Question

A rectangular park 54m x 40m has 2m wide roads in the centre running parallel to the breadth. Find the:
A) Area of the roads
B) Area of the park excluding roads.
C) Cost of laying grass at Rs.5 per m2

Answer 1

Answer:-

Answer:-A) Area of roads = 80m^2

Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2

Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400

Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-

• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54m
• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40m
• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40mLength of road = 2m
• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40mLength of road = 2mBreadth of road = Breadth of park = 40m

Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40mLength of road = 2mBreadth of road = Breadth of park = 40mTo Find:-

• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40mLength of road = 2mBreadth of road = Breadth of park = 40mTo Find:- A) Area of the roads
• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40mLength of road = 2mBreadth of road = Breadth of park = 40mTo Find:- A) Area of the roadsB) Area of the park excluding roads.
• Answer:-A) Area of roads = 80m^2B) Area of park excluding roads=2080 m ^2C) Cost of laying grass at Rs.5 per m^2 = ₹10400 Given:-Length of park = 54mBreadth of park = 40mLength of road = 2mBreadth of road = Breadth of park = 40mTo Find:- A) Area of the roadsB) Area of the park excluding roads.C) Cost of laying grass at Rs.5 per m^2

Now first we find the area ,

Now first we find the area , →Area of park = Length of park × Breadth of park

Now first we find the area , →Area of park = Length of park × Breadth of park = 54 m × 40 m

Now first we find the area , →Area of park = Length of park × Breadth of park = 54 m × 40 m = 54 m × 40 m

Now first we find the area , →Area of park = Length of park × Breadth of park = 54 m × 40 m = 54 m × 40 m = 2160m ^2

Now first we find the area , →Area of park = Length of park × Breadth of park = 54 m × 40 m = 54 m × 40 m = 2160m ^2→Area of road = Length of road × Breadth of road

Now first we find the area , →Area of park = Length of park × Breadth of park = 54 m × 40 m = 54 m × 40 m = 2160m ^2→Area of road = Length of road × Breadth of road→ = 2 m × 40 m

Now first we find the area , →Area of park = Length of park × Breadth of park = 54 m × 40 m = 54 m × 40 m = 2160m ^2→Area of road = Length of road × Breadth of road→ = 2 m × 40 m→ = 80 m ^2

b) Area of park excluding roads= Area of park - Area of roads

b) Area of park excluding roads= Area of park - Area of roads = 2160 – 80 = 2080 m ^2

c) Cost of laying 1m^2 grass = ₹5

Cost of laying 2080m^2 grass = 2080×5 = ₹10400

Answer 2

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

Given that,

A rectangular park 54m x 40m has 2m wide roads in the centre running parallel to the breadth.

Let assume that ABCD be the rectangular park such that AB = 54 m and CD = 40 m

Let assume that EFGH be the 2 m wide road in the centre running parallel to the breadth.

So, Length of road, EF = 40 m

Breadth of road, FG = 2 m

$\rm \: Area_{(road)} = EF \times FG$

$\rm \: Area_{(road)} = 40 \times 2$

$\rm\implies \:Area_{(road)} = 80 \: {m}^{2}$

Now,

$\rm \: Area_{(park)}$

$\rm \: = \: Area_{(ABCD)} - Area_{(road)}$

$\rm \: = \: 54 \times 40 - 80$

$\rm \: = \: 2160 - 80$

$\rm \: = \: 2080 \: {m}^{2}$

Now,

Further given that

$\rm \: Cost\:of\:laying \: grass \: for\: {1 \: m}^{2} = Rs \: 5$

So,

$\rm \: Cost\:of\:laying \: grass \: for\: {2080\: m}^{2} = 2080 \times 5 = Rs \: 10400$

So,

$\begin{gathered}\begin{gathered}\bf\: \bf\implies \:\begin{cases} &\sf{Area_{(road)} = 80 \: {m}^{2} } \\ \\ &\sf{Area_{(park)} = 2080 \: {m}^{2} } \\ \\ &\sf{Cost\:of\:laying \: grass \: in \: park= Rs \: 10400} \end{cases}\end{gathered}\end{gathered}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Base\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}d\sqrt {4a^2-d^2}\\ \\ \star\sf Parallelogram =Base\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}\end{gathered}$