Question

.A school play ground is rectangular in shape wie
length 100 m and breadth 80 m. A cemented pathway
running all around it on its outside of width 3 m is
built. Find the cost of cementing if the rate of
cementing 1 sq. m is 50.​

Answer 1

Correct Question :

A school play ground is rectangular in shape with length 100 m and breadth 80 m. A cemented pathway running all around it on its outside of width 3 m is built. Find the cost of cementing if the rate of cementing 1 sq. m is 50 paise.

AnswEr :

◦ Inner Length = 100 m

◦ Inner Breadth = 80 m

◦ Outer Length = 100 + 2(3) = 106 m

◦ Outer Breadth = 80 + 2(3) = 86 m

⋆ Refrence of Image is in the attachment :

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• Area of the Pathway around Playground :

↠ Area = Outer Area – Inner Area

↠ Area = (L × B) – (l × b)

↠ Area = (106 m × 86 m) – (100 m × 80 m)

↠ Area = 9,116 m² – 8,000 m²

↠ Area of Pathway = 1,116 m²

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• Cost of Cementing the Pathway :

↠ Cost = Area × Rate

↠ Cost = 1,116 m² × 50 paise / m²

↠ Cost = 1,116 × Rs. 0.5

↠ Cost = Rs. 558

∴ Cost of cementing of Pathway is Rs. 558

Answer 2

Answer:

${\boxed{\bold\red{Cost\:of\:cementing=Rs.55800}}}$

Step-by-step explanation:

Correct question:-

A school playground is rectangular in shape with length 100 m and breadth 80 m.A cemented pathway running all around it on its outside of width 3 m is built.Find the cost of cementing if the rate of cementing 1 sq.m is Rs.50.

________________________

$\bf{Given}\begin{cases}\sf{Length\:of\:field=100\:m}\\\sf{Breadth\:of\:field=80\:m}\\\sf {Width\:of\:path = 3\:m}\\\sf{Rate =Rs.50\:per\:sq.m}\end{cases}$

Formula used:-

${\boxed{\bold\green{Area = Length \times Breadth}}}$

$\rightarrow\sf Area = (100\times 80)\:{m}^{2} \\\rightarrow\sf Area = 8000\:{m}^{2}$

$\sf Length\:of\: field\:including\:path:- \\\rightarrow\sf (100 + 2\times 3)\:m \\\rightarrow\sf (100+6)\:m \\\rightarrow\sf 106\:m \\\\\sf Breadth\:of\:field\:including\:path :-\\\rightarrow\sf (80+2\times 3)\:m \\\rightarrow\sf (80+6)\:m \\\rightarrow\sf 86\:m$

$\rightarrow\sf Area = (106\times 86)\:{m}^{2} \\\rightarrow\sf Area = 9116\:{m}^{2}$

$\sf Area\:of\:path = (9116 - 8000)\:{m}^{2} \\\rightarrow\sf Area\:of\:path = 1116\:{m}^{2}$

$\sf Cost\:of\:cementing:- \\\rightarrow\sf Cost = Area \times Rate \\\rightarrow\sf Cost = Rs.(1116\times 50) \\\rightarrow{\boxed{\bold\purple {Cost = Rs.55800}}}$

$\therefore\bold{\underline{Cost \:of\:cementing = Rs.55800}}$