Problems on Area - Solved Examples(Set 4)

Problems on Area - Solved Examples(Set 4)

16. The ratio between the length and the breadth of a rectangular park is $3:2$. If a man cycling along the boundary of the park at the speed of $12$ km/hr completes one round in $8$minutes, then what is the area of the park (in sq. m)?
A. $142000$	B. $112800$
C. $142500$	D. $153600$

answer with explanation

Answer: Option D
Explanation:
Solution 1

Let length $=3x$ km,
breadth $=2x$ km

Distance travelled by the man at the speed of $12$ km/hr in $8$ minutes $=2(3x+2x)=10x$

Therefore,$12\times \frac{8}{60}=10xx=\frac{4}{25}\text{km}=160\text{m}$

Area $=3x\times 2x=6x^2$
$=6\times {160}^2=153600{\text{m}}^2$

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Solution 2

$l:b=3:2\cdots \left(1\right)$

Perimeter of the rectangular park
= Distance travelled by the man at the speed of $12$ km/hr in $8$ minutes
= speed × time $=12\times \frac{8}{60}$(∵ 8 minute = $\frac{8}{60}$ hour)
= $\frac{8}{5}$ km $=\frac{8}{5}\times 1000$ m $=1600$ m

Perimeter $=2(l+b)$

Therefore,
$2(l+b)=1600l+b=\frac{1600}{2}=800\text{m}\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$
$l=800\times \frac{3}{5}=480\text{m}b=800\times \frac{2}{5}=320\text{m}$(Or $b=800-480=320$ m)

Area $=lb=480\times 320=153600$ m²

17. What is the percentage increase in the area of a rectangle, if each of its sides is increased by $20\%?$
A. $45\%$	B. $44\%$
C. $40\%$	D. $42\%$

answer with explanation

Answer: Option B
Explanation:
Solution 1

Change in area
$=(20+20+\frac{20\times 20}{100})\%=44\%$

i.e., area is increased by $44\%$

(This formula is explained in detail here)

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Solution 2

Let original length $=10$
original breadth $=10$

Then, original area
$=10\times 10=100$

Length is increased by $20\%$
⇒ New length $=10+2=12$(∵ $2$ is $20\%$ of $10$)

Breadth is increased by $20\%$
⇒ New breadth $=10+2=12$

New area $=12\times 12=144$

Increase in area
= New area - Original area
$=144-100=44$

Percentage increase in area
$=\frac{\text{increase in area}}{\text{original area}}\times 100=\frac{44}{100}\times 100=44\%$

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Solution 3

Let original length $=l$
original breadth $=b$

Then original area $=lb$

Length is increased by $20\%$
⇒ New length $=l\times \frac{120}{100}=1.2l$

Breadth is increased by $20\%$
⇒ New breadth $=b\times \frac{120}{100}=1.2b$

New area $=1.2l\times 1.2b=1.44lb$

Increase in area = new area - original area
$=1.44lb-lb=0.44lb$

Percentage increase in area
$=\frac{\text{increase in area}}{\text{original area}}\times 100=\frac{0.44lb}{lb}\times 100=44\%$

18. If the difference between the length and breadth of a rectangle is $23$ m and its perimeter is $206$ m, what is its area?
A. $2800$ m2	B. $2740$ m2
C. $2520$ m2	D. $2200$ m2

answer with explanation

Answer: Option C
Explanation:
Solution 1

$l-b=23\cdots \left(1\right)$

perimeter $=206$
$\Rightarrow 2(l+b)=206$
$\Rightarrow l+b=103\cdots \left(2\right)$

$\left(1\right)+\left(2\right)\Rightarrow 2l=23+103=126l=\frac{126}{2}=63\text{metre}$

Substituting the value of $l$ in $\left(1\right)$, we get
$63-b=23b=63-23=40\text{metre}$

Area $=lb=63\times 40=2520$ m²

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Solution 2

length = breadth $+23$. Therefore,
$4\times$ breadth $+2\times 23=206$ m
⇒ breadth $=40$ m

length $=40+23=63$ m

Area $=63\times 40=2520$ m²

19. The ratio between the perimeter and the breadth of a rectangle is $5:1$. If the area of the rectangle is $216$ sq. cm, what is the length of the rectangle?
A. $16$ cm	B. $18$ cm
C. $14$ cm	D. $20$ cm

answer with explanation

Answer: Option B
Explanation:
Solution 1

Given that $\frac{2(l+b)}{b}=5$
$\Rightarrow 2l+2b=5b\Rightarrow 2l=3b$
$\Rightarrow b=\frac{2l}{3}$

Also given that, area $=216$ cm²
$\Rightarrow lb=216$ cm²

Substituting the value of $b$, we get,
$l\times \frac{2l}{3}=216\Rightarrow l^2=\frac{3\times 216}{2}\Rightarrow l^2=3\times 108=(3\times 3)\times 36\Rightarrow l=3\times 6=18\text{cm}$

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Solution 2

Let perimeter $=5x$ cm
breadth $=x$ cm
Then, length $=\frac{5x-2x}{2}=\frac{3x}{2}$ cm

Area $=216$ cm²$\Rightarrow \frac{3x}{2}\times x=216\Rightarrow 3x^2=216\times 2\Rightarrow x^2=72\times 2=36\times 2\times 2\Rightarrow x=6\times 2=12$

length $=\frac{3x}{2}=\frac{3\times 12}{2}=18$ cm

20. What is the least number of square tiles required to pave the floor of a room $15\text{m}17\text{cm}$ long and $9\text{m}2\text{cm}$ broad?
A. $814$	B. $802$
C. $836$	D. $900$

answer with explanation

Answer: Option A
Explanation:
$l=15$ m $17$ cm $=1517$ cm
$b=9$ m $2$ cm $=902$ cm
Area $=1517\times 902$ cm²

Now we need to find out HCF(Highest Common Factor) of $1517$ and $902$
Let's find out the HCF using long division method for quicker results.


Hence, HCF of $1517$ and $902$ $=41$

Therefore, side length of largest square tile $=41$ cm

Area of each square tile $=41\times 41$ cm²

Number of tiles required
$=\frac{1517\times 902}{41\times 41}=37\times 22=407\times 2=814$