Math, asked by assaultsniper07, 8 months ago

Heron's Formula class 9 NCRT Ex. 12.1 Q.3

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Answers

Answered by Anonymous
27

{\bf \green{Q}\blue{u}\pink{e}\purple{s}\gray{t}\red{i}\orange{o}\green{n}}

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN” (see Attachment ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour

{\bf \green{A}\blue{n}\pink{s}\purple{w}\gray{e}\red{r}}

{\purple{\sf Given :-}}

  • 1st Side ⇒ 15 m
  • 2nd Side ⇒ 11 m
  • 3rd Side ⇒ 6 m

{\purple{\sf To \: Find :-}}

  • Area

{\purple{\sf Formula \: Used :- }}

  • \tt s=\dfrac{a+b+c}{2}

  • \tt \sqrt{s(s-a)(s-b)(s-c)}

Note → s here refers to semi perimeter

\implies\sf s=\dfrac{15+11+6}{2}

\implies\sf s=\dfrac{32}{2}

\implies\sf s=16

Now,

 \sf \implies \sqrt{16(16-15)(16-11)(16-6)}

\implies \sf \sqrt{16 \times 1 \times 5 \times 10}

\implies \sf  \sqrt{800}

\implies \sf 20\sqrt{2}

{\green{ \underbrace{\boxed{\underline{\underline{\purple{ \tt \therefore 20\sqrt{2} \: m^{2} \: is \: the \: area  }}}}}}}

Attachments:
Answered by ItzCuteboy8
114

Question :-

There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer with explanation :-

Given :-

  • If the sides of the wall are 15 m, 11 m and 6 m

To Find :-

  • The area painted in colour

Solution :-

Let,

  • 1st side(a) = 15 m
  • 2nd side(b) = 11 m
  • 3rd side(c) = 6 m

We know that,

\boxed{\sf s = \frac{a + b + c}{2}}

Where,

  • s = Semi-perimeter
  • a = 1st side
  • b = 2nd side
  • c = 3rd side

Substitute the values we get,

:\implies\sf s = \frac{15 + 11 + 6}{2}

:\implies\sf s = \frac{\cancel{32}}{\cancel2}

:\implies\sf s = 16

We know that,

\boxed{\sf\sqrt{s(s-a)(s-b)(s-c)}}\: \: (\bf Heron's\:formula)

Where,

  • s = Semi-perimeter
  • a = 1st side
  • b = 2nd side
  • c = 3rd side

Substitute the values we get,

=\sf\sqrt{16(16-15)(16-11)(16-6)}

= \sf\sqrt{16 \times 1 \times 5 \times 10}

= \sf\sqrt{800}

=\underline{\boxed{ \blue{\sf20\sqrt{2}\:m^{2}}}}

\green{\sf\therefore 20\sqrt{2}\:m^{2} \: the  \: area \:  painted  \: in \:  colour.}

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