Question

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

Answer 1

Let A (3,10); B(5,2);C(14,12)
area of triangle ABC = 1/2 [ x1(y2-y3)+x2(y3-y1)+x3(y1-y2)
                                  = 1/2 [ 3(2-12)+5(12-10)+14(10-2)]
                                  = 1/2 [3 (-10)+5(2)+14(8)]
                                  = 1/2 [-30+10+112]
                                  = 1/2 [92]
                                  = 46 sq. units

Answer 2

Answer:

33.74 units.

Step-by-step explanation:

Coordinates of A = (3,10)

Coordinates of B = (5,2)

Coordinates of C = (14,12)

Now find the sides of triangle AB,BC,AC

To find AB use distance formula :

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

$(x_1,y_1)=(3,10)$

$(x_2,y_2)=(5,2)$

Substitute the values in the formula :

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

$AB=\sqrt{(2)^2+(-8)^2}$

$AB=\sqrt{4+64}$

$AB=\sqrt{68}$

$AB=8.246$

To Find BC

$(x_1,y_1)=(5,2)$

$(x_2,y_2)=(14,12)$

Substitute the values in the formula :

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

$BC=\sqrt{(9)^2+(10)^2}$

$BC=\sqrt{81+100}$

$BC=\sqrt{181}$

$BC=13.453$

To Find AC

$(x_1,y_1)=(3,10)$

$(x_2,y_2)=(14,12)$

Substitute the values in the formula :

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

$AC=\sqrt{(11)^2+(2)^2}$

$AC=\sqrt{141+4}$

$AC=\sqrt{145}$

$AC=12.041$

Perimeter of a triangle = AB+BC+AC= 8.246+13.453+12.041=33.74 units

Hence the perimeter of the triangle is 33.74 units.