Exercise 5.5 - 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 $5.5$
Question 1 .
Construct the following angles using ruler and compass only.
(i) $60^{\circ}$
(ii) $120^{\circ}$
(iii) $30^{\circ}$
(iv) $90^{\circ}$
(v) $45^{\circ}$
(vi) $150^{\circ}$
(vii) $135^{\circ}$
Solution:
(i) $60^{\circ}$
Construction:

Step 1: Drawn a line and marked a point 'A' on it.
Step 2: With A as center drawn an arc of convenient radius to meet the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C. Step 4: Joined $\mathrm{AC}$. The $\angle \mathrm{ABC}$ is the required angle with the measure $60^{\circ}$.
(ii) $120^{\circ}$
Construction :

We know that there are two $60^{\circ}$ angles in $120^{\circ}$.
$\therefore$ We can construct two $60^{\circ}$ angles consecutively construct $120^{\circ}$
Step 1: Drawn a line and marked a point 'A' on it.
Step 2: With 'A' as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and C as center, drawn an arc to cut the arc drawn in step 2 at D. Step 5: Joined AD. Then $\angle B A D$ is the required angle with measure $120^{\circ}$.
(iii) $30^{\circ}$
Constructions :

Since $30^{\circ}$ is half of $60^{\circ}$, we can construct $30^{\circ}$ by bisecting the angle $60^{\circ}$.
Step 1: Drawn a line and marked a point $\mathrm{A}$ on it.
Step 2: With A as center drawn an arc of convenient radius to the line to meet at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: Joined $\mathrm{AC}$ to get $\angle \mathrm{BAC}=60^{\circ}$
Step 5: With B as center drawn an arc of convenient radius in the interior of $\angle B A C$
Step 6: With the same radius and $\mathrm{C}$ as center drawn an arc to cut the previous arc at $\mathrm{D}$.
Step 7: Joined AD.
$\therefore \angle B A D$ is the required angle of measure $30^{\circ}$.
(iv) $90^{\circ}$
Construction:

Step 1: Drawn a line and marked a point 'A' on it.
Step 2: With 'A' as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at ' $\mathrm{C}$ '.
Step 4: With the same radius and $\mathrm{C}$ as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. $\angle B A D=120^{\circ}$.
Step 6: With C as center, drawn an arc of convenient radius in the interior of $\angle \mathrm{CAD}$.
Step 7: With the same radius and D as center, drawn an arc to cut the arc at E.
Step 8: Joined AF $\angle B A E=90^{\circ}$.
(v) $45^{\circ}$
Construction :

Step 1: Drawn a line and marked a point $\mathrm{A}$ on it
Step 2: With A as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and $\mathrm{C}$ as center, drawn an arc to cut the arc drawn in step 2 at D.
Step 5: Joined AD. $\angle B A D=120^{\circ}$.
Step 6: With G as center and any convenient radius drawn an arc in the interior of $\angle \mathrm{GAB}$
Step 7: With the same radius and B as center drawn an arc to cut the arc at F.
Step 8: Joined AF. $\angle B A F=45^{\circ}$

(vi) $150^{\circ}$
Construction :

Since $150^{\circ}=60^{\circ}+60^{\circ}+30^{\circ}$; we construct as follows
Step 1: Drawn a line and marked a point $\mathrm{A}$ on it.
Step 2: With 'A' as center, drawn a full arc of convenient radius to the line at a point B and at E the other end.
Step 3: With the same radius and B as center, drawn an arc to cut the previous arc at C.
Step 4: With the same radius and $\mathrm{C}$ as center drawn an arc to cut the already drawn arc at $\mathrm{D}$.
Step 5: With D as center, drawn an arc of convenient radius in the interior of $\angle D A E$
Step 6: With $\mathrm{E}$ as center and with the same radius drawn an arc to cut the previous arc at $\mathrm{F}$.
Step 7: Joined AF, $\angle \mathrm{FAB}=150^{\circ}$.

(vii) $135^{\circ}$
Construction :

Step 1: Drawn a line and marked a point $\mathrm{A}$ on it.
Step 2: With 'A' as center, drawn an arc of convenient radius to the line at a point B.
Step 3: With the same radius and B as center drawn an arc to cut the previous arc at C.
Step 4: With the same radius and $\mathrm{C}$ as center, drawn an arc to cut the arc at $\mathrm{D}$.
Step 5: With C and D as centers drawn arcs of convenient (same) radius in the interior of $\angle \mathrm{CAD}$.
Marked the point of intersection as E.
Step 6: Joined $\mathrm{AE}$, through $\mathrm{G} . \angle \mathrm{BAE}=90^{\circ}$.
Step 7: Drawn angle bisector to $\angle \mathrm{GAH}$ through $\mathrm{F}$.
Now $\angle B A F=135^{\circ}$.