Question

find the area of a triangle two side of which are 16 cm and 12 cm and perimeter is 42 CM​

Answer 1

$\bf\huge\underline{\underline{ Given :- }}$

• 1st side of traingle is 16 cm

• second side of triangle is 12 cm

• Perimeter is 42 cm

$\bf\huge\underline{\underline{ To \: Find:- }}$

• Area of triangle

$\bf\huge\underline{\underline{ Answer :- }}$

3rd side

• Perimeter = sum of 3 sides .

• $\red{\implies}$ 42 = side + 16 + 12

• $\red{\implies}$side = 42 - ( 16 + 12 )

• $\red{\implies}$side = 42 - 28

• $\bf\red{\implies side = 14 cm }$

$\rule{10cm}{0.05cm}$

Area

According to Heron's Formula ,

$\boxed{ \bf \red{a \: = \sqrt{s(s - a)(s - b)(s - c)} }}$

Where , s is semi perimeter and

$\boxed{ \bf \red{s = \frac{a + b + c}{2} }}$

Applying here ,

$\bf \: s \: = \frac{16 + 12 + 14}{2}$

• $\bf\red{\implies} s \: = \frac{42}{2}$

• $\bf\red{\implies \: s \: = 21}$

$\rule{10cm}{0.005cm}$

$\boxed{ \bf \red{a \: = \sqrt{s(s - a)(s - b)(s - c)} }}$

• $\bf\red{\implies} \: \sqrt{21(21 - 16)(21 - 14)(21 - 12)}$

• $\bf\red{\implies} \sqrt{ 21 \times 5 \times 7 \times 9 }$

• $\bf\red{\implies\: 81.3 \: {cm}^{2} }$

$\rule{10cm}{0.05cm}$

Area of triangle = 81.3 cm²