In Text Questions Try These (Textbook Page No.33, 35, 36, 41, 46) - Chapter 2 Measurements Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Exercise $2.1$
Parallelogram
(Try These Text book Page No. 33)
Question $1 .$
Find the missing values for the following:

Solution:
(i) Given length $\mathrm{l}=12 \mathrm{~m}$; Breadth $\mathrm{b}=8 \mathrm{~cm}$
$\therefore$ Area of rectangle $=1 \times$ b sq. units $=12 \times 8 \mathrm{~m}^{2}=96 \mathrm{~m}^{2}$
Perimeter of the rectangle $=2 \times(1+b)$ units $=2 \times(12+8) \mathrm{m}=2 \times 20=40 \mathrm{~m}$
(ii) Given Length $\mathrm{l}=15 \mathrm{~cm}$; Area of the rectangle $=90$ sq. $\mathrm{cm}$
$\mathrm{l} \times \mathrm{b}=90 ; 15 \times 6=90 ; \mathrm{b}=\frac{90}{15}=6 \mathrm{~cm}$
Perimeter of the rectangle $=2 \times(1+b)$ units $=2 \times(15+6) \mathrm{cm}=2 \times 21 \mathrm{~cm}=42 \mathrm{~cm}$
(iii) Given Breadth of rectangle $=50 \mathrm{~mm}$; Perimeter of the rectangle $=300 \mathrm{~mm}$
$2 \times(1+b)=300$
$2 \times(1+50)=300$
$1+50=\frac{300}{2}=150$
$\mathrm{l}=150-50$

$\mathrm{I}=100$
Area $=1 \times$ b sq. untis $=100 \times 50 \mathrm{~mm}^{2}=5000 \mathrm{~mm}^{2}$
(iv) Length of the rectangle $=12 \mathrm{~cm}$; Perimeter $=44 \mathrm{~cm}$ $2(\mathrm{l}+\mathrm{b})=44$
$\begin{aligned}
&2(12+b)=44 \\
&12+b=\frac{44}{2} \\
&12+b=22 ; b=22-12 ; b=10 \mathrm{~cm} \\
&\text { Area }=1 \times b \text { sq. units } \\
&=12 \times 10 \mathrm{~cm}^{2}=120 \mathrm{~cm}^{2}
\end{aligned}$

Question 2.

Solution:
(i) Given side $\mathrm{a}=60 \mathrm{~cm}$
Area of the square $=\mathrm{a} \times$ a sq.units $=60 \times 60 \mathrm{~cm}^{2}=3600 \mathrm{~cm}^{2}$
Perimeter of the square $=4 \times$ a units $=4 \times 60 \mathrm{~cm}=240 \mathrm{~cm}$
(ii) Given area of a square $=64$ sq. $\mathrm{m}$
$\mathrm{a} \times \mathrm{a}=64$
$a \times a=8 \times 8$
$\mathrm{a}=8 \mathrm{~m}$
Perimeter $=4 \times \mathrm{a}$
$=4 \times 8$
$=32 \mathrm{~m}$
(iii) Given perimeter of the square $=100 \mathrm{~mm}$
$4 \times \mathrm{a}=100$
$\mathrm{a}=\frac{100}{4} \mathrm{~mm}$
$\mathrm{a}=25 \mathrm{~mm}$
Area $=\mathrm{a} \times$ a sq. units
$=25 \times 25 \mathrm{~mm}^{2}$
$=625 \mathrm{~mm}^{2}$

Question 3.

Solution:
(i) Given base of the right angled triangle $=13 \mathrm{~m}$; height $=5 \mathrm{~m}$
(i) Given base of the right angled triangle $=13 \mathrm{~m}$; height $=5 \mathrm{~m}$
Area $=\frac{1}{2} \times(b \times h)$ sq. units $=\frac{1}{2} \times(13 \times 5) \mathrm{m}^{2}=\frac{65}{2} \mathrm{~m}^{2}=32.5 \mathrm{~m}^{2}$
(ii) Base $=16 \mathrm{~cm} ;$ Area $=240$ sq. $\mathrm{cm} ; \frac{1}{2} \times b \times h=240$
$\frac{1}{2} \times 16\times h=240 ; h=\frac{240}{8} ; h=30 \mathrm{~cm}$
(iii) Given height $\mathrm{h}=6 \mathrm{~mm}$; Area $=84$ sq. $\mathrm{mm}$
$\frac{1}{2} \times \mathrm{b} \times \mathrm{h}=84 ; \frac{1}{2} \times \mathrm{b} \times 6=84$

$\mathrm{b}=\frac{84 \times 2}{6} ; \mathrm{b}=28 \mathrm{~mm}$

(Try This Textbook Page No. 35)
Question $1 .$
Explain the area of the parallelogram as sum of the areas of the two triangles.
Solution:
ABCD is a parallelogram. It can be divided into two triangles of equal area by drawing the diagonal BD.

Area of the parallelogram $\mathrm{ABCD}=$ base $\times$ height $=\mathrm{AB} \times \mathrm{DE}$
But area of the triangle $\mathrm{ABD}=\frac{1}{2} \times$ base $\times$ height $$ \begin{aligned} &=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\ \text { Area of triangle } \mathrm{CDB} &=\frac{1}{2} \times \mathrm{DC} \times \mathrm{BF}[\because \mathrm{AB}=\mathrm{DC}, \mathrm{DE}=\mathrm{BF}] \\ &=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\ \text { Area of parallelogram } \mathrm{ABCD} &=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE}+\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\ &=\text { Area of } \Delta \mathrm{ABD}+\text { Area of } \Delta \mathrm{CDB} \end{aligned} $$
$\begin{aligned}
&=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\
\text { Area of triangle } \mathrm{CDB} &=\frac{1}{2} \times \mathrm{DC} \times \mathrm{BF}[\because \mathrm{AB}=\mathrm{DC}, \mathrm{DE}=\mathrm{BF}] \\
&=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\
\text { Area of parallelogram } \mathrm{ABCD} &=\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE}+\frac{1}{2} \times \mathrm{AB} \times \mathrm{DE} \\
&=\text { Area of } \Delta \mathrm{ABD}+\text { Area of } \Delta \mathrm{CDB}
\end{aligned}$

Question $2 .$
A rectangle is a parallelogram but a parallelogram is not a rectangle. Why?
Solution:
(i) For both rectangle and parallelogram
(i) opposite sides are equal and parallel.
(ii) For rectangle all angles equal to $90^{\circ} .$ But for parallelogram opposite angles are equal.
$\therefore$ All rectangles are parallelograms. But all parallelograms are not rectan $\neg$ gles as their angles need not be equal to $90^{\circ}$.

(Try These Textbook Page No. 36)
Question $1 .$
Count the squares and find the area of the following parallelograms by converting those into rectangles of the same area. (Without changing the base and height).

(a)___ sq. units
(b) ___sq. units
(c) ___sq. units
(d) ____sq. units
Solution:
Converting the given parallelograms into rectangles we get.
(a) 10 sq. units
(b) 18 sq. units
(c) 16 sq. units
(d) 5 sq. units

Question 2 .
Draw the heights for the given parallelograms and mark the measure of their bases and find the area.
Analyze your result.

Solution:
(a) Area of the parallelogram $=b \times h$ sq. units

$=4 \times 2$ sq. units $=8$ sq. units
By counting the small squares also we get number of full
squares $+$ number of square more than half $=6+2=8$ sq. units.
(b) Area of the parallelogram $=$ base $\times$ height $=4 \times 2=8$ sq. units

(c) Area of the parallelogram $=$ base $\times$ height $=4 \times 2=8$ sq. units
Also area $=$ Number of full squares $+$ Number of squares more than half $+\frac{1}{2}$ Number of half squares $=4+4=8 \mathrm{sq}$. units

(d) Area of the parallelogram $=$ (base $\times$ height) sq. units $=4 \times 2$ sq. units $=8$ sq. units

Also area of the parallelogram $=$ Number of full squares $+\frac{1}{2}$ [Number of half squares] + Number of squares more than half $=4+0+4=8$ sq. units
(e) Area of parallelogram $=$ (base $\times$ height) sq. units $=4 \times 2$ sq. units $=8$ sq. units

Also area of the parallelogram $=$ Number of full squares $+$ Number of squares more than half $+\frac{1}{2}$ [Number of half squares] $=2+6=8$ sq. units

Question $3 .$
Find the area o the following parallelograms by measuring their base and height, using formula.

(a) ___ sq. units
(b) ___ sq. units
(c) ___ sq. units
(d) ___ sq. units
(e) ___ sq. units
Solution:
(a) Area of the rectangle $=($ base $\times$ height $)$ sq. units base $=5$ units
height $=5$ units
$\therefore$ Area $=(5 \times 5)=$ sq. units $=25$ sq. units
(b) Area of the rectangle $=($ base $\times$ height $)$ sq. units base $=4$ units
height $=1$ units
$\therefore$ Area $=(4 \times 1)=$ sq. units $=4$ sq. units
(c) Area of the rectangle $=$ (base $\times$ height) sq. units base $=2$ units
height $=3$ units
$\therefore$ Area $=(2 \times 3)=$ sq. units $=6$ sq. units

(d) Area of the rectangle $=$ (base $\times$ height) sq. units
base $=4$ units
height $=4$ units
$\therefore$ Area $=(4 \times 4)=$ sq. units $=16$ sq. units
(e) Area of the parallelogram $=$ (base $\times$ height) sq. units
base $=7$ units
height $=5$ units
$=7 \times 5=35$ sq. units

Question 4 .
Draw as many parallelograms as possible in a grid sheet with the area 20 square units each.
Solution:

Area of parallelogram (a), (b) or (c) = 20 sq. units

Exercise $2.2$
Rhombus
(Try These Textbook Page No. 41)
Question $1 .$
Observe the figure and answer the following questions.
(i) Name two pairs of opposite sides.
(ii) Name two pairs of adjacent sides.
(iii) Name the two diagonals.

Solution:
(i) (a) $\mathrm{PQ}$ and $\mathrm{RS}$ (b) $\mathrm{QR}$ and $\mathrm{PS}$
(ii) (a) $\mathrm{PQ}$ and $\mathrm{QR}$ (b) $\mathrm{PS}$ and $\mathrm{RS}$
(iii) (a) PR and Question are diagonals.
 

Question $2 .$
Find the area of the rhombus given in (i) and (ii).

Solution:
(i) Area of the rhombus $=\frac{1}{2}\left(\mathrm{~d}_{1}+\mathrm{d}_{2}\right)$ sq. units $=\frac{1}{2}(11+13)$ sq. units $=\frac{1}{2} \times(24) \mathrm{cm}^{2}=12 \mathrm{~cm}^{2}$
(ii) Base $=10 \mathrm{~cm}$; Height $=7 \mathrm{~cm}$
Area of the rhombus $=\mathrm{b} \times \mathrm{h}$ sq. units $=10 \times 7 \mathrm{~cm}^{2}=70 \mathrm{~cm}^{2}$

Question $3 .$
Can you find the perimeter of the rhombus?
Solution:
If we know the length of one side we can find the perimeter using $4 \times$ side units.
 

Question $4 .$
Can diagonals of a rhombus be of the same length?
Solution:
When the diagonals of a rhombus become equal it become a square.

Question $5 .$
A square is a rhombus but a rhombus is not a square. Why?
Solution:
In a square
(i) all sides are equal.
(ii) opposite sides are parallel
(iii) diagonals divides the square into 4 right angled triangles of equal area
(iv) the diagonals bisect each other at right angles.
So it become a rhombus also.
But in a rhombus (i) each angle need not equal to $90^{\circ}$.
(ii) the length of the diagonals need not be equal. Therefore it does not become a square.
 

Question $6 .$
Can you draw a rhombus in such a way that the side is equal to the diagonal.
Solution:
Yes, we can draw a rhombus with one of its diagonals equal to its side length. In such case the diagonal will divide the rhombus into two congruent equilateral triangles.

Exercise $2.3$
(Try These Textbook Page No. 46)
Question $1 .$
Can you find the perimeter of the trapezium? Discuss.
Solution:
If all sides are given, then by adding all the four lengths we can find the perimeter of a trapezium.
 

Question $2 .$
In which case a trapezium can be divided into two equal triangles?
Solution:
If two parallel sides are equal in length. Then it can be divided into two equal triangles.
 

Question $3 .$
Mention any three life situations where the isosceles trapeziums are used?
Solution:
(i) Glass of a car windows.
(ii) Eye glass (glass in spectacles)
(iii) Some bridge supports.
(iv) Sides of handbags.