^2)=6\text{ unit}


BC=√((1-7)^2+(9-1)^2)=√(6^2+8^2)=√(36+64)=√(100)=10\text{ unit}


CA=√((1-1)^2+(1-9)^2)=√(8^2)=8\text{ unit}

Thus, the perimeter of the triangle ABC = AB + BC + CA = 6 + 10 + 8 = 24 unit

answered
User Eyoeldefare
by
7.8k points

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