Question

Geeta having a toy which is in the shape of a triangle, two sides of which are 16cm and 14cm and its perimeter is 42m. find the area of triangular toy. ​

Answer 1

Given:

Geeta having a toy which is in the shape of a triangle, two sides of which are 16cm and 14cm and its perimeter is 42cm.

To Find:

Area of triangular toy?

Solution:

Since, shape of the toy is triangular and its two sides are 16cm and 14cm and its perimeter is 42cm.

let, third side of triangular toy be x.

Since, perimeter of triangle = sum of all sides of triangle.

  therefore,

        $( 14 + 16 + x ) = 42\\30 + x = 42\\x = 12$

Thus, 3rd side of triangular toy is 12 cm and its all sides are of different lengths.

so, using Heron's formula area of the triangle whose side have lengths a , b and c is   $A = \sqrt{s(s-a)(s-b)(s-c)}$

where, $s = (a + b + c)/2$

   Here, a = 16cm , b = 14cm and c = 12cm

            $s = (16+14+12)/2\\ = 21$  cm

So, Area,

               $A = \sqrt{21(21-16)(21-14)(21-12)} \\ = \sqrt{6615} \\ = 21\sqrt{15}$ $cm^{2}$

Hence, Area of the triangular toy is $21\sqrt{15}$ $cm^{2}$.