Additional Questions - 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

Additional Questions
Question $1 .$
The length of a rectangular field is twice its breadth. Ramu runs around it four times and covered a distance of $6 \mathrm{~km}$. What is the length of the field.
Solution:
Breadth of rectangular field $=b \mathrm{~m}$
Length of the field $=2 \mathrm{~b} \mathrm{~m}$
Distance covered in one round $=$ perimeter
Distance covered in 4 rounds $=4 \times$ perimeter
$\begin{aligned}
4(2 \times(l+b)) &=6 \mathrm{~km} \\
8(l+b) &=6000 \mathrm{~m} \\
8(2 b+b) &=6000 \\
8(3 b) &=6000 \\
24 b &=6000 \\
b &=\frac{6000}{24} \mathrm{~m}=250 \mathrm{~m}
\end{aligned}$
Length of the field $=2 \times 250=500 \mathrm{~m}$

Question 2 .
How many tiles whose length and breadth are $12 \mathrm{~cm}$ and $5 \mathrm{~cm}$ respectively will be needed to fit in a rectangular region whose length and breadth are $70 \mathrm{~cm}$ and $36 \mathrm{~cm}$ ?
Solution:

Given length of the rectangular region $=70 \mathrm{~cm}$
$\begin{aligned}
\text { breadth } &=36 \mathrm{~cm} \\
\text { Area }=l \times b &=70 \times 36=2520 \mathrm{sq} . \mathrm{cm} \\
\text { Area of one tile } &=12 \times 5 \mathrm{~cm}^{2}=60 \mathrm{~cm}^{2} \\
\text { Number of the required tiles } &=\frac{\text { Area of Rectangle }}{\text { Area of } 1 \text { square }}=\frac{2520}{60}=42
\end{aligned}$
42 tiles are needed

Question $3 .$
A room is $4 \mathrm{~m}$ long and $3 \mathrm{~m} 50 \mathrm{~cm}$ wide, How many square metres of carpet is needed to cover the floor of the room?
Solution:
Given length of the room $=4 \mathrm{~m}$
Breadth $=3 \mathrm{~m} 50 \mathrm{~cm}=3 \mathrm{~m}+50 \mathrm{~cm}$
$\begin{aligned}
&=3 \mathrm{~m}+50 \times \frac{1}{100} \mathrm{~m}=3 \mathrm{~m}+\frac{50}{100} \mathrm{~m} \\
&=3 \mathrm{~m}+0.50 \mathrm{~m}=3.50 \mathrm{~m}
\end{aligned}$
Carpet needed = Area of the floor $=$ length $\times$ breadth
$=4 \times 3.5 \mathrm{~m}=14$ sq. $\mathrm{m}$

Question $4 .$
Find the area of the square Whose side is $14 \mathrm{~cm}$.
Solution:
Side of the square $=14 \mathrm{~cm}$
Area of the square $=$ side $\times$ side $=(14 \times 14) \mathrm{cm}^{2}$ = $196 \mathrm{sq} \mathrm{cm} .$
 

Question 5 .
Two sides of a triangle are $12 \mathrm{~cm}$ and $14 \mathrm{~cm}$. The perimeter of the triangle is $36 \mathrm{~cm}$. What is its third side.
Solution:
Given two sides of a triangle $=12 \mathrm{~cm}$ and $14 \mathrm{~cm}$
$\therefore$ Perimeter $=36 \mathrm{~cm}$
$12 \mathrm{~cm}+14 \mathrm{~cm}+$ third side $=36 \mathrm{~cm}$
Third side $=36-26=10 \mathrm{~cm}$
$\therefore$ Length of the third side