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Additional Questions - Chapter 2 Measurements Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]


Updated By SaraNextGen
On April 24, 2024, 11:35 AM

Additional Questions and Answers
Exercise $2.1$
Question $1 .$

In the following figure, $P Q R S$ is a parallelogram find $x$ and $y$.

Solution:
We know that in a parallelogram opposite sides are equal.
$\begin{aligned}
&\therefore 3 \mathrm{x}=18 \\
&\mathrm{x}=\frac{18}{3} \\
&\mathrm{x}=6 \text { and } \\
&3 \mathrm{y}-1=26 \\
&3 \mathrm{y}=26+1=27 \\
&\mathrm{y}=\frac{27}{9} \\
&\mathrm{y}=9
\end{aligned}$

 

Question $2 .$
Two adjacent sides of a parallelogram are $5 \mathrm{~cm}$ and $7 \mathrm{~cm}$ respectively. Find its perimeter.

Solution:

Perimeter $=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}+\mathrm{AD}[\because \mathrm{AB}=\mathrm{DC} \& \mathrm{AD}=\mathrm{BC}]$
$=7 \mathrm{~cm}+5 \mathrm{~cm}+7 \mathrm{~cm}+5 \mathrm{~cm}=24 \mathrm{~cm}$

 

Question $3 .$
The perimeter of a parallelogram is $150 \mathrm{~cm}$. One of its sides is greater than the other by $25 \mathrm{~cm}$. Find the length of the sides of the parallelogram.
Solution:

Given perimeter $=150 \mathrm{~cm}$
Let one side of the parallelogram be ' $\mathrm{b}$ ' $\mathrm{cm}$ Then the other side $=b+25 \mathrm{~cm}$
$\mathrm{b}+(\mathrm{b}+25)+\mathrm{b}+(\mathrm{b}+25)=150$
$b+b+25+b+b+25=150$
$4 \mathrm{~h}+50=150=4 \mathrm{~b}=100$
$\mathrm{b}=\frac{100}{4}=25$
$\therefore$ One side $\mathrm{b}=25 \mathrm{~cm}$
Other side $\mathrm{b}+25=50 \mathrm{~cm}$

 

Exercise 2.2

Question 1.

ABCD is a rhombus. Find x, y, and z.

Solution:
We know that all sides of rhombus are equal and its diagonals bisect each other. $\therefore \mathrm{x}=5, \mathrm{y}=12$ and $\mathrm{z}=13$.
 

Question $2 .$
Find the altitude of the rhombus whose area is $315 \mathrm{~cm}^{2}$ and its perimeter is $180 \mathrm{~cm}$.
Solution:
Given perimeter of the rhombus $=180 \mathrm{~cm}$
$\therefore$ One side of the rhombus $=\frac{180}{4}=45 \mathrm{~cm}$
Given area of the rhombus $=315 \mathrm{~cm}^{2}$
$\mathrm{b} \times \mathrm{h}=315$
$45 \times \mathrm{h}=315=\frac{315}{45}$
h $-7 \mathrm{~cm}$
Altitude of the rhombus $=7 \mathrm{~cm}$
 

Question $3 .$
The floor of a building consists of 2000 titles which are rhombus shaped and each of its diagonals are $40 \mathrm{~cm}$ and $25 \mathrm{~cm}$. Find the total cost of polishing the floor, if the cost per $\mathrm{m}^{2}=₹ 5$.
Solution:
Area of each title $=\frac{1}{2} \times \mathrm{d}_{1} \times \mathrm{d}_{2}$ sq. units
$=\frac{1}{2} \times 40 \times 25 \mathrm{~cm}^{2}=500 \mathrm{~cm}^{2}$
$\therefore$ Area of 2000 titles $=500 \times 20,000=10,000 \mathrm{~cm}^{2}=100 \mathrm{~m}^{2}$
Cost of polishing $1 \mathrm{~m}^{2}=₹ 5$
$\therefore$ Cost of polishing $100 \mathrm{~m}^{2}=5 \times 100=₹ 500$

Also Read : Exercise-3.1-Chapter-3-Algebra-Term-1-7th-Maths-Guide-Samacheer-Kalvi-Solutions

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