Question

Find the perimeter of triangle PQR if the vertices are P(3,0), Q(-1,3) and R(11,6).​

Answer 1

Answer :

$\dag\bf\:P(x_1,y_1)=(3,0)$

$\dag\bf\:Q(x_2,y_2)=(-1,3)$

$\dag\bf\:R(x_3,y_3)=(11,6)$

❅ Distance between P and Q :

$\sf\:PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$\sf\:PQ=\sqrt{(-1-3)^2+(3-0)^2}$

$\sf\:PQ=\sqrt{(-4)^2+(3)^2}$

$\sf\:PQ=\sqrt{16+9}$

$\sf\:PQ=\sqrt{25}$

$\bf\:PQ=5\:unit$

❅ Distance between Q and R :

$\sf\:QR=\sqrt{(x_3-x_2)^2+(y_3-y_2)^2}$

$\sf\:QR=\sqrt{(11-(-1))^2+(6-3)^2}$

$\sf\:QR=\sqrt{(12)^2+(3)^2}$

$\sf\:QR=\sqrt{144+9}$

$\sf\:QR=\sqrt{153}$

$\bf\:QR=12.37\:unit$

❅ Distance between P and R :

$\sf\:PR=\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}$

$\sf\:PR=\sqrt{(3-11)^2+(0-6)^2}$

$\sf\:PR=\sqrt{(-8)^2+(-6)^2}$

$\sf\:PR=\sqrt{64+36}$

$\sf\:PR=\sqrt{100}$

$\bf\:PR=10\:unit$

◈ Perimeter of Triangle :

➝ perimeter = PQ + QR + PR

➝ perimeter = 5 + 12.37 + 10

➝ Perimeter = 27.37 unit