Problems on Area - Solved Examples(Set 6)

Problems on Area - Solved Examples(Set 6)

26. Two diagonals of a rhombus are $72$ cm and $30$ cm respectively. What is its perimeter?
A. $136$ cm	B. $156$ cm
C. $144$ cm	D. $121$ cm

answer with explanation

Answer: Option B
Explanation:

Remember the following two properties of a rhombus which will be useful in solving this question
1. All the sides of a rhombus are congruent.
2. The diagonals of a rhombus bisect each other at right angles.

Let the diagonals be PR and SQ
Let PR $=72$ cm, SQ $=30$ cm

PO = OR = $\frac{72}{2}=36$ cm

SO = OQ = $\frac{30}{2}=15$ cm

PQ = QR = RS = SP
$=\sqrt{{36}^2+{15}^2}=\sqrt{1296+225}=\sqrt{1521}=39\text{cm}$

perimeter $=4\times 39=156$ cm

27. The base of a parallelogram is $(p+4)$, altitude to the base is $(p-3)$ and the area is $(p^2-4)$,find out its actual area.
A. $40$ sq. units	B. $54$ sq. units
C. $36$ sq. units	D. $60$ sq. units

answer with explanation

Answer: Option D
Explanation:

Area of a parallelogram, A $=bh$
where $b$ is the base and $h$ is the height of the parallelogram

Hence, we have
$p^2-4=(p+4)(p-3)\Rightarrow p^2-4=p^2+p-12\Rightarrow -4=p-12\Rightarrow p=12-4=8$

Hence, actual area $=(p^2-4)$
$=8^2-4=64-4=60$ sq. units

28. A circle is inscribed in an equilateral triangle of side $24$ cm, touching its sides. What is the area of the remaining portion of the triangle?
A. $144\sqrt{3}-48\pi$ cm2	B. $121\sqrt{3}-36\pi$ cm2
C. $144\sqrt{3}-36\pi$ cm2	D. $121\sqrt{3}-48\pi$ cm2

answer with explanation

Answer: Option A
Explanation:
Solution 1

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}{a}^2$
where $a$ is length of one side of the equilateral triangle

Area of the equilateral Δ ABC
$=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times {24}^2=144\sqrt{3}{\text{cm}}^2\cdots \left(1\right)$

Area of a triangle $=\frac{1}{2}bh$
where $b$ is the base and $h$ is the height of the triangle

Let $r$ = radius of the inscribed circle. Then,
Area of Δ ABC
= Area of Δ OBC + Area of Δ OCA + area of Δ OAB
$=(\frac{1}{2}\times r\times \text{BC})+(\frac{1}{2}\times r\times \text{CA})$ $+(\frac{1}{2}\times r\times \text{AB})$
$=\frac{1}{2}\times r\times \text{(BC + CA + AB)}=\frac{1}{2}\times r\times (24+24+24)=\frac{1}{2}\times r\times 72=36r{\text{cm}}^2\cdots \left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,
$144\sqrt{3}=36r\Rightarrow r=\frac{144\sqrt{3}}{36}=4\sqrt{3}\cdots \left(3\right)$

Area of a circle $=\pi r^2$
where $r$ = radius of the circle

From $\left(3\right)$, area of the inscribed circle
$=\pi r^2=\pi {\left(4\sqrt{3}\right)}^2=48\pi \cdots \left(4\right)$

Hence, area of the remaining portion of the triangle
= Area of Δ ABC – Area of inscribed circle
$=144\sqrt{3}-48\pi {\text{cm}}^2$

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Solution 2

Radius of incircle of an equilateral triangle of side $a$
$=\frac{a}{2\sqrt{3}}$

Radius of the circle inscribed
$=\frac{24}{2\sqrt{3}}=\frac{12}{\sqrt{3}}\text{cm}$

Area of the circle inscribed
$=\pi {\left(\frac{12}{\sqrt{3}}\right)}^2=\pi \times \frac{144}{3}=48\pi {\text{cm}}^2$

Area of the equilateral triangle
$=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\times {24}^2=144\sqrt{3}{\text{cm}}^2$

Area of the remaining portion of the triangle
= Area of the equilateral triangle – Area of inscribed circle
$=144\sqrt{3}-48\pi {\text{cm}}^2$

29. A rectangular plot measuring $90$ metres by $50$ metres needs to be enclosed by wire fencing such that poles of the fence will be kept $5$ metres apart. How many poles will be needed?
A. $30$	B. $44$
C. $56$	D. $60$

answer with explanation

Answer: Option C
Explanation:

Perimeter of a rectangle $=2(l+b)$
where $l$ is the length and $b$ is the breadth of the rectangle.

Length of the wire fencing
= perimeter $=2(90+50)=280$ metre

Two poles will be kept $5$ metres apart. The poles will be placed along the perimeter of the rectangular plot, not in a single straight line which is very important.

Hence, number of poles required
$=\frac{280}{5}=56$

30. If the diagonals of a rhombus are $24$ cm and $10$ cm, what will be its perimeter?
A. $42$ cm	B. $64$ cm
C. $56$ cm	D. $52$ cm

answer with explanation

Answer: Option D
Explanation:


Let the diagonals be PR and SQ.
Let PR $=24$ cm and SQ $=10$ cm

PO = OR $=\frac{24}{2}=12$ cm
SO = OQ $=\frac{10}{2}=5$ cm

PQ = QR = RS = SP
$=\sqrt{{12}^2+5^2}=\sqrt{144+25}=\sqrt{169}=13\text{cm}$

perimeter $=4\times 13=52$ cm

31. What will be the length of the longest rod which can be placed in a box of $80$ cm length,$40$ cm breadth and $60$ cm height?
A. $\sqrt{11600}$ cm	B. $\sqrt{14400}$ cm
C. $\sqrt{10000}$ cm	D. $\sqrt{12040}$ cm

answer with explanation

Answer: Option A
Explanation:
Solution 1

Length of the longest rod that can be placed in a box
of length $l$, breadth $b$ and height $h$

$=\sqrt{l^2+b^2+h^2}$

Therefore, length of the longest rod
$=\sqrt{{80}^2+{40}^2+{60}^2}=\sqrt{11600}$ cm

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Solution 2

The longest road which can fit into the box will have one end at A and other end at G (or any other similar diagonal).
Hence the length of the longest rod = AG

Initially let's find out AC. Consider the right angled triangle ABC

AC² = AB² + BC²
$\Rightarrow$ AC² $={40}^2+{80}^2=1600+6400=8000$
$\Rightarrow \text{AC =}\sqrt{8000}\text{cm}$

Consider the right angled triangle ACG

AG² = AC² + CG²
$={\left(\sqrt{8000}\right)}^2+{60}^2=8000+3600=11600$

⇒ AG = $\sqrt{11600}$ cm
⇒ Length of the longest rod $=\sqrt{11600}$ c
m