Exercise 2.3 - Chapter 2 Measurements Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $2.3$
Question 1 .
Find the missing values.

Solution:
(i) Given Height $\mathrm{h}=10 \mathrm{~m}$; Parallel sides $\mathrm{a}=12 \mathrm{~m} ; \mathrm{b}=20 \mathrm{~m}$
Area of the Trapezium $=\frac{1}{2} \mathrm{~h}(\mathrm{a}+\mathrm{b})$ sq. units $=\frac{1}{2} \times 10 \times(12+20) \mathrm{m}^{2}$ $=(5 \times 32) \mathrm{m}^{2}=160 \mathrm{~m}^{2}$
(ii) Given the parallel sides a $=13 \mathrm{~cm} ; 6=28 \mathrm{~cm}$
Area of the trapezium $=492$ sq. $\mathrm{cm}$
$\begin{aligned}
&\frac{1}{2} \mathrm{~h}(\mathrm{a}+\mathrm{b})=492 \\
&\frac{1}{2} \times \mathrm{h} \times(13+28)=492 \\
&\mathrm{~h} \times 41=492 \times 2 \\
&\mathrm{~h}=\frac{492 \times 2}{41} \\
&\mathrm{~h}=24 \mathrm{~cm}
\end{aligned}$
(iii) Given height ' $\mathrm{h}$ ' $=19 \mathrm{~m}$; Parallel sides $\mathrm{b}=16 \mathrm{~m}$
Area of the trapezium $=323$ sq. $\mathrm{m}$
$\begin{aligned}
&\frac{1}{2} h(a+b)=323 \\
&\frac{1}{2} \times h \times(a+16)=323
\end{aligned}$

$\begin{aligned}
&\mathrm{a}+16=\frac{323 \times 2}{19}=34 \\
&\mathrm{a}=34-16=18 \mathrm{~m} \\
&\mathrm{a}=18 \mathrm{~m}
\end{aligned}$
(iv) Given the height $\mathrm{h}-16 \mathrm{~cm}$; Parallel sides $\mathrm{a}=15 \mathrm{~cm}$ Area of the trapezium $=360$ sq. $\mathrm{cm}$
$\begin{aligned}
&\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})=360 \\
&\frac{1}{2} \times 16 \times(15+6)=360 \\
&15+\mathrm{b}=\frac{360}{8}=45 \\
&\mathrm{~b}=45-15=30 \\
&\mathrm{~b}=30 \mathrm{~cm}
\end{aligned}$

Question $2 .$
Find the area of a trapezium whose parallel sides are $24 \mathrm{~cm}$ and $20 \mathrm{~cm}$ and the distance between them is $15 \mathrm{~cm}$.

Solution:
Given the parallel sides $\mathrm{a}=24 \mathrm{~cm} ; \mathrm{b}=20 \mathrm{~cm}$

Distance between $\mathrm{a}$ and $\mathrm{b}$ is ' $\mathrm{h}$ ' $=15 \mathrm{~cm}$
Area of the trapezium $=\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})$ sq. units $=\frac{1}{2} \times 15 \times(24+20) \mathrm{cm}^{2}$
$=\frac{1}{2} \times 15 \times 44=330 \mathrm{~cm}^{2}$
Area of the trapezium $=330 \mathrm{~cm}^{2}$

Question $3 .$
The area of a trapezium is 1586 sq. $\mathrm{cm}$. The distance between its parallel sides is $26 \mathrm{~cm}$. If one of the parallel sides is $84 \mathrm{~cm}$ then find the other side.
Solution:
Given one parallel side $=84 \mathrm{~cm}$. Let the other parallel side be ' $\mathrm{b}$ ' $\mathrm{cm}$.
Distance between $\mathrm{a}$ and $\mathrm{b}$ is $\mathrm{h}=26 \mathrm{~cm}$.
Area of the trapezium $=1586$ sq. $\mathrm{cm}$

Question $4 .$
The area of a trapezium is 1080 sq. $\mathrm{cm}$. If the lengths of its parallel sides are $55.6 \mathrm{~cm}$ and $34.4 \mathrm{~cm}$. Find the distance between them.
Solution:
Length of the parallel sides $\mathrm{a}=55.6 \mathrm{~cm} ; \mathrm{b}=34.4 \mathrm{~cm}$
Area of the trapezium $=1080$ sq. $\mathrm{cm}$
$\frac{1}{2} \times h \times(a+b)=1080$
$\begin{aligned}
\frac{1}{2} \times h \times(55.6+34.4) &=1080 \\
\frac{1}{2} \times h \times 90.0 &=1080 \\
h &=\frac{1080}{45}=24 \mathrm{~cm}
\end{aligned}$
Distance between parallel sides $=24 \mathrm{~cm}$.

Question 5.

The area of a trapezium is 180 sq. cm and its height is 9 cm. If one of the parallel sides is longer than the other by $6 \mathrm{~cm}$. Find the length of the parallel sides.
Solution:
Let one of the parallel side be 'a' $\mathrm{cm}$. Given one parallel sides is longer than the other by $6 \mathrm{~cm}$.
i.e. $b=a+6 \mathrm{~cm}$ Also given height ' $h$ ' $=9 \mathrm{~cm}$
Area of trapezium $=180$ sq. $\mathrm{cm}$
$\begin{aligned}
&\frac{1}{2} \times h \times(a+b)=180 \mathrm{~cm}^{2} \\
&\frac{1}{2} \times 9 \times(a+a+6)=180 \\
&\frac{1}{2} \times 9 \times(2 a+6)=180 \\
&\qquad 2 a+6=\frac{180 \times 2}{9}=20 \times 2 \\
&2 \mathrm{a}+6=40 \\
&2 \mathrm{a}=40-6=34 \\
&\mathrm{a}=\frac{34}{2}=17 \mathrm{~cm} \\
&\mathrm{~b}=\mathrm{a}+6=17+6=23 \mathrm{~cm} \\
&\therefore \text { The parallel sides are } \mathrm{a}=17 \mathrm{~cm} \text { and } \mathrm{b}=23 \mathrm{~cm}
\end{aligned}$

Question 6 .
The sunshade of a window is in the form of isoceles trapezium whose parallel sides are $81 \mathrm{~cm}$ and 64 $\mathrm{cm}$ and the distance between them is $6 \mathrm{~cm}$. Find the cost of painting the surface at the rate of $₹ 2$ per sq. $\mathrm{cm}$.
Solution:
Given the parallel sides $\mathrm{a}=81 \mathrm{~cm} ; \mathrm{b}=64 \mathrm{~cm}$
Distance between ' $a$ ' and ' $b$ ' is height $h=6 \mathrm{~cm}$
Area of the trapezium $=\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b})$ sq. units
$=\frac{1}{2} \times 6^{3} \times(81+64)=3 \times 145 \mathrm{~cm}^{2}=435 \mathrm{~cm}^{2}$
Cost of painting $1 \mathrm{~cm}^{2}=₹ 2$
Cost of painting $435 \mathrm{~cm}^{2}=₹ 435 \times 2=₹ 870$
Cost of painting $=₹ 870$.
 

Question $7 .$
A window is in the form of trapezium whose parallel sides are $105 \mathrm{~cm}$ and $50 \mathrm{~cm}$ respectively and the distance between the parallel sides is $60 \mathrm{~cm}$. Find the cost of the glass used to cover the window at the rate of $₹ 15$ per 100 sq. $\mathrm{cm}$.
Solution:
Given the parallel sides $\mathrm{a}=105 \mathrm{~cm} ; \mathrm{b}=50 \mathrm{~cm} ;$ Height $=60 \mathrm{~cm}$
Area of the trapezium $=\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b}) \mathrm{sq} .$ units $=\frac{1}{2} \times 60 \times(105+50) \mathrm{cm}^{2}$

$=30 \times 155 \mathrm{~cm}^{2}=4650 \mathrm{~cm}^{2}$
For $100 \mathrm{~cm}^{2}$ cost of glass used $=₹ 15$
$\therefore$ For $4650 \mathrm{~cm}^{2}$ cost of glass $=₹ \frac{4650}{100} \times 15=₹ 697.50$
Cost of the glass used $=₹ 697.50$

Objective Type Questions
Question 8 .
The area of the trapezium, if the parallel sides are measuring $8 \mathrm{~cm}$ and $10 \mathrm{~cm}$ and the height $5 \mathrm{~cm}$ is
(i) $45 \mathrm{sq} . \mathrm{cm}$
(ii) 40 sq. cm
(iii) 18 sq. $\mathrm{cm}$
(iv) $50 \mathrm{sq} . \mathrm{cm}$
Solution:
(i) $45 \mathrm{sq} \cdot \mathrm{cm}$
Hint: $\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})=\frac{1}{2} \times 5 \times(10+8)=45$

Question $9 .$
In a trapezium if the sum of the parallel sides is $10 \mathrm{~m}$ and the area is 140 sq.m, then the height is
(i) $7 \mathrm{~cm}$
(ii) $40 \mathrm{~cm}$
(iii) $14 \mathrm{~cm}$
(iv) $28 \mathrm{~cm}$
Solution:
(iv) $28 \mathrm{~cm}$
Hint: Area $=\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})=140=\frac{1}{2} \times \mathrm{h} \times 10 \Rightarrow \mathrm{h}=28$

Question 10 .
When the non-parallel sides of a trapezium are equal then it is known as
(i) a square
(ii) a rectangle
(iii) an isoceles trapezium
(iv) a parallelogram
Solution:
(iii) an isoceles trapezium