Question

The sides of the triangular ground are 22 m, 120 m and 122 m. Find the area a
cost of levelling the ground at the rate of 20 per m2.​

Answer 1

cost=Rs.26400

as the area =1320m2

and when we multiply by Rs.20 we get this answer

Answer 2

$\sf{\underline{\boxed{\large{\blue{\mathsf{Solution}}}}}}$

$\sf{\bold{\green{\underline{\underline{Given}}}}}$

• Sides of a field = 22m , 120m , 122m
• Cost of levelling $\sf{\bold{ 1m^2 }}$ = Rs. 20

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$\sf{\bold{\green{\underline{\underline{To\:Find}}}}}$

• Cost of levelling the field = ??

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$\sf{\bold{\green{\underline{\underline{Solution}}}}}$

$\sf{\red{\boxed{\bold{Semi\: Perimeter = \dfrac{ a + b + c }{ 2 }}}}}$

$\sf{\red{\boxed{\bold{area = \sqrt{ s ( s - a ) ( s - b ) ( s - c )}}}}}$

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• Finding semi perimeter

: $\sf\implies\: {\bold{ s = \dfrac{ 22 + 120 + 122 }{2}}}$

⠀⠀⠀⠀

: $\sf\implies\: {\bold{ s = \dfrac{ 264 }{2}}}$

⠀⠀⠀⠀

: $\sf\implies\: {\bold{ s = 132 m}}$

⠀⠀⠀⠀

• Finding area

: $\sf\implies\: {\bold{ area = \sqrt{ 132 ( 132 - 22 ) ( 132 - 120) ( 132 - 122 )}}}$

⠀⠀⠀⠀

: $\sf\implies\: {\bold{ area = \sqrt{ 132\times 110 \times 12 \times 10}}}$

⠀⠀⠀⠀

: $\sf\implies\: {\bold{ area = \sqrt{ 2\times 2 \times 3 \times 11 \times 2 \times 5 \times 10 \times 2 \times 2 \times 3 \times 2 \times 5 }}}$

⠀⠀⠀⠀

: $\sf\implies\: {\bold{ area = 2 \times 2 \times 2 \times 3 \times 5 \times 11}}$

⠀⠀

: $\sf\implies\: {\bold{ area = 1320 m^2}}$

⠀⠀

$\sf{\pink{\boxed{\bold{area = 1320 m^2}}}}$

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Cost of levelling $\sf{\bold{ 1m^2 }}$ = Rs. 20

⠀⠀⠀⠀

Cost of levelling $\sf{\bold{ 1320m^2 }}$ = Rs. 20 x 1320

⠀⠀⠀⠀

Cost of levelling $\sf{\bold{ 1320m^2 }}$ = Rs. 26400

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$\sf{\bold{\green{\underline{\underline{Answer}}}}}$

• Cost of levelling the ground at Rs. 20 per meter is Rs. 26400