Question

plzz solve kardo.....​

Answer 1

$\implies \huge \sf\underline \red{Answer}$

$\sf\therefore \underline \purple{area = 21 \sqrt{11} {cm}^{2}}$

To find:-

Area of triangle

solution:-

we know that area of triangle formula,

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

$\sf \underline{so}$

s is semi perimeter

a,b,c is sides of triangle

$\sf \underline{Given}$

a = 18 cm

b = 10 cm

perimeter = 42cm

$\sf \underline{we \: know \: that \: semi \: perimeter \: formula}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{semi \: perimeter = \dfrac{perimeter}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{s = \dfrac{42}{2}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{semi = 21cm}$

we have to find c

$\sf \underline{now}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{perimeter = 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{a + b + c = 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{18cm + 10cm + c = 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{28cm + c= 42}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{c = 42 - 28cm}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \blue{c = 14cm}$

$\sf \underline{take \: the \: formula \: again : }$

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

a = 18 cm

b = 10 cm

c = 14 cm

s = 21 cm

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

$\: \: \: \: \: \: \sf{Area \: of \: traingle = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21(3)(11)(7)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21(7 \times 3)(11)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21(21)(11)}}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{21 \times 21 \times \sqrt{11}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \sqrt{ {21}^{2} \times \sqrt{11}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 21 \times \sqrt{11} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = 21 \sqrt{11} }$

$\sf \purple\therefore \underline{area = 21 \sqrt{11} {cm}^{2}}$

Answer 2

Answer : $\sf{21\sqrt{11}cm^2}$

Given :

Two sides of a triangle measure 18 cm and 10 cm

The perimeter of the triangle is 42 cm

To find :

The area of the traingle

Solution :

As we know that perimeter is equal to the sum of all three sides of a triangle ,

= a + b + c = 42 cm

= 18 + 10 + c = 42 cm

= 28 + c = 42 cm

= c = 42 - 28

= c = 14 cm

Also the semi perimeter of a triangle ,

$\sf\implies\:\dfrac{Perimeter}{2}$

$\sf\implies\:\dfrac{42}{2}$

$\sf\implies\:21\:cm$

Here we can use Heron's formula to find the area of the triangle .

Formula :

$\sf{Area\:=\:\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}$

Where ,

s = semi perimeter

a = length of side a

b = length of side b

c = length of side c

According to the Statement we have ,

s = 21 cm

a = 18 cm

b = 10 cm

c = 14 cm

Solving for area ,

$\sf{Area\:=\:\sqrt{s(s\:-\:a)(s\:-\:b)(s\:-\:c)}}$

$\sf{Area\:=\:\sqrt{21(21\:-\:18)(21\:-\:10)(21\:-\:14)}}$

$\sf{Area\:=\:\sqrt{21(3)(11)(7)}}$

$\sf{Area\:=\:\sqrt{4851}}$

$\sf{Area\:=\:21\sqrt{11}cm^2}$