Exercise 5.4 - 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 $5.4$
Question $1 .$
Construct the following angles using protractor and draw a bisector to each of the angle using ruler and compass.
(a) $60^{\circ}$
(b) $100^{\circ}$
(c) $90^{\circ}$
(d) $48^{\circ}$
(e) $110^{\circ} .$
Solution:
(a) $60^{\circ}$

Construction:

Step 1: Drawn the given angle $\angle \mathrm{ABC}$ with the measure $60^{\circ}$ using protractor.
Step 2: With B as centre and convenient radius, drawn an are to cut BA and BC. Marked the points of intersection as $\mathrm{E}$ on BA and F on BC.
Step 3: With the same radius and $\mathrm{E}$ as centre drawn an arc in the interior of $\angle \mathrm{ABC}$ and another arc of same measure with centre at $\mathrm{F}$ to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given $\angle \mathrm{ABC}$
Now $\lcm{A B G}=\angle C B G=30^{\circ}$

(b) $100^{\circ}$
Construction :

Step 1: Drawn the given angle $\angle \mathrm{ABC}$ with the measure $100^{\circ} \mathrm{c}$ using protractor.
Step 2: With B as centre and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as $\mathrm{E}$ on BA and F on BC.
Step 3: With the same radius and $\mathrm{E}$ as centre drawn an arc in the interior of $\angle \mathrm{ABC}$ and another arc of the same measure with centre at $\mathrm{F}$ to cut the previous arc.
Step 4: Marked the point of intersection at G. Drawn a ray BX through G.
$\mathrm{BG}$ is the required bisector of angle $\angle \mathrm{ABC}$
$\angle \mathrm{ABG}=\angle \mathrm{GBC}=50^{\circ}$
(c) $90^{\circ}$
Construction :

Step 1: Drawn the given angle $\angle \mathrm{ABC}$ with the measure $90^{\circ}$ using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked the points of intersection as $\mathrm{E}$ on BA and F on BC.
Step 3: With the same radius and $E$ as center drawn an arc in the interior of $\angle A B C$ and another arc of same measure with center at $F$ to cut the previous arc.
Step 4: Mark the point of interaction as G. Drawn a ray BX through G. BG is the required bisector of the given angle $\angle \mathrm{ABC}$
$\angle \mathrm{ABG}=\angle \mathrm{GBC}=45^{\circ}$

(d) $48^{\circ}$
Construction

:
Step 1: Drawn the given angle $\angle \mathrm{ABC}$ with the measure $48^{\circ}$ using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and to cut BA and BC. Marked the points of intersection as $\mathrm{E}$ on $\mathrm{BA}$ and $\mathrm{F}$ on $\mathrm{BC}$.
Step 3: With the same radius and $\mathrm{E}$ as center drawn an arc in the interior of $\angle \mathrm{ABC}$ and another arc of the same measure with center at $F$ to cut the previous arc.
Step 4: Marked the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle $\angle \mathrm{ABC}$
Now $\angle \mathrm{ABC}=\angle \mathrm{GBC}=24^{\circ}$
(e) $110^{\circ}$
Construction:

Step 1: Drawn the given angle $\angle \mathrm{ABC}$ with the measure $110^{\circ}$ using protractor.
Step 2: With B as center and convenient radius, drawn an arc to cut BA and BC. Marked points of intersection as $\mathrm{E}$ on BA and F BC.
Step 3: With the same radius and $E$ as center, drawn an arc in the interior of $\angle A B C$ and another arc of same measure with center at $F$ to cut the previous arc.
Step 4: Mark the point of intersection as G. Drawn a ray BX through G. BG is the required bisector of the given angle $\angle \mathrm{ABC}$
$\angle \mathrm{ABG}=\angle \mathrm{GBC}=55^{\circ}$