Question

two sides of a triangle are 13 cm and 14 cm and its perimeter is 36 cm find the area of the triangle

Answer 1

Finding sides of the triangle :-

13 + 14 + x = 36

27 + x = 36

x = 36 - 27

x = 9

→ Sides are 13cm, 14cm and 9cm

--------------------

Finding area of triangle :-

$\sqrt{s(s-a)(s-b)(s-c)}$

Here,

s = 36/2 = 18

• s = 18

• a = 13

• b = 14

• c = 9

--------------------

Substituting in formula :-

→ $\sqrt{18(18-13)(18-14)(18-9)}$

→ $\sqrt{18*5*4*9}$

→ $\sqrt{3240}$

→ $56.9m^{2}$

Answer 2

Answer:

Given that the two sides of triangle are 13cm and 14cm and its perimeter is 36cm.

First we have to find the third side of triangle.

Given

a = 13 cm

b = 14 cm

Perimeter = 36cm

=> a + b + c = p

=> 13 + 14 + c = 36

=> 27 + c = 36

=> c = 36 - 27

=> c = 9 cm.

The length of third side of triangle is 9 cm.

As the semi-perimeter is the half of the sum of sides of triangle.

$s \: = \: \frac{a + b + c}{2}$

$s \: = \: \frac{13 + 14 + 9}{2}$

$s \: = \: \frac{38}{2}$

s = 19

Therefore area of triangle =

$= \sqrt{s \: ( \: s - a \: )( \: s - b \: )( \: s - c \: )} \\ = \sqrt{19 \: ( \: 19 - 13 \: )( \: 19 - 14 \: )( \: 19 - 9 \: ) } \\ = \sqrt{19 \times 6 \times 5 \times 10 \: } \\ = \sqrt{19 \times 2 \times 3 \times 5 \times 2 \times 5} \\ = 2 \times 5 \sqrt{19 \times 3} \\ = 10 \sqrt{57 {cm}^{2} } \\ = or \: 56.9 {m \: }^{2} ( \: \: approx \: )$