Question

Find the perimeter of the triangle defined by the coordinates (7, 1), (-6, 1), and (10, 6). (Round to nearest tenth)

Answer 1

Answer:

$35.6\ units$

Step-by-step explanation:

we know that

The perimeter of triangle is equal to the sum of the length of the three sides

Let

$A(7, 1), B(-6, 1), C(10, 6)$

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

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

Find the distance AB

$A(7, 1), B(-6, 1)$

substitute in the formula

$d=\sqrt{(1-1)^{2}+(-6-7)^{2}}$

$d=\sqrt{(0)^{2}+(-13)^{2}}$

$dAB=13\ units$

Find the distance BC

$B(-6, 1), C(10, 6)$

substitute in the formula

$d=\sqrt{(6-1)^{2}+(10+6)^{2}}$

$d=\sqrt{(5)^{2}+(16)^{2}}$

$dBC=16.8\ units$

Find the distance AC

$A(7, 1), C(10, 6)$

substitute in the formula

$d=\sqrt{(6-1)^{2}+(10-7)^{2}}$

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

$dAC=5.8\ units$

Find the perimeter

$P=dAB+dBC+dAC$

$P=13+16.8+5.8=35.6\ units$