Problems on Area - Solved Examples(Set 3)

Problems on Area - Solved Examples(Set 3)

11. The length of a rectangle is twice its breadth. If its length is decreased by $5$ cm and breadth is increased by $5$ cm, the area of the rectangle is increased by $75$ sq.cm. What is the length of the rectangle?
A. $18$ cm	B. $16$ cm
C. $40$ cm	D. $20$ cm

answer with explanation

Answer: Option C
Explanation:
Let breadth $=x$ cm
Then, length $=2x$ cm
Area $=x\times 2x=2x^2$ sq.cm.

New length $=(2x-5)$ cm
New breadth $=(x+5)$ cm
New area $=(2x-5)(x+5)$ sq.cm.

Given that, new area = initial area $+75$ sq.cm.
$\Rightarrow (2x-5)(x+5)=2x^2+75\Rightarrow 2x^2+10x-5x-25=2x^2+75\Rightarrow 5x-25=75\Rightarrow 5x=75+25=100\Rightarrow x=\frac{100}{5}=20\text{cm}$
Length $=2x=2\times 20=40$ cm

12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus?
A. equal to $\frac{1}{2}$	B. equal to $\frac{3}{4}$
C. greater than $1$	D. equal to $1$

answer with explanation

Answer: Option C
Explanation:

If a square and a rhombus lie on the same base, area of the square will be greater than area of the rhombus (In the special case when each angle of the rhombus is $90°$, rhombus is also a square and therefore areas will be equal)

Hence greater than $1$ is the more suitable choice from the given list

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Note : Proof


Consider a square and rhombus standing on the same base $a$. All the sides of a square are of equal length. Similarly all the sides of a rhombus are also of equal length.

Since both the square and rhombus stands on the same base $a,$
length of each side of the square $=a$
length of each side of the rhombus $=a$

Area of the square $=a^2\cdots \left(1\right)$

From the diagram, $\sin \theta =\frac{\text{h}}{a}$
$\Rightarrow h=a\sin \theta$

Area of the rhombus
$=ah=a\times a\sin \theta =a^2\sin \theta \cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$\frac{\text{area of the square}}{\text{area of the rhombus}}=\frac{a^2}{a^2\sin \theta }=\frac{1}{\sin \theta }$

Since $0°<\theta <90°,0<\sin \theta <1$.
Therefore, area of the square is greater than that of rhombus, provided both stands on same base.

(Note that, when each angle of the rhombus is $90°$, rhombus is also a square (can be considered as special case) and in that case, areas will be equal.

13. The breadth of a rectangular field is $60\%$ of its length. If the perimeter of the field is $800$ m, find out the area of the field.
A. $37500$ m2	B. $30500$ m2
C. $32500$ m2	D. $40000$ m2

answer with explanation

Answer: Option A
Explanation:
Solution 1

Given that breadth of the rectangular field is $60\%$ of its length.
$\Rightarrow b=\frac{60l}{100}=\frac{3l}{5}$

perimeter of the field $=800$ m
$\Rightarrow 2(l+b)=800\Rightarrow 2(l+\frac{3l}{5})=800\Rightarrow l+\frac{3l}{5}=400\Rightarrow \frac{8l}{5}=400\Rightarrow \frac{l}{5}=50\Rightarrow l=5\times 50=250\text{m}$

$b=\frac{3l}{5}=\frac{3\times 250}{5}=3\times 50=150\text{m}$

Area $=lb=250\times 150=37500{\text{m}}^2$

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Solution 2

breadth $=60\%$ of length

perimeter $=800$ m
⇒ $2$(length + breadth) $=800$
⇒ $2$(length + $60\%$ of length) $=800$
⇒ $2$($160\%$ of length) $=800$
⇒ $160\%$ of length $=400$
⇒ length $=\frac{400\times 100}{160}=250\text{m}$

breadth $=\frac{250\times 60}{100}=150\text{m}$

Area $=250\times 150=37500{\text{m}}^2$

14. A room $5\text{m}$ $44\text{cm}$ long and $3\text{m}$ $74\text{cm}$ broad needs to be paved with square tiles. What will be the least number of square tiles required to cover the floor?
A. $176$	B. $124$
C. $224$	D. $186$

answer with explanation

Answer: Option A
Explanation:
length $=5$ m $44$ cm $=544$ cm
breadth $=3$ m $74$ cm $=374$ cm

Area $=544\times 374$ cm²

Now we need to find out HCF(Highest Common Factor) of $544$ and $374$.
Let's find out the HCF using long division method for quicker results.
Hence, HCF of $544$ and $374$ $=34$

Therefore, side length of largest square tile $=34$ cm

Area of each square tile $=34\times 34$ cm²

Number of tiles required
$=\frac{544\times 374}{34\times 34}=16\times 11=176$

15. The length of a rectangular plot is $20$ metres more than its breadth. If the cost of fencing the plot @Rs.$26.50$ per metre is $\text{Rs.}5300$, what is the length of the plot in metres?
A. $60$ m	B. $100$ m
C. $75$ m	D. $50$ m

answer with explanation

Answer: Option A
Explanation:
Solution 1

Length of the fence $=\frac{5300}{26.50}=200\text{m}$

⇒ $2$(length + breadth)$=200\text{m}$
⇒ $2$(breadth + $20$ + breadth)$=200\text{m}$(∵ length = breadth$+20$)
⇒ breadth + $20$ + breadth $=100\text{m}$
⇒ breadth $=40$ m

length $=40+20=60$ m

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Solution 2

Length of the plot is $20$ metres more than its breadth.
Hence, let's take the length as $l$ metre and breadth as $(l-20)$ metre.

Length of the fence = perimeter
$=2$(length + breadth)
$=2[l+(l-20\left)\right]$
$=2(2l-20)$ metre

Cost per meter = Rs.$26.50$
Total cost $=2(2l-20)\times 26.50$

Total cost is given as Rs.$5300$
$\Rightarrow 2(2l-20)\times 26.50=5300\Rightarrow (2l-20)\times 26.50=2650\Rightarrow (l-10)\times 26.50=1325\Rightarrow (l-10)=\frac{1325}{26.50}=50\Rightarrow l=50+10=60\text{metre}$