Question

38. A rectangular field 242m long has got an area of 4840sq.
m. The cost of fencing that field on all the four sides, if
1m of fencing costs Rs. 10, in rupees is :
(a) 524
(b) 262
(C) 2620
(d) 5240​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{Area = 4840 \: {m}^{2} } \\ &\sf{length \: = 242 \: m}\\ &\sf{cost \: of \: fencing \: Rs \: 10 \: per \: m.} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{cost \: of \: fencing \: the \: field}  \end{cases}\end{gathered}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

Area of Rectangular field = 4840 m^2.

Length of Rectangular field = 240 m

We know,

$\boxed{ \bf \: Area\:of\:Rectangle=Length × Breadth}$

$\tt \: :\implies \: 4840 = 242 \times breadth$

$\tt \: :\implies \: breadth \: = \dfrac{4840}{242}$

$\tt \: :\implies \: breadth \: = \: 20 \: m$

Now,

To find the cost of fencing, we have to first find the Perimeter of Rectangular field.

We have,

Length of Rectangular field = 242 m

Breadth of Rectangular field = 20 m

We know,

$\boxed{ \pink{ \bf \: Perimeter_{(rectangle)} = 2(Length + Breadth)}}$

$\tt \: :\implies \: Perimeter_{(rectangle)} = 2(242 + 20)$

$\tt \: :\implies \: Perimeter_{(rectangle)} = 2 \times 262$

$\tt \: :\implies \: Perimeter_{(rectangle)} = 524 \: m$

Now,

$\begin{gathered}\rm\red{According \: to \: statement}\end{gathered}$

$\tt \: :\implies \: Cost \: of \: 1 \: m = Rs \: 10$

$\tt \: :\implies \: Cost \: of \: 524 \: m = Rs \: 524 \times 10 = Rs5240$

$\large{\boxed{\boxed{\bf{Option \: (d) \: is \: correct}}}}$

Additional Information :-

More info:

Perimeter of rectangle = 2(length× breadth)

Diagonal of rectangle = √(length² + breadth²)

Area of square = side²

Perimeter of square = 4× side

Volume of cylinder = πr²h

T.S.A of cylinder = 2πrh + 2πr²

Volume of cone = ⅓ πr²h

C.S.A of cone = πrl

T.S.A of cone = πrl + πr²

Volume of cuboid = l × b × h

C.S.A of cuboid = 2(l + b)h

T.S.A of cuboid = 2(lb + bh + lh)

C.S.A of cube = 4a²

T.S.A of cube = 6a²

Volume of cube = a³

Volume of sphere = 4/3πr³

Surface area of sphere = 4πr²

Volume of hemisphere = ⅔ πr³

C.S.A of hemisphere = 2πr²

T.S.A of hemisphere = 3πr²