Question

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

Answer 1

Answer:

here, a=120m,b=80m,c=50m

and, s=2a+b+c=2250=125

Area she need to plant grass =s(s−a)(s−b)(s−c)=125(125−120)(125−80)(125−50)=37515m2

cost of fencing =[(a+b+c)−3].20=[250−3].20=Rs4940

Step-by-step explanation:

Hope it will help you

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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