Question

Q3) Solve the following examples. -(12) (i)Find the area of triangle of which two sides are 14 cm and 12 cm and the perimeter is 42 cm.​

Answer 1

Given :

• First side of the triangle = 14cm
• Second side of the triangle = 12cm
• Perimeter of the triangle = 42cm

To Find :

• Area of the Triangle.

Solution :

In this question, two sides and perimeter of a triangle are given and we have to find the area of the triangle. So firstly we will find the third side of the triangle, after that we will find the semi perimeter of the triangle after finding semi perimeter we will use Heron's Formula to find the area of the triangle.

Finding the third side of the triangle :

• Sum of all sides = Perimeter
• 14 + 12 + Third side = 42
• 26 + Third side = 42
• Third Side = 42 - 26
• Third Side = 16cm

Finding the semi perimeter of the triangle :

• Semi perimeter = Sum of all sides/2
• Semi perimeter = 14 + 12 + 16/2
• Semi perimeter = 26 + 16/2
• Semi perimeter = 42/2
• Semi perimeter = 21cm

Finding the area of the triangle :

• Area = √s (s - a) (s - b) (s - c)
• Area = √21 (21 - 14) (21 - 12) (21 - 16)
• Area = √21 × 7 × 9 × 5
• Area = √3 × 7 × 7 × 3 × 3 × 5
• Area = √7 × 7 × 3 × 3 × 3 × 5
• Area = √7 × 3 × 3 × 5
• Area = 3 × 5√7 × 3
• Area = 15√21 cm²

Therefore :

• Area of the triangle is 15√21 cm².

━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Question:-

Find the area of triangle of which two sides are 14 cm and 12 cm and the perimeter is 42 cm.

Given:-

• Two sides of the triangle = 14cm and 12cm.

• Perimeter = 42cm.

To Find:-

• Area of triangle.

Solution:-

By using Heron's Formula, we can calculate the area of triangle.

Heron's Formula for the area of a triangle is:

${ \boxed{ \sf \large \red{Area = \sqrt{{s} (s - a)(s - b)(s - c)}.}}}$

where a,b and c are the sides of the triangle and

s = semi ‐ perimeter = Half the perimeter of the traingle.

The sides of triangle given:

• a = 14cm,

• b = 12cm.

Perimeter of the traingle = (a + b + c).

42 = 14 + 12 + c

42 = 26 + c

c = 42 – 26

c = 16cm.

Semi - Perimeter:

s = (a + b + c) = 42/2 = 21cm.

By using Heron's Formula:

${ \boxed { \sf \large \red{{Area \: of \: a \: triangle = \sqrt{s(s - a)(s - b)(s - c)}. }}}}$

$\sf \large Area \: of \: a \: triangle = \sqrt{21(21 - 14)(21 - 12)(21 - 16)} .$

$\sf \large Area \: of \: a \: triangle = \sqrt{21 \times 7 \times 9 \times 5} .$

$\sf \large Area \: of \: a \: triangle = \sqrt{3 \times 7 \times 7 \times 3 \times 3 \times 5} .$

$\sf \large Area \: of \: a \: triangle = \sqrt{7 \times 7 \times 3 \times 3 \times 3 \times 5} .$

$\sf \large Area \: of \: a \: triangle = \sqrt{7 \times 3 \times 3 \times 5} .$

$\sf \large Area \: of \: a \: triangle = 3 \times 5 \sqrt{7} \times 3.$

$\sf \large Area \: of \: a \: triangle = 15 \sqrt{21} cm {}^{2} .$

Answer:-

${ \boxed{ \sf \large \red{ \therefore \: Area \: of \: a \: triangle = 15 \sqrt{21} cm {}^{2}. }}}$

Hope you have satisfied. ⚘