Mathematics: Find the perimeter of abc with vertices a (1,1), b (7,1), and c (1,9)

Find the perimeter of abc with vertices a (1,1),

hchvvhhv

12.03.2022

24 unit

Step-by-step explanation:

Given,

The vertices of the triangle ABC are,

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

By the distance formula,

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

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

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

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

0,0(0 оценок)

emilianacassandra

26.04.2023

The answer is 2,

If the find the area of one side on the end (3.5 in ² )

You take 3.5 and dived 70 by that and get 20

70/3.5 = 10x

10x = 20

x = 2

0,0(0 оценок)