Question

Find the area of the triangle whose coordinates are (1, 2) , (3, 4) and (5, 10)

Answer 1

Answer:

Step-by-step explanation:

A=(1,2)

B=(3,4)

C=(5,10)

We will use distance formula to find the sides of triangle

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

$A=(x_1,y_1)=(1,2)\\B=(x_2,y_2)=(3,4)$

Substitute the values

$AB=\sqrt{(3-1)^2+(4-2)^2}$

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

$AB=\sqrt{4+4}$

$AB=\sqrt{8}$

$AB=2.82$

$B=(x_1,y_1)=(3,4)\\C=(x_2,y_2)=(5,10)$

Substitute the values

$BC=\sqrt{(5-3)^2+(10-4)^2}$

$BC=\sqrt{(2)^2+(6)^2}$

$BC=\sqrt{4+36}$

$BC=\sqrt{40}$

$BC=6.32$

$A=(x_1,y_1)=(1,2)\\C=(x_2,y_2)=(5,10)$

Substitute the values

$AC=\sqrt{(5-1)^2+(10-2)^2}$

$AC=\sqrt{(4)^2+(8)^2}$

$AC=\sqrt{16+64}$

$AC=\sqrt{80}$

$AC=8.94$

Now to find the area we will use heron's formula:

$Area=\sqrt{s(s-a)(s-b)(s-c)}$

Where $s=\frac{a+b+c}{2}$

a=2.82

b=6.32

c=8.94

Substitute the values

$s=\frac{2.82+6.32+8.94}{2}$

$s=9.04$

$Area=\sqrt{9.04(9.04-2.82)(9.04-6.32)(9.04-8.94)}$

$Area=3.910 units^2$

Hence the area of the triangle is 3.910 sq.units.