Question

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

Answer 1

`\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=√((x_2-x_1)^2+(y_2-y_1)^2)`

`\sf\:PQ=√((-1-3)^2+(3-0)^2)`

`\sf\:PQ=√((-4)^2+(3)^2)`

`\sf\:PQ=√(16+9)`

`\sf\:PQ=√(25)`

`\bf\:PQ=5\:unit`

❅ Distance between Q and R :

`\sf\:QR=√((x_3-x_2)^2+(y_3-y_2)^2)`

`\sf\:QR=√((11-(-1))^2+(6-3)^2)`

`\sf\:QR=√((12)^2+(3)^2)`

`\sf\:QR=√(144+9)`

`\sf\:QR=√(153)`

`\bf\:QR=12.37\:unit`

❅ Distance between P and R :

`\sf\:PR=√((x_1-x_3)^2+(y_1-y_3)^2)`

`\sf\:PR=√((3-11)^2+(0-6)^2)`

`\sf\:PR=√((-8)^2+(-6)^2)`

`\sf\:PR=√(64+36)`

`\sf\:PR=√(100)`

`\bf\:PR=10\:unit`

◈ Perimeter of Triangle :

➝ perimeter = PQ + QR + PR

➝ perimeter = 5 + 12.37 + 10

➝ Perimeter = 27.37 unit