Calculate the area of a triangle who's vertices are the points (0,16) , (0,5) and (8,2)

neela | Student

To calculate the area of the triangle whose vertices are (0,16),(0,5) and(8,2).

We know that if (x,y1), (x2,y2) and (x3,y3) are the vertices of the triangle, then the area of the triangle is given by:

Area of the triangle = (1/2) |{(x2-x1)(y2+y1) +(x3-x2)(y3+y1)+(x1-x3)(y1+y2)}|

Therefore the area of the triangle whose vertices are (0,16), (0,5), (8 ,2) is given by:

Area of the triangle = (1/2) {(0-0)(16+5) +(8-0)((2+5) +(0-8)(2+16)}

Area of triangle = (1/2) |{0 + 56 - 144}|

Area of the given triangle = (1/2) * 88 = 44 sq units.

W can calculate the area by Heron's formula also:

Area of the triangle = sqrt{s(s-a)(s-b)(s-c)}, a, b ,c are sides of triangle, s = (a+b+c)/2.

a = sqrt(0+(16-5)^2 )= 11

b = sqrt[8^2+ (2-14)^2] = sqrt260 = 16.125..

c =sqrt[(8-0)*2+(2-5)^2] = sqrt73 = 8.544..

Therefore s = (11+sqrt260+sqrt73)/2 = 17.834..

Area of the triangle = sqrt{(17.834..)(6.834...)(1.7097..)(9.2902...)}

Area of the triangle = sqrt1935.999998

Area of the triangle = 43.999...sq units

 

giorgiana1976 | Student

We can calculate the area of the triangle, using Heron's formula.

In Heron's formula, we have to know the lengths of the sides of the triangle. Since we know the coordinates of the vertices of the triangle, we can calculate the lengths of the sides.

We'll use the following formula to determine the lenght of the side:

AB = sqrt[(xB-xA)^2 + (yB-yA)^2]

We'll calculate the length of the side, AB, whose vertices are the points  A(0,5) and B( 8,2).

(0,16) , (0,5) and (8,2)

AB = sqrt[(8-0)^2 + (2-5)^2]

AB = sqrt(64 + 9)

AB = sqrt 73

We'll calculate the length of the side, AC, whose vertices are the points  A(0,5) and C(0,16).

AC = sqrt[(0-0)^2 + (16-5)^2]

AC = 11

We'll calculate the length of the side, BC, whose vertices are the points  B( 8,2) and C(0,16).

BC = sqrt[(0-8)^2 + (16-2)^2]

BC =sqrt(64 + 196)

BC = sqrt260

We'll apply Heron's formula:

S = sqrt[p(p-AB)(p-AC)(p-BC)]

p = (AB+AC+BC)/2

p = (sqrt73+11+sqrt260)/2

william1941 | Student

We are given the triangle with vertices (0, 16), (0, 5) and (8, 2).

Now the points (0, 16) and (0, 5) lie on the y-axis. Therefore we can find the length of this side by just subtracting 5 from 16, 16-5 = 11.

Now the point (8, 2) is a distance 8 away from the y-axis.

If we take the distance 11 as the base, 8 is the height.

Now the area of a triangle is given by the relation (1/2)* base * height.

Here base is equal to 11.

And height is equal to 8.

Therefore the area is equal to (1/2)* 11*8 = 11*4 = 44.

The required area of the triangle is 44.