Question

Find area of a triangle whose perimeter is 32 cm . One of its side is 11cm and the difference of other two sides is 5cm.

Answer 1

Solutions :-


We have,

Perimeter of triangle = 32 cm
One of its side = 11 cm

Let the second side be x
And third side be x + 5


Perimeter of triangle = sum of three sides

A/q

=> 11 + x + x + 5 = 32
=> 2x = 32 - 16
=> 2x = 16
=> x = 16/2 = 8


So, second side = x = 8 cm
Third side = x + 5 = 8 + 5 = 13 cm


$Semi \: perimeter = \frac{a + b + c}{2} \\ \\ = \frac{11 + 8 + 13}{2} = \frac{32}{2} = 16 \: cm$


Now,

By using heron's formula,

Find the area of a triangle :-

$area \: = \sqrt{s(s - a)(s - b)(s - c) } \\ \\ = \sqrt{16(16 - 11)(16 - 8)(16 - 13) } \\ \\ = \sqrt{16 \times 5 \times 8 \times 3} \\ \\ = \sqrt{1920} = 43.81 \: approx.$

Answer : Area of triangle = 43.81 cm²

Answer 2

Solutions :-


There are three sides in a triangle.

Suppose
First side = a
Second side = b
Third side = c
Semi perimeter = s


Given :

Perimeter of triangle = 32 cm
Side a = 11 cm

Let the side b be x
So, side c = x + 5


We know,
Perimeter of triangle = sum of all sides of triangle


A/q

=> a + b + c = 32
=> 11 + x + x + 5 = 32
=> 2x + 16 = 32
=> 2x = 32 - 16
=> 2x = 16
=> x = 16/2 = 8


Therefore, side b = x = 8 cm
Side c = x + 5 = 8 + 5 = 13 cm


Find the semi perimeter of triangle :-

$s = ( \frac{a + b + c}{2}) cm \\ \\ = ( \frac{11 + 8 + 13}{2} )cm \\ \\ = ( \frac{32}{2} )cm \: = 16 \: cm$



Now,

By heron's formula,

Area of triangle =>

$area \: = \sqrt{s(s - a)(s - b)(s - c) } \\ \\ = \sqrt{16(16 - 11)(16 - 8)(16 - 13) } \\ \\ = \sqrt{16 \times 5 \times 8 \times 3} \\ \\ = \sqrt{2 \times 2 \times 2 \times 2 \times 5 \times 2 \times 2 \times 2 \times 3} \\ \\ = 2 \times 2 \times 2 \times \sqrt{5 \times 2 \times 3} \\ \\ = 8 \times \sqrt{30} \\ \\ = 8 \sqrt{30} \: \: \: \: or \: \: 43.81 \: \: \: approx.$


Hence,
Area of triangle is 8√30 cm² or 43.81 cm²