Question

a triangular park ABC has sides 120m,80m and 15m a gardener dhaniya was as put a fence all around it also plant grass inside how much area does she need to plant find the cost of fencing at the barbed wire at rate of 20 rupees per metre living space 3 metre wide for a gate

Please don't spam ​

Answer 1

GIVEN

• 120 M
• 80M
• 5O M

FOR FINDING THE AREA OF THE PARK .

• 2s = 50 m + 80 m + 120 m = 250 m
• S = 125 m

NO

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

THEREFORE THE AREA OF PARK

$\small\bf\underline\pink{ \sqrt{s \: (s - a) \:s - b) \: (s - c) } }$

• By using above formula .

YOU GOT

$\small\bf\underline\green{ \sqrt{125 \times 5 \times 45 \times 75 \: {m}^{2} } }$

• 375√15 m²

ALSO WE HAVE TO FIND PERIMETER OF PARK

• AB + BC + CA = 250 m
• Therefore , length of the wire needed for fencing = 250 m - 3m ( it is for left of gate ) = 247 m

And so cost of fencing = Rs 20 × 247 = 4940

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.

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

To Find :

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

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 :

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

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

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

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

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

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

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

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

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

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

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

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

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

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

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬