Exercise 2.4 - 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 $2.4$
Miscellaneous Practice Problems
Question $1 .$
The base of the parallelogram is $16 \mathrm{~cm}$ and the height is $7 \mathrm{~cm}$ less than its base. Find the area of the parallelogram.
Solution:
In a parallelogram
Given base $\mathrm{b}=16 \mathrm{~cm}$; height $\mathrm{h}=$ base $-7 \mathrm{~cm}=16-7=9 \mathrm{~cm}$
Area of the parallelogram $=$ (base $\times$ height) sq. units
$=16 \times 9 \mathrm{~cm}^{2}=144 \mathrm{~cm}^{2}$
Area of the parallelogram $=144 \mathrm{~cm}^{2}$

Question $2 .$
An agricultural field is in the form of a parallelogram, whose area is $68.75 \mathrm{sq} . \mathrm{hm}$. The distance between the parallel sides is $6.25 \mathrm{~cm}$. Find the length of the base.
Solution:
Height of the parallelogram $=6.25 \mathrm{hm}$
Area of the parallelogram $=68.75$ sq. $\mathrm{hm}$
$\mathrm{b} \times \mathrm{h}=68.75$
$\mathrm{b} \times 6.25=68.75$
$\mathrm{b}=\frac{68.75}{6.25}=\frac{6875}{625}=11 \mathrm{~km}$
Length of the base $=11 \mathrm{~km}$.
 

Question $3 .$
A square and a parallelogram have the same area. If the side of the square is $48 \mathrm{~m}$ and the height of the parallelogram is $18 \mathrm{~m}$. Find the length of the base of the parallelogram.
Solution:
Given side of the square is $48 \mathrm{~m}$

Area of the square $=($ side $\times$ side $)$ sq. unit $=48 \times 48 \mathrm{~m}^{2}$
Height of the parallelogram $=18 \mathrm{~m}$
Area of the parallelogram $=$ 'bh' sq. units $=b \times 18 \mathrm{~m}^{2}$
Also area of the parallelogram = Area of the square
$\mathrm{b} \times 18=48 \times 48$
$\mathrm{b}=\frac{48 \times 48}{18}=8 \times 16=128 \mathrm{~m}$
Base of the parallelogram $=128 \mathrm{~m}$

Question $4 .$
The height of the parallelogram is one fourth of its base. If the area of the parallelogram is 676 sq. $\mathrm{cm}$, find the height and the base.
Solution:
Let the base of the parallelogram be 'b' $\mathrm{cm}$
Given height $=\frac{1}{4} \times$ base ; Area of the parallelogram $=676 \mathrm{sq} . \mathrm{cm}$
$\mathrm{b} \times \mathrm{h}=676$
$\mathrm{b} \times \frac{1}{4} \mathrm{~b}=676$
$\mathrm{b} \times \mathrm{b}=676 \times 4$
$\mathrm{b} \times \mathrm{b}=13 \times 13 \times 4 \times 4$
$\mathrm{b}=13 \times 4 \mathrm{~cm}=52 \mathrm{~cm}$
Height $=\frac{1}{4} \times 52 \mathrm{~cm}=13 \mathrm{~cm}$
Height $=13 \mathrm{~cm}$, Base $52 \mathrm{~cm}$

Question $5 .$
The area of the rhombus is 576 sq. $\mathrm{cm}$ and the length of one of its diagonal is half of the length of the other diagonal then find the length of the diagonal.
Solution:
Let one diagonal of the rhombus $=\mathrm{d}_{2} \mathrm{~cm}$
The other diagonal $\mathrm{d}_{2}=\frac{1}{2} \times \mathrm{d}_{1} \mathrm{~cm}$
Area of the rhombus $=576 \mathrm{sq} . \mathrm{cm}$
$\frac{1}{2} \times\left(\mathrm{d}_{1} \times \mathrm{d}_{2}\right)=576$
$\frac{1}{2} \times\left(\mathrm{d}_{1} \times \frac{1}{2} \mathrm{~d}_{1}\right)=576$
$\mathrm{d}_{1} \times \mathrm{d}_{1}=576 \times 2 \times 2=6 \times 6 \times 4 \times 4 \times 2 \times 2$
$\mathrm{d}_{1} \times \mathrm{d}_{1}=\underline{6 \times 4 \times 2} \times \underline{6 \times 4 \times 2}$
$\mathrm{d}_{1}=6 \times 4 \times 2$
$\mathrm{d}_{1}=48 \mathrm{~cm}$
$\mathrm{d}_{2}=\frac{1}{2} \times 48=24 \mathrm{~cm}$
$\therefore$ Length of the diagonals $\mathrm{d}_{1}=48 \mathrm{~cm}$ and $\mathrm{d}_{2}=24 \mathrm{~cm}$.

Question $6 .$
A ground is in the form of isoceles trapezium with parallel sides measuring $42 \mathrm{~m}$ and $36 \mathrm{~m}$ long. The distance between the parallel sides is $30 \mathrm{~m}$. Find the cost of levelling it at the rate of $₹ 135$ per sq. $\mathrm{m}$.
Solution:
Parallel sides of the trapezium $\mathrm{a}=42 \mathrm{~m} ; \mathrm{b}=36 \mathrm{~m}$
Also height $\mathrm{h}=30 \mathrm{~m}$
Area of the trapezium $=\frac{1}{2} \times \mathrm{h} \times(\mathrm{a}+\mathrm{b})$ sq. unit
$=\frac{1}{2} \times 30 \times(42+36) \mathrm{m}^{2}$
$=\frac{1}{2} \times 30 \times 78 \mathrm{~m}^{2}$
Area $=1,170 \mathrm{~m}^{2}$
Cost of levelling $1 \mathrm{~m}^{2}=₹ 135$
$\therefore$ Cost of levelling $1170 \mathrm{~m}^{2}=₹ 1170 \times 135=₹ 1,57,950$
Cost of levelling the ground $=₹ 1,57,950$

Challenge Problems
Question $7 .$
In a parallelogram PQRS (See the diagram) PM and PN are the heights corresponding to the sides $\mathrm{QR}$ and RS respectively. If the area of the parallelogram is $900 \mathrm{sq.} \mathrm{cm}$ and the length of $\mathrm{PM}$ and PN are $20 \mathrm{~cm}$ and $36 \mathrm{~cm}$ respectively, find the length of the sides QR and SR.

Solution:
Considering $Q R$ as base of the parallelogram height $h_{1}=20 \mathrm{~cm}$
Area of the parallelogram $=900 \mathrm{~cm}^{2}$
$\mathrm{b}_{1} \times \mathrm{h}_{1}=900 ; \mathrm{b}_{1} \times 20=900$
$\mathrm{b}_{1}=\frac{900}{20}=45 \mathrm{~cm}$
Again considering SR as base height $=36 \mathrm{~cm} ;$ Area $=900 \mathrm{~cm}^{2}$
$\mathrm{b}_{2} \times \mathrm{h}_{2}=900 ; \mathrm{b}_{2} \times 36=900$
$\mathrm{b}_{2}=\frac{900}{36}$
$\mathrm{b}_{2}=25 \mathrm{~cm}$
$\mathrm{SR}=25 \mathrm{~cm} ; \mathrm{QR}=45 \mathrm{~cm} ; \mathrm{SR}=25 \mathrm{~cm}$

Question $8 .$
If the base and height of a parallelogram are in the ratio $7: 3$ and the height is $45 \mathrm{~cm}$, then fixed the area of the parallelogram.
Solution:
Given base; height $=7: 3$
Let base $=7 \mathrm{x} \mathrm{cm}$
height $=3 x \mathrm{~cm}$
also given height $=45 \mathrm{~cm}$
$3 x=45 \mathrm{~cm} 45$.
$x=\frac{45}{3}=15$
Now base $=7 \mathrm{x} \mathrm{cm}=7 \times 15 \mathrm{~cm}=105 \mathrm{~cm}$
Area of the parallelogram $=\mathrm{b} \times \mathrm{h}$ sq. unit
$=105 \times 45=4725 \mathrm{~cm}^{2}$
$=4725 \mathrm{~cm}^{2}$

Question $9 .$
Find the area of the parallelogram $\mathrm{ABCD}$ if $\mathrm{AC}$ is $24 \mathrm{~cm}$ and $\mathrm{BE}=\mathrm{DF}=8 \mathrm{~cm}$.

Solution:
Area of the parallelogram $\mathrm{ABCD}$ Area of the triangle $=$ Area of the triangle $\mathrm{ABC}+$ Area of the triangle ADC
Area of the triangle $=\frac{1}{2} \times($ base $\times$ height $)$ sq. units
Area of $\triangle \mathrm{ABC}=\frac{1}{2} \times 24 \times 8 \mathrm{~cm}^{2}=96 \mathrm{~cm}^{2}$
Area of $\triangle \mathrm{ADC}=\frac{1}{2} \times 24 \times 8 \mathrm{~cm}^{2}=96 \mathrm{~cm}^{2}$
Area of the parallelogram $\mathrm{ABCD}=96+96=192 \mathrm{~cm}^{2}$

Question 10 .
The area of the parallelogram $\mathrm{ABCD}$ is 1470 sq. $\mathrm{cm}$. If $\mathrm{AB}=49 \mathrm{~cm}$ and $\mathrm{AD}=35 \mathrm{~cm}$ then, find the height, DF and BE.

Solution:
Area of the parallelogram $=1470$ sq. $\mathrm{cm}$
Considering $\mathrm{AB}=$ base $=49 \mathrm{~cm}$
height $=\mathrm{DF}$
Area $=$ base $\times$ height
$49 \times \mathrm{DF}=1470$
$\mathrm{DF}=\frac{1470}{49}$
$\mathrm{DF}=30 \mathrm{~cm}$
Now considering AD as base
Base $=\mathrm{AD}=35 \mathrm{~cm}$; height $=\mathrm{BE}$
Base $\times$ Height $=1470$
$35 \times \mathrm{BE}=1470: \mathrm{BE}=\frac{1470}{35}$
$\mathrm{BE}=42 \mathrm{~cm} ; \mathrm{DF}=30 \mathrm{~cm} ; \mathrm{BE}=42 \mathrm{~cm}$

Question $11 .$
One of the diagonals of a rhombus is thrice as the other. If the sum of the length of the diagonals is $24 \mathrm{~cm}$, then find the area of the rhombus.
Solution:
Let one of the diagonals of rhombus be ' $\mathrm{d}_{1}$ ' $\mathrm{cm}$ and the other be $\mathrm{d}_{2} \mathrm{~cm}$.
Give $\mathrm{d}_{1}=3 \times \mathrm{d}_{2}$
Also $\mathrm{d}_{1}+\mathrm{d}_{2}=24 \mathrm{~cm}$
$\Rightarrow 3 \mathrm{~d}_{2}+\mathrm{d}_{2}=24$
$4 \mathrm{~d}_{2}=24$
$\mathrm{d}_{2}=\frac{24}{4}$
$\mathrm{d}_{2}=6 \mathrm{~cm}$
$\mathrm{d}_{1}=3 \times \mathrm{d}_{2}=3 \times 6$
$\mathrm{d}_{1}=18 \mathrm{~cm}$
$\therefore$ Area of the rhombus $=\frac{1}{2} \times \mathrm{d}_{1} \times \mathrm{d}_{2}$ sq. units
$=\frac{1}{2} \times 18 \times 6 \mathrm{~cm}^{2}=54 \mathrm{~cm}^{2}$
Area of the rhombus $=54 \mathrm{~cm}^{2}$

Question $12 .$
A man has to build a rhombus shaped swimming pool. One of the diagonal is $13 \mathrm{~m}$ and the other is twice the first one. Then find the area of the swimming pool and also find the cost of cementing the
floor at the rate of $₹ 15$ per sq. $\mathrm{cm}$.
Solution:
Let the first diagonal $\mathrm{d}_{1}=13 \mathrm{~m}$
$\mathrm{d}_{2}=2 \times 13 \mathrm{~m}=26 \mathrm{~m}$
Area of the rhombus $=\frac{1}{2} \times \mathrm{d}_{1} \times \mathrm{d}_{2}$ sq. units
$=\frac{1}{2} \times 13 \times 26 \mathrm{~m}^{2}=169 \mathrm{~m}^{2}$
Cost of cementing $1 \mathrm{~m}^{2}=₹ 15$
Cost of cementing $169 \mathrm{~m}^{2}=₹ 169 \times 15=₹ 2,535$
Cost of cementing $=₹ 2,535$
 

Question $13 .$
Find the height of the parallelogram whose base is four times the height and whose area is 576 sq. $\mathrm{cm}$.

Solution:
Let the height be 'A' and base be 'h' units
Given $b=4 \times h$
Area of the parallelogram $=576$ sq. $\mathrm{cm}$
$\mathrm{b} \times \mathrm{h}=576$
$4 \mathrm{~h} \times \mathrm{h}=576$
$\mathrm{h} \times \mathrm{h}=\frac{576}{4}=144$
$\mathrm{h} \times \mathrm{h}=12 \times 12$
$\mathrm{h}=12 \mathrm{~cm}$
Height $=12 \mathrm{~cm}$; base $=4 \times 12=48 \mathrm{~cm}$
 

Question $14 .$
The table top is in the shape of a trapezium with measurements given in the figure. Find the cost of the glass used to cover the table at the rate of $₹ 6$ per 10 sq. $\mathrm{cm}$.

Solution:
Length of the parallel sides $\mathrm{a}=200 \mathrm{~cm}$
$\mathrm{b}=150 \mathrm{~cm}$
Height $\mathrm{h}=50 \mathrm{~cm}$
Area of the trapezium $=\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b})$ sq. units
$=\frac{1}{2} \times 50(200+150) \mathrm{cm}^{2}$

$\begin{aligned}
&=\frac{1}{2} \times 50 \times 350 \mathrm{~cm}^{2}=8750 \mathrm{~cm}^{2} \\
&\text { Cost for } 10 \mathrm{sq} . \mathrm{cm} \text { glass }=₹ 68 \\
&\therefore \text { Cost of } 8750 \mathrm{~cm}^{2} \text { glass }=\frac{8750}{10} \times 6=₹ 5250 \\
&\text { Cost of glass used }=₹ 5,250
\end{aligned}$

Question $15 .$
Arivu has a land $\mathrm{ABCD}$ with the measurements given in the figure. If a portion $\mathrm{ABED}$ is used for cultivation (where $E$ is the midpoint of $D C$ ). D Find the cultivated area.

Solution:
From the figure given ABED is a trapazium with height $\mathrm{h}=18 \mathrm{~m}$
One of the parallel side $a=24 \mathrm{~m}$
Since $E$ is the midpoint of $D$.
Other parallel side $\mathrm{b}=\frac{24}{2}=12 \mathrm{~m}$
Area of the cultivated $\mathrm{ADEB}=\frac{1}{2} \times \mathrm{h}(\mathrm{a}+\mathrm{b}) \mathrm{m}^{2}=\frac{1}{2} \times 18(24+12)$
$=9 \times 36 \mathrm{~m}^{2}=324 \mathrm{~m}^{2}$
Area of cultivation $=324 \mathrm{~m}^{2}$