Question

find the area of a triangle whose vertices are (-5,-1),(3,-5),(5,2)

Answer 1

therefore area =32 units^2.
i have also given u the formula to calculate area formed by points in the photo.


hope this helped u : )

Answer 2

Answer:

33.2624 sq.units

Step-by-step explanation:

A=(-5,-1)

B = (3,-5)

C = (5,2)

Now find the sides of triangle AB,BC,AC

To find AB use distance formula :

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

[tex](x_1,y_1)=(-5,-1) [/tex]

[tex](x_2,y_2)=(3,-5) [/tex]

Substitute the values in the formula :

$AB=\sqrt{(3+5)^2+(-5+1)^2}$

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

$AB=\sqrt{64+16}$

$AB=\sqrt{80}$

$AB=8.944$

To Find BC

[tex](x_1,y_1)= (3,-5) [/tex]

[tex](x_2,y_2)=(5,2) [/tex]

Substitute the values in the formula :

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

$BC=\sqrt{(3)^2+(7)^2}$

$BC=\sqrt{9+49}$

$BC=\sqrt{58}$

$BC=7.615$

To Find AC

[tex](x_1,y_1)=(-5,-1) [/tex]

[tex](x_2,y_2)=(5,2) [/tex]

Substitute the values in the formula :

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

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

$AC=\sqrt{100+9}$

$AC=\sqrt{109}$

$AC=10.44$

So, sides of triangle :

a =8.94

b=7.615

c=10.44

Now to find area:

$Area = \sqrt{s(s-a)(s-b)(s-c)}$  

Where $s = \frac{a+b+c}{2}$  

a,b,c are the side lengths of triangle  

Now substitute the values :  

$s = \frac{8.94+7.615+10.44}{2}$  

$s =13.4975$  

$Area = \sqrt{13.4975(13.4975-8.94)(13.4975-7.615)(13.4975-10.44)}$  

$Area =33.2624$  

Hence the area of the given triangle is 33.2624 sq. units.