Question

11. 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 :-

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

Aim :-

• To find the area of the triangle

Answer :-

Concept :

To find the area of the triangle, we first find the third side of the triangle. By using heron's Formula, the area is hence found

Formula :

$\boxed{ \sf perimeter \: of \: a \: triangle = sum \: of \: the \: three \: sides}$

$\boxed {\sf semiperimeter = \dfrac{perimeter}{2} }$

$\boxed {\sf area \: of \: the \: triangle = \sqrt{s(s - a)(s - b)(s - c)} }$

• s here represents the semi perimeter of the triangle.
• a, b and c respectively represent the 3 sides of the triangle

Solution :

Let us first find the third side of the triangle.

We know that,

Sum of the 3 sides = Perimeter.

Let the third side be x.

→ 18 + 10 + x = 42

→ 28 + x = 42

Transposing 28 to the RHS (Right Hand Side of the Equation),

→ x = 42 - 28

→ x = 14cm

Now that we have the value of the third side, the semi perimeter will be :

$\implies \sf semiperimeter = \dfrac{42}{2}$

Reducing to the lowest terms,

$\implies \sf semiperimeter = 21cm$

Substituting these values for s, a, b and c respectively :

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

$\implies \sf \sqrt{21 \times 3 \times 11 \times 7}$

$\implies \sf \sqrt{21 \times 21 \times 11}$

$\implies \sf \sqrt{ {21}^{2} \times 11 }$

$\implies \sf 21 \times \sqrt{11}$

$\implies \sf 21 \sqrt{11} cm^{2}$

The area of the triangle is thus 21√11cm²

More formulas :

• Area of the triangle = ½ × base × height
• Area of a rectangle = Length × breadth
• Area of a square = (side)²
• Area of a parallelogram = Base × height
• Area of rhombus = ½ × Diagonal1 × Diagonal2