# If the coordinates of the triangle are (2,3) , (5,3) and (5,7) ?calculate the perimeter and the area.

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The coordinates of the vertex of the triangle are (2,3) , (5, 3) and (5, 7)

The lengths of the sides are

sqrt [ (2 - 5)^2 + 0] = 3

sqrt [(5 - 5)^2 + (3 - 7)^2] = 4

sqrt [(5 - 2)^2 + (7 - 3)^2] = sqrt 25 = 5

We see that 3^2 + 4^2 = 5^2

This is a right triangle, so the area is half the product of the shorter sides

Area = (1/2)*3*4 = 6

The perimeter is the sum of the lengths of the side = 4+3+5 = 12

**The area of the triangle is 6 square units and the the perimeter is 12 units.**

First we will calculate the length of the sides.

We will use the distance formula to find the length.

==> (2,3) and (5.3)

==> D1 = sqrt( 2-5)62 + (3-3)62 = sqrt(9 = 3

==> (2,3)(5.7)

==> D2 = sqrt(2-5)^2 + (3-7)^2 = sqrt(9+16) = sqrt25 = 5

==> (5.3) and (5,7)

==> D = sqrt(5-5)^2 + (3-7)^2 = sqrt(16 = 4

Then the length of the sides are 3, 4, and 5.

Now we will calculate the perimeter.

==> P = 3 + 4 + 5 = 12

==> Now we will calculate the area.

==> A = sqrt ( s(s-a)(s-b) (s-c) such that s = p/2 and a, b, c are the length of the sides.

==> A = sqrt( 6( 6-3)(6-4)(6-5) = sqrt(6*3*2*1) = sqrt36 = 6

**Then the perimeter of the triangle is 12 units and the area is 6 square units.**