Question

Find the perimeter of triangle ABC with vertices (2,4) (8,12) and (24,0)

Answer 1

The perimeter of triangle ABC is 30 + 10sqrt(5)

Step-by-step explanation:

To find the perimeter of triangle ABC, we need to find the lengths of its three sides.

Using the distance formula, we can calculate the lengths of the sides:

Side AB = sqrt((8-2)^2 + (12-4)^2) = sqrt(6^2 + 8^2) = sqrt(36 + 64) = sqrt(100) = 10

Side AC = sqrt((24-2)^2 + (0-4)^2) = sqrt(22^2 + 4^2) = sqrt(484 + 16) = sqrt(500) = 10sqrt(5)

Side BC = sqrt((24-8)^2 + (0-12)^2) = sqrt(16^2 + 12^2) = sqrt(256 + 144) = sqrt(400) = 20

Therefore, the perimeter of triangle ABC is AB + AC + BC = 10 + 10sqrt(5) + 20 = 30 + 10sqrt(5).

Answer 2

Let
A(2,4)
B(8,12)
C (24,0)

we know that
[perimeter of triangle]=AB+BC+AC

step 1
find the distance AB
d=√[(12-4)²+(8-2)²]-----> dAB=√100------> dAB=10 units

step 2
find the distance BC
d=√[(0-12)²+(24-8)²]-----> dBC=√400------> dBC=20 units

step 3
find the distance AC
d=√[(0-4)²+(24-2)²]-----> dAC=√500------> dAC=22.36 units

[perimeter of triangle]=AB+BC+AC------> 10+20+22.36-----> 52.36 units

the answer is
the perimeter of triangle is 52.36 units