Question

pls answer correctly question number 17​

Answer 1

Step-by-step explanation:

Perimeter=2(l+b)

2(300+200)

2(500)

1000m²

cost of fencing 1 m= 40

cost of fencing 1000m²=1000x40

₹40000

Answer 2

Question :

⠀⠀⠀ Find the cost of fencing of a rectangular Park 300 m long and 200 m wide at the rate of Rs. 40 per metre .

⠀⠀⠀⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Given : The length and breadth Rectangular park is 300 m and 200m , respectively and The cost of fencing is at the rate of Rs. 40 per metre.

Need To Find : The cost of Fencing of Rectangular Park.

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\sf{\bf{Solution \: of\:Question \::}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀We know that if we are given with Length , Breadth and cost of fencing @ of per m of Rectangular Park and we have to find the cost of Fencing of a Rectangular Park. We have to fence the boundary of Rectangular park and we know that Boundary of Rectangular Park is equal to the Perimeter of Rectangular Park and with the help of Required Formula we can find the Perimeter of Rectangular Park that is :

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀ $\implies {\underline {\sf{ Perimeter _{(Rectangle)} = 2 ( l + b ) \:units .}}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀ Here l is the length of Rectangular park & b is the Breadth of Rectangular Park and we have given with the value of length and breadth. And Length (l) = 300 m & Breadth (b) = 200 m . So by using the given Values we can find Perimeter of Rectangular Park :

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\sf{\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\longmapsto {\sf{ Perimeter _{( Rectangle)} = 2 ( 300 + 200 ) }}$

⠀⠀⠀⠀⠀⠀⠀⠀$\longmapsto {\sf{ Perimeter _{( Rectangle)} = 2 ( 500 ) }}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed{\pink{\sf{ Perimeter _{( Rectangle)} = 1000\:m }}}}$

⠀⠀⠀

Therefore, ⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\underline {\therefore{\pink{\sf{ Hence,\:Perimeter\:of\: Rectangular \:Park\:is\:\: 1000\:m }}}}$

⠀⠀⠀⠀⠀ Now As We have found the Perimeter of Rectangular Park and we need to fence the Boundary of Rectangular Park at the rate of Rs . 40 per metre .

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\sf{\bf{\star\:Now \: \: Finding \: cost\:of \: Fencing \: \::}}}$

As , We know that ,

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀ $\implies {\underline {\sf{ Cost\:of\:Fencing = Rate \:of\:Fencing \: \times Perimeter}}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀Here Cost of Fencing of Rectangular park is at the rate of Rs. 40 per metre and Perimeter of Rectangular park is 1000 m . So by using given Values we can find Cost of Fencing of Rectangular Park :

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀⠀$\underline {\sf{\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}}$

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\longmapsto {\sf{ Cost\:of\:Fencing = 40 \times 1000 }}$

⠀⠀⠀⠀⠀⠀⠀⠀$\underline {\boxed{\pink{\sf{ Cost\:of\:Fencing \: = Rs.40,000 }}}}$

⠀⠀⠀⠀

Therefore,

⠀⠀⠀⠀⠀

⠀⠀⠀⠀⠀$\underline {\therefore{\pink{\sf{ Hence,\:Cost\:of\:Fencing \:of\:Rectangular\:Park\:is\:Rs.40,000\:}}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

$\large {\boxed{\mathrm |\:\:{\underline { More \:to\:Know:\:}}\:\:|}}$

⠀⠀⠀⠀⠀Some Formulas of Areas of Shapes :

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

• Area of Rectangle = Length × Breadth sq. units

• Area of Square = Side² sq.units

• Area of Triangle = ½ × Base × Height sq.units

• Area of Parallelogram = Base × Height sq.units

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀