Question

△ABC

Answer 1

Answer:

The approximate perimeter of  △ABC is 12.34 units

Step-by-step explanation:

we know that

The perimeter of a triangle is the sum of its three length sides

so

$P=AB+BC+AC$

the formula to calculate the distance between two points is equal to

$d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}$

step 1

Find the distance AB

we have

A(3,4), B(−1,1)

substitute the values

$d=\sqrt{(1-4)^{2}+(-1-3)^{2}}$

$d=\sqrt{(-3)^{2}+(-4)^{2}}$

$d=\sqrt{25}$

$d_A_B=5\ units$

step 2

Find the distance BC

we have

B(−1,1),C(−2,5)

substitute the values

$d=\sqrt{(5-1)^{2}+(-2+1)^{2}}$

$d=\sqrt{(4)^{2}+(-1)^{2}}$

$d=\sqrt{5}$

$d_B_C=2.24\ units$

step 3

Find the distance AC

we have

A(3,4),C(−2,5)

substitute the values

$d=\sqrt{(5-4)^{2}+(-2-3)^{2}}$

$d=\sqrt{(1)^{2}+(-5)^{2}}$

$d=\sqrt{26}$

$d_A_C=5.10\ units$

step 4

Find the perimeter

$P=5+2.24+5.10=12.34\ units$