Question

25. Find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.​

Answer 1

Given :

• Two sides of the triangle

• a = 8 cm
• b = 10 cm

• Perimeter of the triangle = 42 cm

To calculate :

• Area of the triangle.

Calculation :

~ Calculating measure of the third side :

Perimeter of ∆ = a + b + c

a = first side

b = second side

c = third side

According to the question,

$\longrightarrow$ 42 cm = 18 cm + 10 cm + c

$\longrightarrow$ 42 cm = 28 cm + c

$\longrightarrow$ 42 cm - 28 cm = c

$\longrightarrow$ 14 cm = c

» Third side = 14 cm

By using heron's formula :

• Area of ∆ = √[ s (s - a)(s - b)(s - c)]

→ s = Semi-perimeter

$\longrightarrow$ Semi-perimeter = $\sf { \dfrac{Perimeter }{2} }$

$\longrightarrow$ Semi-perimeter = $\sf { \dfrac{42 \: cm }{2} }$

$\longrightarrow$ Semi-perimeter = 21 cm

Substituting values in the heron's formula :

$\implies$ Area of ∆ = √[ s (s - a)(s - b)(s - c)]

$\implies$ Area of ∆ = √[ 21 (21 - 18)(21 - 10)(21 - 14)] cm²

$\implies$ Area of ∆ = √(21 × 3 × 11 × 7) cm²

$\implies$ Area of ∆ = √(21 × 21 × 11) cm²

$\implies$ Area of ∆ = √(21² × 11) cm²

$\implies$ Area of ∆ = √21² × √11 cm²

$\implies$ Area of ∆ = 21 × √11 cm²

$\implies$ Area of ∆ = 21√11 cm²

Hence, area of the triangle is 21√11 cm².

More about triangles :

★ Angle sum property of a triangle :

Sum of interior angles of a triangle = 180°

★ Exterior angle property of a triangle :

Sum of two interior opposite angles = Exterior angle

★ Perimeter of triangle :

Sum of all sides

★ Area of triangle :

$\sf { \dfrac{1}{2} \times Base \times Height }$

★ Area of an equilateral triangle:

$\sf { \dfrac{\sqrt{3}}{4} \times {Side}^{2} }$

★ Area of a triangle when its sides are given :

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

Where,

S= Semi-perimeter or $\sf {\dfrac{a+b+c}{2} }$

Answer 2

Given:

• Sides = 18cm and 10cm
• Perimeter = 42 cm

To Find

• Find the Area of a Triangle?

Solution:

Firstly, We have to Find the Length of the third side of the triangle as;

$\circ \: \: \: {\boxed{\tt\red{ Perimeter_{( \Delta )} = a + b + c }}}$

Where as

• a,b,c = Sides

Putting values,

$\leadsto{\tt{ Perimeter_ { \Delta} = a + b + c}} \\ \\ \leadsto{\tt{ 42 = 18 + 10 + c }} \\ \\ \leadsto{\tt{ 42 = 28 + c }} \\ \\ \leadsto{\tt{ 42 - 28 = c }} \\ \\ \leadsto{\tt\purple{ c = 14 \ cm }}$

• Third side of the triangle is 14 cm

Now, We can find Area of the Triangle Using Heron's Formula as;

$\\ \circ \: {\boxed{\tt\gray{ Area_{( \Delta )} = \sqrt{ s(s-a)(s-b)(s-c)} }}}$

Where as

$\implies{\tt{s = \dfrac{(a+b+c)}{2} }} \\ \\ \implies{\tt{ s = \dfrac{18 + 10 + 14}{2} }} \\ \\ \implies{\tt{ s = \cancel{ \dfrac{42}{2} } }} \\ \\ \implies{\tt{ s = 21 \ cm }}$

After substituting values in Suitable positions to find area of the triangle,

$\\ \colon\implies{\tt{ Area_{( \Delta )} = \sqrt{ s(s-a)(s-b)(s-c)} }} \\ \\ \\ \colon\implies{\tt{ Area_{( \Delta )} = \sqrt{ 21(21-18)(21-10)(21-14)} }} \\ \\ \\ \colon\implies{\tt{ Area_{( \Delta )} = \sqrt{ 21(3)(11)(7)} }} \\ \\ \\ \colon\implies{\tt{ Area_{( \Delta )} = \sqrt{ 21 \times 3 \times 11 \times 7} }} \\ \\ \\ \colon\implies{\tt{ Area_{( \Delta )} = \sqrt{ 3 \times 7 \times 3 \times 11 \times 7} }} \\ \\ \\ \colon\implies{\boxed{\tt\large\pink{ Area_{( \Delta )} = 21 \sqrt{11} \ cm^2 }}}$

Hence,

• The Area of the triangle is ${\tt{ 21 \sqrt{11} \ cm^2 }}$