Exercise 5.3 - Chapter 5 Geometry Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $5.3$
Question $1 .$
Draw a line segment of given length and construct a perpendicular bisector to each line segment using scale and compass
(a) $8 \mathrm{~cm}$
(b) $7 \mathrm{~cm}$
(c) $5.6 \mathrm{~cm}$
(d) $10.4 \mathrm{~cm}$
(e) $58 \mathrm{~cm}$
Solution:
(a) $8 \mathrm{~cm}$
Construction :

Step 1: Drawn a line. Marked two points $\mathrm{A}$ and $\mathrm{B}$ on it so that $\mathrm{AB}=8 \mathrm{~cm}$
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above $\mathrm{AB}$ and one below $\mathrm{AB}$
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 . Marked the points of intersection of the arcs as C and D.
step 4: Joined $\mathrm{C}$ and $\mathrm{D}, \mathrm{CD}$ intersect $\mathrm{AB}$. Marked the point of intersection as ' $\mathrm{O}$ '.

$\mathrm{CD}$ is the required perpendicular bisector of $\mathrm{AB}$.
$\mathrm{AO}=\mathrm{OB}=\frac{8}{2}=4 \mathrm{~cm} ; \angle \mathrm{AOC}=90^{\circ}$
(b) $7 \mathrm{~cm}$
Construction :

step 1: Drawn a line and marked points $\mathrm{A}$ and $\mathrm{B}$ on it so that $\mathrm{AB}=7 \mathrm{~cm}$.
step 2: Using compass with A as centre and radius more than half of the length of AB drawn two arcs of same length one above $\mathrm{AB}$ and one below $\mathrm{AB}$.
step 3: With the same radius and B as centre drawn two arcs to cut the already drawn arcs in step $2 .$ Marked the intersection of the arcs as $C$ and $D$
step 4: Joined $\mathrm{C}$ and $\mathrm{D}, \mathrm{CD}$ is the required perpendicular bisector of $\mathrm{AB}$.
$\mathrm{AO}=\mathrm{OB}=\frac{7}{2}=3.5 \mathrm{~cm} ; \angle \mathrm{AOC}=90^{\circ}$

(c) $5.6 \mathrm{~cm}$
Construction :

Step 1: Drawn a line and marked two points $\mathrm{A}$ and $\mathrm{B}$ on it so that $\mathrm{AB}=5.6 \mathrm{~cm}$
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length, one above $A B$ and one below $A B$ Step 3: With the same radius and B as centre drawn two ares to cut the ares drawn in step 2 and marked the points of intersection of the arcs as C and D
Step 4: Joined $\mathrm{C}$ and $\mathrm{D}$. $\mathrm{CD}$ intersects $\mathrm{AB}$. Marked the point of intersection as ' $\mathrm{O}$ ' $\mathrm{CD}$ is the required perpendicular bisector of $\mathrm{AB}$.
Now $\angle \mathrm{AOC}=90^{\circ} \mathrm{AO}=\mathrm{BO}=2.8 \mathrm{~cm}$

(d) $10.4 \mathrm{~cm}$
Construction :

Step 1: Drawn a line and marked two points $\mathrm{A}$ and $\mathrm{B}$ on it so that $\mathrm{AB}=10.4 \mathrm{~cm}$.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of same length one above $\mathrm{AB}$ and one below $\mathrm{AB}$.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs drawn in step 2 and marked the points of intersection of the ares as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the points of intersection as O. CD is the required perpendicular bisector.
Now $\angle \mathrm{AOC}=90^{\circ} ; \mathrm{AO}=\mathrm{BO}=5.2 \mathrm{~cm}$
(e) $58 \mathrm{~mm}$
$58 \mathrm{~mm}=58 \times \frac{1}{10} \mathrm{~cm}=5.8 \mathrm{~cm}$
Construction :

Step 1: Drawn a line. Marked two points $\mathrm{A}$ and $\mathrm{B}$ on it so that
$\mathrm{AB}=5.8 \mathrm{~cm}=58 \mathrm{~mm}$.
Step 2: Using compass with A as centre and radius more than half of the length of AB, drawn two arcs of the same length one above $\mathrm{AB}$ and one below $\mathrm{AB}$.
Step 3: With the same radius and B as centre drawn two arcs to cut the arcs of drawn in step 2 . Marked the points of intersection of the ares as C and D.
Step 4: Joined C and D. CD intersects AB. Marked the point of intersection as O. CD is the required perpendicular bisector. $\angle \mathrm{AOC}=90^{\circ}$
$\mathrm{AO}=\mathrm{BO}=2.9 \mathrm{~cm}$