Question

The sides of the triangle are 122 m, 22 m, 120 m. Find out the area of the triangle?

Answer 1

⚝ Given:-    

• Side a = 122 m
• Side b = 22 m
• Side c = 120 m

⚝ To find:-      

• The area of the triangle.

⚝ Formula Used:-      

~Semi-Perimeter

• $\Large\boxed{\underline{\tt{Semi\:-\:Perimeter\:=\:\dfrac{a\:+\:b\:+\:c}{2}}}}$

Where,

• a = length of side a
• b = Length of side b
• c = length of side c

~Area of the triangle

• $\Large\boxed{\underline{\tt{Area\:_{(\:TRIANGLE\:)}\:=\sqrt{s\:(\:s\:-\:a\:)\:(\:s\:-\:b\:)\:(\:s\:-\:c\:)} }}}$

Where,

• s = semi - perimeter
• a = length of side a
• b = Length of side b
• c = length of side c

                       _________________

⚝ Solution:-    

Here we have sides of the triangle,

To find out the area of the triangle we need semi - perimeter of the triangle. So firstly we will find out the semi - perimeter of the triangle.

~Semi-Perimeter

On applying the formula we get,

$\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:\dfrac{a\:+\:b\:+\:c}{2}}$

$\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:\dfrac{122\:+\:22\:+\:120}{2}}$

$\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:\dfrac{264}{2}}$

$\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:132\:m}$

. ° .Thus, The semi - perimeter is 132 m.

                           ___________________

~Area of the triangle

Now we will find out the area of the triangle by applying the formula,

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{s\:(\:s\:-\:a\:)\:(\:s\:-\:b\:)\:(\:s\:-\:c\:)} }$

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:(\:132\:-\:122\:)\:(\:132\:-\:22\:)\:(\:132\:-\:120\:)} }$

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:(\:10\:)\:(\:110\:)\:(\:12\:)} }$

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:\times\:10\:\times\:110\:\times\:12\:} }$

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:\times\:1100\:\times\:12\:} }$

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{\:1742400\:} }$

$\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\:1320\:m^{2}}$

. ° . Thus the area of the triangle is 1320 m².

                 _______________

Answer 2

⚝ Given:-

Side a = 122 m

Side b = 22 m

Side c = 120 m

⚝ To find:-

The area of the triangle.

⚝ Formula Used:-

~Semi-Perimeter

\Large\boxed{\underline{\tt{Semi\:-\:Perimeter\:=\:\dfrac{a\:+\:b\:+\:c}{2}}}}

Semi−Perimeter=

2

a+b+c

Where,

a = length of side a

b = Length of side b

c = length of side c

~Area of the triangle

\Large\boxed{\underline{\tt{Area\:_{(\:TRIANGLE\:)}\:=\sqrt{s\:(\:s\:-\:a\:)\:(\:s\:-\:b\:)\:(\:s\:-\:c\:)} }}}

Area

(TRIANGLE)

=

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

_________________

⚝ Solution:-

Here we have sides of the triangle,

To find out the area of the triangle we need semi - perimeter of the triangle. So firstly we will find out the semi - perimeter of the triangle.

~Semi-Perimeter

On applying the formula we get,

\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:\dfrac{a\:+\:b\:+\:c}{2}}:⟹Semi−Perimeter=

2

a+b+c

\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:\dfrac{122\:+\:22\:+\:120}{2}}:⟹Semi−Perimeter=

2

122+22+120

\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:\dfrac{264}{2}}:⟹Semi−Perimeter=

2

264

\qquad\tt{:\implies\:Semi\:-\:Perimeter\:=\:132\:m}:⟹Semi−Perimeter=132m

. ° .Thus, The semi - perimeter is 132 m.

___________________

~Area of the triangle

Now we will find out the area of the triangle by applying the formula,

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{s\:(\:s\:-\:a\:)\:(\:s\:-\:b\:)\:(\:s\:-\:c\:)} }:⟹Area

(TRIANGLE)

=

s(s−a)(s−b)(s−c)

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:(\:132\:-\:122\:)\:(\:132\:-\:22\:)\:(\:132\:-\:120\:)} }:⟹Area

(TRIANGLE)

=

132(132−122)(132−22)(132−120)

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:(\:10\:)\:(\:110\:)\:(\:12\:)} }:⟹Area

(TRIANGLE)

=

132(10)(110)(12)

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:\times\:10\:\times\:110\:\times\:12\:} }:⟹Area

(TRIANGLE)

=

132×10×110×12

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{132\:\times\:1100\:\times\:12\:} }:⟹Area

(TRIANGLE)

=

132×1100×12

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\sqrt{\:1742400\:} }:⟹Area

(TRIANGLE)

=

1742400

\qquad\tt{:\implies\:Area\:_{(\:TRIANGLE\:)}\:=\:1320\:m^{2}}:⟹Area

(TRIANGLE)

=1320m

2

. ° . Thus the area of the triangle is 1320 m².

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