Question

find the area of triangle two sides of which are 18cm and 10 cm and perimeter is 42cm​

Answer 1

Answer :-

Let third side be x

then,

x + 18 + 10 = 42 ( Perimeter = sum of all sides )

=> x + 28 = 42

=> x = 42 - 28

=> x = 14

• a = 18

• b = 10

• c = 14

S = (a + b + c)/2

=> 18 + 10 + 14 / 2

=> 21

area of ∆ = √{s (s - a) (s - b) (s - c)}

=> √ {21 (21 - 18) (21 - 10) (21 - 14) }

=> √ 21 × 3 × 11 × 7

=> 1058.57cm2

Answer 2

Given :–

• Two sides of a triangle = 18cm and 10cm.
• Perimeter of a triangle = 42cm.

To Find :–

• Area of a triangle.

Solution :–

We know that,

Area of a triangle = $\sqrt{s(s-a)(s-b)(s-c)}$

Where,

s = Semi perimeter and a, b & c = sides of a triangle.

• s = $\dfrac{a + b + c}{2}$

Given, Perimeter = a + b + c = 42.

Semi-Perimeter = $\dfrac{\cancel{42}}{\cancel2}$

⟹ 21cm

• Value of s = 21

Now, we need to find the value of c,

Perimeter of a triangle = a + b + c

a + b + c = 42

Where,

• a = 18cm.
• b = 10cm.

Now, put the given values in the formula of perimeter of a triangle.

⟹ 18 + 10 + c = 42

⟹ 28 + c = 42

⟹ c = 42 – 28

⟹ c = 14cm

• Value of c = 14cm.

Now, we have to find the area of a triangle.

• a = 18cm
• b = 10cm
• c = 14cm
• s = 21

Put all the values in the formula of the area of a triangle.

⟹ $\sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

⟹ $\sqrt{21(3)(11)(7)}$

⟹ $\sqrt{(3 \times 7) \times 3 \times 11 \times 7}$

⟹ ${(\sqrt {3})}^{2} \times {(\sqrt{7})}^{2} \times \sqrt{11}$

⟹ $3 \times 7 \sqrt{11}$

⟹ $21 \sqrt{11} \: {cm}^{2}$

Hence,

The area of a triangle is 2√11 cm².