# What is the area of the triangle which has sides of lengths 9 cm, 3 cm and 7 cm?

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We know that the area of the triangle if given the length of all sides is:

A = sqrt ( s ( s - a) ( s-b) ( s- c) where :

s = perimeter / 2

a , b, and c are the sides of the trinagle:

Given the sides are:

a= 3,

b= 7

c = 9

Let us find the perimeter:

p = ( 3 + 7 + 9) = 19\

Then S = p / 2 = 19/ 2 = 9.5

Now let us substitute in the area formula:

A = sqrt[ s ( s- a) ( s-b) ( s-c)]

= sqrt[ 9-5) ( 9.5 - 3) ( 9.5 - 7) ( 9.5 - 9)

= sqrt( 9.5)*6.5*2.5 ( 0.5)

= sqrt(77. 1875)

= 8.79 cm^2( approx,)

Then the area of the triangle is:

**A = 8.79 cm^2.**

We are given a triangle with sides of length 9 cm, 7 cm and 3 cm.

To find the area we use Heron’s formula which states that for a triangle with sides equal to a, b and c the area is equal to sqrt[ s*(s-a)*(s-b)*(s-c)], where s is the semi perimeter.

Here, the semi perimeter s is equal to (9+7+3)/2 = 19/2 = 9.5. The lengths of the sides are 9, 7 and 3.

Therefore the area = sqrt [ 9.5*(9.5-9)*(9.5-7)*(9.5-3)] = sqrt( 9.5*0.5*2.5*6.5) = sqrt 77.1875 = 8.78 approximately.

**Therefore the area of the triangle is 8.78 cm^2.**