Question

Find the perimeter of a triangle whose vertices have coordinates (3,10) (5,2) and (14,12)

Answer 1

area of triangle = 1/2( x1(y2-y3) +x2(y3-y1)+x3(y1-y2)
= 1/2(3(2-12)+5(12-10)+14(10-2)
=1/2(-30+10+112)
=1/2×92=46

Answer 2

Answer:

(√68 + √181 + √125) unit

Step-by-step explanation:

Let,  A≡(3,10), B≡(5,2) and C≡(14,12),

Then, the perimeter of triangle ABC = AB + BC + CA,

By the distance formula,

$AB=\sqrt{(5-3)^2+(2-10)^2}$

$=\sqrt{2^2+8^2}$

$=\sqrt{4+64}$

$=\sqrt{68}\text{ unit}$

$BC=\sqrt{(14-5)^2+(12-2)^2}$

$=\sqrt{9^2+10^2}$

$=\sqrt{81+100}$

$=\sqrt{181}\text{ unit}$

$CA=\sqrt{(3-14)^2+(10-12)^2}$

$=\sqrt{11^2+2^2}$

$=\sqrt{121+4}$

$=\sqrt{125}\text{ unit}$

Hence, the perimeter of the triangle ABC = (√68 + √181 + √125) unit