Question

A reatangular park is 40 m long and 25, wide.

A path 2.0 m wide is constructed within the park -

i) Remaining area
ii)Cost of payment if the rate is
RS.45/m2
${m}^{2}$



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Answer 1

Answer:

The length of the park is 40 m

the width of the park is 25m

then the area is=40×25 m2

=1000 m2

The path is 2 m wide.

The the length of the remaining part is 40-2×2

=40-4=36m

As well as the width is 25-4=21m

the the remaining area is 36×21 m2=756m2

cost of payment is 756×45 rs. =34020 rs.

Answer 2

$\large\underline{\sf{Solution-}}$

Let us suppose that rectangular park is ABCD such that,

• Length, AB= 40 m

• Breadth, BC = 25 m

We know,

$\red{\boxed{ \bf \: Area_{(rectangle)} = \:Length \times Breadth}} \:$

So,

$\rm :\longmapsto\:Area_{(ABCD)} = 40 \times 25$

$\rm :\longmapsto\:Area_{(ABCD)} = 1000 \: {m}^{2}$

Now,

• Path of width 2 meter is built inside the rectangular park ABCD.

• Let the inner rectangular park is represented as PQRS.

So,

• Length, PQ = 40 - 2 × 2 = 40 - 4 = 36 m

• Breadth, QR = 25 - 2 × 2 = 25 - 4 = 21 m

So,

$\rm :\longmapsto\:Area_{(rectangle \: PQRS)} = 36 \times 21$

$\bf :\longmapsto\:Area_{(rectangle \: PQRS)} =756 \: {m}^{2}$

Hence,

Remaining area of Rectangular Park PQRS = 756 sq. m.

Cost of payment

$\rm :\longmapsto\:Cost \: of \: 1 \: {m}^{2} \: is \: Rs \: 45$

$\rm :\longmapsto\:Cost \: of \: 756 \: {m}^{2} \: is \: Rs \: 45 \times 756$

$\bf :\longmapsto\:Cost \: of \: 756 \: {m}^{2} \: is \: Rs \: 34020$

Additional Information :-

$\rm :\longmapsto\:Area_{(square)} = {(side)}^{2}$

$\rm :\longmapsto\:Area_{(circle)} = \pi{(r)}^{2}$

$\rm :\longmapsto\:Perimeter_{(circle)} = 2 \pi{r}$

$\rm :\longmapsto\:Perimeter_{(rectangle)} = 2(Length + Breadth)$

$\rm :\longmapsto\:Perimeter_{(square)} = 4 \times side$

$\rm :\longmapsto\:Perimeter_{(rhombus)} = 4 \times side$