Exercise 3.2 - Chapter 3 - Perimeter and Area - Term 3 - 6th Maths Guide Samacheer Kalvi Solutions

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Miscellaneous Practice Problems
Question $1 .$
A piece of wire is $36 \mathrm{~cm}$ long. What will be the length of each side if we form
i) a square
ii) an equilateral triangle.
Solution:
Given the length of the wire $=36 \mathrm{~cm}$
i) When a square is formed out of it
The perimeter of the square $=36 \mathrm{~cm}$
$4 \times$ side $=36$
side $=\frac{36}{4}=9 \mathrm{~cm}$
Side of the square
ii) When an equilateral triangle is formed out of it, its perimeter $=36 \mathrm{~cm}$
i.e., side $+$ side $+$ side $=36 \mathrm{~cm}$.
$3 \times$ side $=36 \mathrm{~cm}$
side $=\frac{36}{3}=12 \mathrm{~cm}$
One side of an equilateral triangle $=12 \mathrm{~cm}$
 

Question $2 .$
From one vertex of an equilateral triangle with side $40 \mathrm{~cm}$, an equilateral triangle with $6 \mathrm{~cm}$ side is removed. What is the perimeter of the remaining portion?
Solution:

If an equilateral triangle of side $6 \mathrm{~cm}$ is removed the perimeter $=(40+34+6+34) \mathrm{cm}=114 \mathrm{~cm}$ Perimeter of the remaining portion $=114 \mathrm{~cm}$
 

Question $3 .$
Rahim and peter go for a morning walk, Rahim walks around a square path of side $50 \mathrm{~m}$ and Peter walks around a rectangular path with length $40 \mathrm{~m}$ and breadth $30 \mathrm{~m}$. If both of them walk 2 rounds each, who covers more distance and by how much?
Solution:
Distance covered by Rahim
$=50 \times 4 \mathrm{~m}$
$=200 \mathrm{~m}$
If he walks 2 rounds, distance covered $=2 \times 200 \mathrm{~m}$
$=400 \mathrm{~m}$
Distance covered by peter
$=2(40+30) \mathrm{m}$
$=2(70) \mathrm{m}$
$=140 \mathrm{~m}$
If he walks 2 rounds, distance covered $=2 \times 140 \mathrm{~m}$
$=280 \mathrm{~m}$
$\therefore$ Rahim covers more distance by $(400-280)=120 \mathrm{~m}$
 

Question $4 .$
The length of a rectangular park is $14 \mathrm{~m}$ more than its breadth. If the perimeter of the park is 200 $\mathrm{m}$, what is its length? Find the area of the park?
Solution:
Given length of rectangular park is $14 \mathrm{~m}$ more than its breadth.
Let the breadth be b $\mathrm{m}$.
$\therefore$ Length of the park will be $\mathrm{l}=\mathrm{b}+14 \mathrm{~m}$
Given perimeter $=200 \mathrm{~m}$
$2 \times(1+b)=200 \mathrm{~m}$
$2 \times(b+14+b)=200 m[\because 1=b+14]$
$2 \times(2 b+14)=200 m$

$\begin{aligned}
&2 \mathrm{~b}+14=\frac{200}{2} \mathrm{~m} \\
&2 \mathrm{~b}+14=100 \mathrm{~m} \\
&2 \mathrm{~b}=100-14 \mathrm{~m} \\
&2 \mathrm{~b}=86 \mathrm{~m} \\
&\mathrm{~b}=\frac{86}{2} \mathrm{~m} \\
&\mathrm{~b} 43 \mathrm{~m} \\
&\text { Length Length of the park }=57 \mathrm{~m} \\
&\text { Area of a rectangle }=(\text { length } \times \text { breadth }) \text { unit }^{2} \\
&=(57 \times 43) \mathrm{m}^{2}=2,451 \mathrm{~m}^{2} \\
&\text { Area of the park }=2,451 \mathrm{~m}^{2}
\end{aligned}$

Question 5 .
Your garden is in the shape of a square of side $5 \mathrm{~m}$. Each side is to be fenced with 2 rows of wire. Find how much amount is needed to fence the garden at Rs 10 per meter.
Solution:
$\mathrm{a}=5 \mathrm{~m}$
Perimeter of the garden
$=4$ a units
$=4 \times 5 \mathrm{~m}$
$=20 \mathrm{~m}$
For 1 row
Amount needed to fence $1 \mathrm{~m}=$ Rs 10
Amount needed to fence $20 \mathrm{~m}$
$=\operatorname{Rs} 10 \times 20$
$=\operatorname{Rs} 200$
For 2 rows
Total amount needed $=2 \times$ Rs 200
$=\operatorname{Rs} 400$

Question $6 .$
A closed shape has 20 equal sides and one of its sides is $3 \mathrm{~cm}$. Find its perimeter.
Solution:
Number of equal sides in the shape $=20$
One of its side $=3 \mathrm{~cm}$
Perimeter $=$ length of one side $\times$ Number of equal sides
$\therefore$ Perimeter $=(3 \times 20) \mathrm{cm}=60 \mathrm{~cm}$
$\therefore$ Perimeter $=60 \mathrm{~cm}$
 

Question $7 .$
A rectangle has length $40 \mathrm{~cm}$ and breadth $20 \mathrm{~cm}$. How many squares with side $10 \mathrm{~cm}$ can be
formed from it.
Solution:
$\mathrm{l}=40 \mathrm{~cm}, \mathrm{~b}=20 \mathrm{~cm}$
Area of the rectangle $=1 \times b$ sq units
$=40 \times 20 \mathrm{~cm}^{2}$
$=800 \mathrm{~cm}^{2}$
$\mathrm{a}=10 \mathrm{~cm}$
Area of the square $=a \times a$ sq. units
$=10 \times 10 \mathrm{~cm}^{2}$
$=100 \mathrm{~m}^{2}$
No of squares formed $=\frac{800}{100} \mathrm{~cm}^{2}$
$=8$

Question $8 .$
The length of a rectangle is three times its breadth. If its perimeter is $64 \mathrm{~cm}$, find the sides of the rectangle.
Solution:
Given perimeter of a rectangle $=64 \mathrm{~cm}$
Also given length is three times its breadth.
Let the breadth of the rectangle $=\mathrm{b} \mathrm{cm}$
$\therefore$ Length $=3 \times \mathrm{bcm}$
Perimeter $=64 \mathrm{~m}$
i.e., $2 \times(1+b)=64 \mathrm{~m}$
$2 \times(3 b+b)=64 \mathrm{~m}$
$2 \times 4 b=64 m$
$4 \mathrm{~b}=\frac{64}{2}=32 \mathrm{~m}$
$\mathrm{b}=\frac{32}{4}=8 \mathrm{~m}$
$\mathrm{l}=3 \times \mathrm{b}=3 \times 8=24 \mathrm{~m}$
$\therefore$ Breadth of the rectangle $=8 \mathrm{~m}$
Length of the rectangle $=24 \mathrm{~m}$
 

Question $9 .$
How many different rectangles can be made with a $48 \mathrm{~cm}$ long string? Find the possible pairs of length and breadth of the rectangles.
Solution:
12 rectangles
$(1,23),(2,22),(3,21),(4,20),(5,19),(6,18),(7,17),(8,16),(9,15),(10,14),(11,13),(12,12)$

Question $10 .$
Draw a square $\mathrm{B}$ whose side is twice of the square $\mathrm{A}$. Calculate the perimeters of the squares $\mathrm{A}$ and B.
Solution:

Perimeter of $\mathrm{A}=\mathrm{s}+\mathrm{s}+\mathrm{s}+\mathrm{s}$ units $=4 \mathrm{~s}$ units
Perimeter of B $=(2 \mathrm{~s}+2 \mathrm{~s}+2 \mathrm{~s}+2 \mathrm{~s})$ units
$=8 \mathrm{~s}$ units $=2$ (4 $\mathrm{s}$ ) units.
$\therefore$ Perimeter of B is twice perimeter of A

Question $11 .$
What will be the area of a new square formed if the side of a square is made one - fourth?
Solution:
Area of the new square is reduced to $\frac{1}{16}$ th times to that of the original area.
 

Question $12 .$
Two plots have the same perimeter. One is a square of side $10 \mathrm{~m}$ and another is a rectangle of breadth $8 \mathrm{~m}$. Which plot has the greater area and by how much?
Solution:

Given perimeter of square = perimeter of rectangle
$4 \times$ side $=2$ (length + breadth)
$(4 \times 10) \mathrm{m}=2(\mathrm{l}+8) \mathrm{m}$
$\frac{4 \times 10}{2}=1+8$
$20=1+8$
$\begin{aligned}
&\mathrm{I}=20-8 \\
&\mathrm{I}=12 \mathrm{~m}
\end{aligned}$
$\begin{aligned}
&\mathrm{I}=12 \mathrm{~m} \\
&\therefore \text { length of }
\end{aligned}$

Area of the square plot $-$ side $\times$ side $=10 \times 10 \mathrm{~m}^{2}=100 \mathrm{~m}^{2}$
Area of the rectangular plot $=$ length $\times$ breadth $=(12 \times 8) \mathrm{m}^{2}=96 \mathrm{~m}^{2}$ $100 \mathrm{~m}^{2}>96 \mathrm{~m}^{2}$
$\therefore$ Square plot has greater area by $4 \mathrm{~m}^{2}$
 

Question $13 .$
Look at the picture of the house given and find the total area of the shaded portion.

Solution:
Total area of the shaded region = Area of a right triangle + Area of a rectangle $=\left(\frac{1}{2} \times \mathrm{b} \times \mathrm{h}\right)+(\mathrm{l} \times \mathrm{b}) \mathrm{cm}^{2}$
$\begin{aligned}
&=\left(\frac{1}{2} \times b \times h\right)+(1 \times b) \mathrm{cm}^{2} \\
&=\left[\left(\frac{1}{2} \times 3 \times 4\right)+(9 \times 6)\right] \mathrm{cm}^{2} \\
&=(6+54) \mathrm{cm}^{2}=60 \mathrm{~cm}^{2}
\end{aligned}$
 

Question $14 .$
Find the approximate area of the flower in the given square grid.

Solution:
Approximate area $=$ Number of full squares $+$ Number of more than half squares $+\frac{1}{2} \times$ Number of half squares.
$=10+5+\left(\frac{1}{2} \times 1\right) \mathrm{Sq}$ units. $=10+5+\frac{1}{2}$ sq. units
$=15 \frac{1}{2} \mathrm{sq} .$ units $=15.5 \mathrm{sq}$. units.