Question

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

Answer 1

Answer:

a = 18 cm

b = 10 cm

Perimeter = a + b + c

⇒ 42 = ( 18 + 10 ) + c

⇒ c = 42 - 28

⇒ c = 14 cm

Now,

$s = \frac{perimeter}{2} \\ \\ s = \frac{42}{2} = 21$

Area =

$\sqrt{s(s - a)(s - b)(s - c)} \\ \\ \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ \\ \sqrt{21 \times 3 \times 11 \times 7} \\ \\ \sqrt{3 \times 7 \times 3 \times 11 \times 7} \\ \\ 3 \times 7 \times \sqrt{11} \\ \\ \bf 21 \sqrt{7} \: cm {}^{2}$

Thanks!

Answer 2

Solution:-

Given:-

Two sides of the Triangle are 18cm and 10cm.

Perimeter = 42 cm.

To Find:-

Area of the Triangle = ?

Find :-

Perimeter = a + b + c

=) 42 = 10 + 18 + c

=) c = 14.

Semi-perimeter = ( a + b + c)/2

=) s = ( 10 + 18 + 14)/2

=) s = 21

Now,

Area of Triangle = √{ s ( s- a)(s-b)(s-c)}

=) Area of Triangle = √ { 21 ( 21-10)(21-18)(21-14)

=) Area of Triangle = √ { 21 × 11 × 3 × 7}

=) Area of Triangle = √ { 7 × 3 × 11 × 3 × 7}

=) Area of Triangle = 7 × 3 √11

=) Area of Triangle = 21√11 cm²

Hence,

Area of Triangle is 21√11 cm².