Question

a triangular park ABC has side 120 8 and 15 a gardener has to put the fence all around it and also plant grass inside how much area does he need to plant find the cost of fencing with barbed wire at rate of rupees 20 per metre leaving a space 3 metre wide for a gate on one side ​

Answer 1

Correct question :-

A triangular park ABC has side 120m , 80m and 50m .a gardener how to put the fence all around it and also plant grass inside how much area does it need to plant find the cost of fencing with barbed wire at the rate of ₹20 per meter leaving a space 3 metre wide for a gate on one side ?

Solution

Given ,

• A triangular park is having sides 120m, 80m and 50m .
• A gardener has to but offence all around it and also plant grass inside .
• barbed wire at the rate of rupees 20 per metre living space 3 metre wide for a gate on one side .

To find ,

• how much area does she need to plant ?
• find the cost of the fencing ?

So ,

• For for finding the area of Park we have ;

2x = 50m +80m + 120m = 250m .

I.e.

=> s =125 m .

Now ,

• s-a = (125-120)m = 5m
• s-b = (125-80)m = 45m
• s-c = (125-50)m = 75m

Therefore , the area of the park ;

= √s(s-a)(s-b)(s-c)

• now putting the value and solving it we get ;

= √125×5×45×75 m^2

= 375√15m^2

Also , perimeter of the park = AB + BC + CA =250M

Therefore ,

• length of the wore needed for the fencing = 250m - 3m(left for gate ) = 247m

And ,

• The cost of the fencing = ₹20×247=₹4940

The cost of the fencing is ₹4940 and the she will need to plants the plant in 247 m .

Answer 2

Given :

A triangular park ABC has the sides 120 m, 80 m, 15 m. The gardener Dahiya has to put a fence around the field and has to leave 3 meter wide space for gate. He has to plant grass as well in the area of the field.

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To Find :

The area of the field and the cost of fencing the field.

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Solution :

~ Formula Used :

$\large{\color{cyan}{\bigstar}} \: \: \: {\underline{\overline{\boxed{\red{\sf{ Area{\small_{(Triangle) }} = \sqrt{s (s - a) (s - b) (s - c) }}}}}}}$

$\large{\color{cyan}{\bigstar}} \: \: \: {\underline{\overline{\boxed{\red{\sf{ Perimeter{\small_{(Triangle) }} = a + b + c}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━}$

~ Calculating the Perimeter of field :

${\longmapsto{\qquad{\sf{ Perimeter{\small_{(Triangle) }} = a + b + c}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Perimeter{\small_{(Triangle) }} = 120 + 80 + 50 }}}} \\ \\ \ {\qquad{\sf{ Perimeter \: of \: Field \: = {\underline{\underline{\color{darkblue}{\sf{ 250 \: m}}}}}}}}$

~ Calculating Cost of fencing :

${:\longmapsto{\pink{\qquad{\sf{ Cost \: of \: fencing = Perimeter \: of \: fencing \times Rate}}}}}$

Here :

➳ Area of fencing = 250 - 3 = 247 m

➳ Rate = ₹ 20

Calculation Starts :

${\longmapsto{\qquad{\sf{ Cost{\small_{(Fencing) }}\: = Perimeter \times Rate}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Cost{\small_{(Fencing) }} = 247 \times 20}}}} \\ \\ \ {\qquad{\sf{ Cost \: of \: Fencing \: the \: Field \: = {\underline{\underline{\orange{\sf{₹ \: 4940 }}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━}$

~ Calculating the Area of Field :

Semi - Perimeter :

${\longmapsto{\qquad{\sf{ S = \dfrac{a + b + c}{2}}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ S = \dfrac{120 + 80 + 50}{2}}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ S = \dfrac{250}{2}}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ S = \cancel\dfrac{250}{2}}}}} \\ \\ \ {\qquad{\sf{ Semi - Perimeter \: = {\underline{\underline{\red{\sf{ 125 \: m}}}}}}}}$

Area :

${\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{s (s - a) (s - b) (s - c) }}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{125 (125 - 120)(125 - 80)(125 - 50)}}}}} \\ \\ \ {\longmapsto{\qquad{\sf{ Area{\small_{(Triangle) }} = \sqrt{125 \times 5 \times 45 \times 75}}}}} \\ \\ \ {\qquad{\sf{ Area \: of \: Field \: = {\underline{\underline{\pink{\sf{ 375 \sqrt{15} m²}}}}}}}}$

$\qquad{━━━━━━━━━━━━━━━━━━━━━}$

Therefore :

❝ Cost of fencing the field is ₹ 4940. ❞

❝ Area of the field in which he will lay grass is 300√15 m². ❞

$\pink{\underline{\blue{▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬}}}$

✴ Know More :

Formulas :

$\twoheadrightarrow{\sf{ Perimeter{\small_{(Rectangle)}} = 2(Length + Breadth) }}$

$\twoheadrightarrow{\sf{ Perimeter{\small_{(Square)}} = 4 \times Side }}$

$\twoheadrightarrow{\sf{ Perimeter{\small_{(Triangle)}} = a + b + c}}$

$\twoheadrightarrow{\sf{ Semi - Perimeter{\small_{(Triangle)}} = \dfrac{a + b + c}{2}}}$

$\twoheadrightarrow{\sf{ Area{\small_{(Rectangle)}} = Length \times Breadth}}$

$\twoheadrightarrow{\sf{ Area{\small_{(Square)}} = Side \times Side}}$

$\twoheadrightarrow{\sf{ Area{\small_{(Triangle)}} = \dfrac{1}{2} \times Base \times Height }}$

$\twoheadrightarrow{\sf{ Area{\small_{(Rhombus)}} =\dfrac{1}{2} \times D_1 \times D_2}}$

$\twoheadrightarrow{\sf{ Area{\small_{(Triangle)}} = \sqrt{s (s - a) (s - b)(s - c) }}}$

$\twoheadrightarrow{\sf{ CSA{\small_{(Cylinder)}} = 2πrh }}$

$\twoheadrightarrow{\sf{ TSA{\small_{(Cylinder)}} = 2πr(h + r) }}$

$\twoheadrightarrow{\sf{ Volume{\small_{(Cylinder )}} = πr²h}}$

$\twoheadrightarrow{\sf{ Volume{\small_{(Cuboid)}} = Length \times Breadth \times Height}}$

$\twoheadrightarrow{\sf{ Volume{\small_{(Cube)}} = Side³}}$

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