Exercise 3.1 - Chapter 3 Perimeter and Area Term 3 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.1$
Question $1 .$
The table given below contains some measures of the rectangle. Find the unknown values.

Solution:
(i) Area of the rectangle $=($ length $\times$ breadth $)$ sq unit. Perimeter of a rectangle $=2(1+b)$ units.
$\mathrm{l}=5 \mathrm{~cm}$
$\mathrm{b}=8 \mathrm{~cm}$
$\therefore \mathrm{p}=2(\mathrm{l}+\mathrm{b}) \mathrm{cm}=2(5+8) \mathrm{cm}=2 \times 13 \mathrm{~cm}$
$\mathrm{p}=26 \mathrm{~cm}$
Area $=(1 \times b) \mathrm{cm}^{2}=(5 \times 8) \mathrm{cm}^{2}$
$A=40 \mathrm{~cm}^{2}$
(ii) $\mathrm{l}=13 \mathrm{~cm}$
$\mathrm{p}=54 \mathrm{~cm}$
Perimeter $=2(1+b)$ units
$54=2(13+b) \mathrm{cm}$
$\frac{54}{2}=13+b$
$27=13+b$

\begin{aligned}
&\mathrm{b}=27-13 \\
&\mathrm{~b}=14 \mathrm{~cm} \\
&\text { Area }=1 \times \mathrm{b} \text { sq. unit }=13 \times 14 \mathrm{~cm}^{2} \\
&\mathrm{~A}=182 \mathrm{~cm}^{2} \\
&\text { (iii) } \mathrm{b}=15 \mathrm{~cm} \\
&\mathrm{p}=60 \mathrm{~cm} \\
&\mathrm{p}=2(\mathrm{l}+\mathrm{b}) \text { units } \\
&60=2(\mathrm{l}+15) \mathrm{cm} \\
&\frac{60}{2}=1+15 \\
&30=1+15 \\
&\mathrm{l}=30-15 \\
&\mathrm{l}=15 \mathrm{~cm} \\
&\text { Area }=1 \times \mathrm{b} \text { unit }^{2}=15 \times 15 \mathrm{~cm}^{2}=225 \mathrm{~cm}^{2} \\
&\mathrm{~A}=225 \mathrm{~cm}^{2} \\
&\text { (iv) } \mathrm{l}=10 \mathrm{~m} \\
&\text { Area }=120 \mathrm{sq} \text { metre } \\
&\text { Area }=1 \times \mathrm{b} \text { sq.m } \\
&120=10 \times 6 \\
&\mathrm{~b}=\frac{120}{10} \\
&\mathrm{~b}=12 \mathrm{~m} \\
&\text { Perimeter }=2(\mathrm{l}+\mathrm{b}) \text { units }=2(10+12) \text { units }=2 \times 22 \mathrm{~m} \\
&\mathrm{~A}=44 \mathrm{~m}
\end{aligned}

(v) $\mathrm{b}=4$ feet.
Area $=20$ sq. feet
Area $=1 \times$ b sq .feet
$20=1 \times 4$
$\mathrm{l}=\frac{20}{4}$ feet
$1=5$ feet
Perimeter $=2(1+b)$ units.
$\mathrm{p}=2(5+4)$ feet $=2 \times 9$
$\mathrm{p}=18$ feet

Question $2 .$
The table given below contains some measures of the square. Find the unknown values.

Solution:
Perimeter of a square $=(4 \times$ side $)$ units
Area of a square $=($ side $\times$ side $)$ unit $^{2}$
(i) $\mathrm{s}=6 \mathrm{~cm}$
Perimeter $=4 \mathrm{~s}$ units $=4 \times 6 \mathrm{~cm}=24 \mathrm{~cm}$
$\mathrm{P}=24 \mathrm{~cm}$
Area $=\mathrm{s} \times \mathrm{s}$ unit $^{2}=6 \times 6 \mathrm{~cm}^{2}=36 \mathrm{~cm}^{2}$ $\mathrm{~A}=36 \mathrm{~cm}^{2}$
(ii) Perimeter $=4 \times \mathrm{s}$ unit $100=(4 \times \mathrm{s}) \mathrm{m}$ $\frac{100}{4}=\mathrm{s}$ $\mathrm{s}=25 \mathrm{~m}$ Area $=\mathrm{s} \times \mathrm{s}$ unit $^{2}=25 \times 25 \mathrm{~m}^{2}=625 \mathrm{~m}^{2}$ $\mathrm{~A}=625 \mathrm{~m}^{2}$
$100=(4 \times \mathrm{s}) \mathrm{m}$
$\begin{aligned}
&\frac{100}{4}=\mathrm{s} \\
&\mathrm{s}=25 \mathrm{~m} \\
&\text { Area }=\mathrm{s} \times \mathrm{s} \text { unit }^{2}=25 \times 25 \mathrm{~m}^{2}=625 \mathrm{~m}^{2} \\
&\mathrm{~A}=625 \mathrm{~m}^{2}
\end{aligned}$

(iii) Area $=\mathrm{s} \times \mathrm{s}$ unit ${ }^{2}$
$49=\mathrm{s} \times \mathrm{s}$ square feet
$\begin{aligned}
&\mathrm{s}^{2}=7^{2} \\
&\mathrm{~s}=7 \text { feet }
\end{aligned}$
Perimeter $=4 \times \mathrm{s}$ unit $=4 \times 7 \mathrm{feet}=28 \mathrm{feet}$
Perimeter $=28 \mathrm{feet}$

Question $3 .$
The table given below contains some measures of the triangle. Find the unknown values.

Solution: Area of the right triangle $=\frac{1}{2} \times($ base $\times$ height $) \mathrm{unit}^{2}$
(i) $\mathrm{b}=20 \mathrm{~cm}$
$\mathrm{h}=40 \mathrm{~cm}$
Area $=\frac{1}{2}(b \times h) \mathrm{cm}^{2}=\frac{1}{2} \times 20 \times 40=400 \mathrm{~cm}^{2}$
$\mathrm{A}=400 \mathrm{~cm}^{2}$
(ii) $b=5$ feet
Area $=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}$ unit $^{2}$
$=20=\frac{1}{2} \times 5 \times \mathrm{h}$ sq. feet
$\frac{20 \times 2}{5}=\mathrm{h}$
$\mathrm{h}=8$ feet
(iii) Area $=\frac{1}{2} \times($ base $\times$ height $)$ unit $^{2}$
$24=\frac{1}{2} \times \mathrm{b} \times 12 \mathrm{~m}^{2}$

base $=\frac{24 \times 2}{12} \mathrm{~m}=4 \mathrm{~m}$
Base $=4 \mathrm{~m}$

Question $4 .$
The table given below contains some measures of the triangles. Find the unknown values.

Solution:

Perimeter of a triangle $=$ sum of three sides.
(i) Perimeter $=6+5+2 \mathrm{~cm}=13 \mathrm{~cm}$ $\mathrm{p}=13 \mathrm{~cm}$
(ii) Perimeter $=($ side $1+$ side $2+$ side 3$) \mathrm{m}$ $17=($ side $1+8+3) \mathrm{m}$ $17 \mathrm{~m}=($ side $1+11) \mathrm{m}$ side $1=17-11=6 \mathrm{~m}$
(iii) Perimeter $=$ side $1+$ side $2+$ side 3 28 feet $=11$ feet $+$ side $2+9$ feet $28 \mathrm{ft}=20 \mathrm{feet}+$ side 2
$28-20=$ side 2
side $=8$ feet

Question $5 .$
Fill in the blanks.
i) $5 \mathrm{~cm}^{2}=\mathrm{mm}^{2}$
Hint: $1 \mathrm{~cm}^{2}=100 \mathrm{~mm}^{2}$
ii) $26 \mathrm{~m}^{2}=\mathrm{cm}^{2}$
Hint: $1 \mathrm{~m}^{2}=10000$
iii) $8 \mathrm{~km}^{2}=\mathrm{m}^{2}$
Hint $41 \mathrm{~km}^{2}-1000000 \mathrm{~m}^{2}$
Solution:
(i) 500
(ii) $2,60,000$
(iii) $80,00,000$

Question $6 .$
Find the perimeter and area of the following shapes.

Solution:

(i) Perimeter $=(4+4+4+4+4+4+4+4+4+4+4+4) \mathrm{cm}=48 \mathrm{~cm}$ Perimeter $=48 \mathrm{~cm}$
Area of 5 squares of side $4 \mathrm{~cm}$
Area of a square $=($ side $\times$ side $)$ unit $^{2}$
$\therefore \mathrm{A}=5 \times(4 \times 4) \mathrm{cm}^{2}=5 \times 16 \mathrm{~cm} 2=80 \mathrm{~cm}^{2}$
$80 \mathrm{~cm}^{2}$
(ii) Perimeter $=(4+5+4+5+4+5+4+5) \mathrm{cm}=36 \mathrm{~cm}$
Perimeter $=36 \mathrm{~cm}$
Area of a square of side $3 \mathrm{~cm}+$ Area of 4 right triangles
$=(3 \times 3)+\left[4 \times \frac{1}{2} \times 4 \times 3\right] \mathrm{cm}^{2}=(9+24) \mathrm{cm}^{2}=33 \mathrm{~cm}^{2}$
Area $=33 \mathrm{~cm}^{2}$
(iii) Perimeter $=(50+12+13+40+10+10+10+5) \mathrm{cm}=150 \mathrm{~cm}$
Perimeter $=150 \mathrm{~cm}$
Area $=$ Area of a rectangle $+$ Area of a square $+$ Area of a right triangle. $=(1 \times b)+(s \times s)+\left(\frac{1}{2} \times b \times h\right) \mathrm{cm}^{2}$
$\left.=(50 \times 5)+(10 \times 10)+\frac{1}{2} \times 12 \times 5\right) \mathrm{cm}^{2}$
$=(250+100+30) \mathrm{cm}^{2}=380 \mathrm{~cm}^{2}$
Area $=380 \mathrm{~cm}^{2}$

Question $7 .$
Find the perimeter and area of the rectangle whose length is $6 \mathrm{~m}$ and breadth $4 \mathrm{~m}$. Solution:
$\mathrm{I}=6 \mathrm{~m}, \mathrm{~b}=4 \mathrm{~m}$ Perimeter of the rectangle
$=2(1+b)$ units
$=2(6+4) \mathrm{m}$
$=2(10) \mathrm{m}$
$=20 \mathrm{~m}$
Area of the rectangle $=1 \times \mathrm{b}$ sq units
$=4 \times 6 \mathrm{~m}^{2}$
$=24 \mathrm{~m}^{2}$
 

Question $8 .$
Find the perimeter and the area of the square whose side is $8 \mathrm{~cm}$.
Solution:
Perimeter of a square $=(4 \times$ side $)$ units
Side $=8 \mathrm{~cm}$
$\therefore$ Perimeter $=4 \times 8 \mathrm{~cm}=32 \mathrm{~cm}$
Perimeter $=32 \mathrm{~cm}$
Area of a square $=($ side $\times$ side $)$ unit $^{2}=(8 \times 8) \mathrm{cm}^{2}=64 \mathrm{~cm}^{2}$
Area $=64 \mathrm{~cm}^{2}$

Question $9 .$
Find the perimeter and area of the right angled triangle whose sides are 6 feet, 8 feet and 10 feet. Solution:
Perimeter of the triangle
$=(\mathrm{a}+\mathrm{b}+\mathrm{c})$ units
$=(6+8+10)$ feet
$=24$ feet
Area of the triangle $=\frac{1}{2} \times \mathrm{b} \times \mathrm{h}$ squits
$\frac{1}{2} \times 6^{3} \times 8$ feet square
$=24 \mathrm{sq} . \mathrm{feet}$
 

Question 10 .
Find the perimeter of
i) A scalene triangle with sides $7 \mathrm{~m}, 8 \mathrm{~m}, 10 \mathrm{~m}$
ii) An isosceles triangle with equal sides $10 \mathrm{~cm}$ each and third side is $7 \mathrm{~cm}$.
iii) An Equilateral triangle with side $6 \mathrm{~cm}$.
Solution:
i) Perimeter of a scalene triangle $=(7+8+10) \mathrm{m}=25 \mathrm{~m}$
ii) The three sides of the isosceles triangle are $10 \mathrm{~cm}, 10 \mathrm{~cm}$ and $7 \mathrm{~cm}$
$\therefore$ Perimeter $=(10+10+7) \mathrm{cm}=27 \mathrm{~cm}$
iii) An equilateral triangle with side $6 \mathrm{~cm}$.
The sides of equilateral triangle are $6 \mathrm{~cm}, 6 \mathrm{~cm}$ and $6 \mathrm{~cm}$
$\therefore$ Perimeter $=(6+6+6) \mathrm{cm}=18 \mathrm{~cm}$

Question $11 .$
The area of a rectangular shaped photo is $820 \mathrm{sq} . \mathrm{cm}$. and its width is $20 \mathrm{~cm}$. What is its length?
Also find its perimeter.
Solution:
Given Area $=820 \mathrm{~cm}^{2}$
Width $=20 \mathrm{~cm}$
Area of the rectangle
$=1 \times$ b sq. units
$820=1 \times 20$
$\frac{820}{20}=1$
$41=1$ length $1=41 \mathrm{~cm}$
Perimeter $=2(1+b)$ units
$=2(41+20) \mathrm{cm}$
$=2(61) \mathrm{cm}$
$=122 \mathrm{~cm}$
 

Question $12 .$
A square park has $40 \mathrm{~m}$ as its perimeter. What is the length of its side? Also find its area.
Solution:
Given perimeter $=40 \mathrm{~m}$
Perimeter of a square $=4 \times$ Length of a side
$40=4 \times$ Length of a side
$\therefore$ Length of its side $=\frac{40}{4} \mathrm{~m}=0 \mathrm{~m}$
$\therefore$ Side of the park $=10 \mathrm{~m}$
Area of a square $=($ Side $\times$ side $)$ unit $^{2}=(10 \times 10) \mathrm{m}^{2}=100 \mathrm{~m}^{2}$
$\therefore$ Area of the Park $=100 \mathrm{~m}^{2}$

Question $13 .$
The scalene triangle has $40 \mathrm{~cm}$ as its perimeter and whose two sides are $13 \mathrm{~cm}$ and $15 \mathrm{~cm}$, find the third side.
Solution:
Let the third side be $\mathrm{C}$
perimeter $=(a+b+c)$ units
$40=13+15+\mathrm{C}$
$40=28+\mathrm{C}$
$\mathrm{C}=40-28$
$\mathrm{C}=12$ units
$\mathrm{C}=12 \mathrm{~cm}$
 

Question $14 .$
A field is in the shape of right angled triangle whose base is $25 \mathrm{~m}$ and height $20 \mathrm{~m}$. Find the cost of
levelling the field at the rate of $₹ 45 /-$ per sq. $\mathrm{m}$.
Solution:
Area of a right angled triangle $=\frac{1}{2} \times($ base $\times$ height $) u n i t^{2}$
base $=25 \mathrm{~m}$
height $=20 \mathrm{~m}$
$\therefore$ Area $=\frac{1}{2} \times(25 \times 20)$
Area $=250 \mathrm{~m}^{2}$
Cost of levelling per $\mathrm{m}^{2}=₹ 45$.
$\therefore$ Cost of levelling $250 \mathrm{~m}^{2}=250 \times 45=₹ 11,250$
Cost of levelling $=₹ 11,250$

Question 15 .
A square of side $2 \mathrm{~cm}$ is joined with a rectangle of length $15 \mathrm{~cm}$ and breadth $10 \mathrm{~cm}$. Find the perimeter of the combined shape.
Solution:
Perimeter of the combined shape $=$ Lengths of the outer boundaries
$=(15+10+2+2+2+13+10) \mathrm{cm}=54 \mathrm{~cm}$
Perimeter $=54 \mathrm{~cm}$

Objective Type Questions
Question $16 .$
The following figures are of equal area. Which figure has the least perimeter?

Solution:

(b)

Question $17 .$
If two identical rectangles of perimeter $30 \mathrm{~cm}$ are joined together, then the perimeter of the new shape will be
(a) equal to $60 \mathrm{~cm}$
(b) less than $60 \mathrm{~cm}$
(c) greater than $60 \mathrm{~cm}$
(d) equal to $45 \mathrm{~cm}$
Solution:
(b) less than $60 \mathrm{~cm}$
 

Question $18 .$
If every side of a rectangle is doubled, then its area becomes times.
(a) 2
(b) 3
(c) 4
(d) 6
Solution:
(c) 4
 

Question $19 .$
The side of a square is $10 \mathrm{~cm}$. If its side is tripled, then by how many times will its perimeter increase?
(a) 2 times
(b) 4 times
(c) 6 times

(d) 3 times
SolutionL
(d) 3 times
$30 \times 4=120=3 \times 40$
 

Question 20 .
The length and breadth of a rectangular sheet of a paper are $15 \mathrm{~cm}$ and $12 \mathrm{~cm}$ respectively. A rectangular piece is cut from one of its corners. Which of the following statement is correct for the remaining sheet?
(a) Perimeter remains the same but the area changes
(b) Area remains the same but the perimeter changes
(c) There will be a change in both area and perimeter
(d) Both the area and perimeter remains the same
Solution:
(c) There will be a change in both area and perimeter