NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.1 – Part 1
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Construction of point angle at

Construction of Point Angle at 120 Degree
Q-1 Construct an angle of at the initial point of a given of ray and justify the construction.
Solution:
Ray as with the Initial Point A
To construct an angle at point A
Steps of construction:
Draw a ray AS.
With its initial point A as centre and any radius, draw an arc BCD, cutting AS at B.
With same radius and center at B, cut the arc intersecting the circle with center A at C
With centre C and same radius, draw an arc, cutting the circle with center A at D.
With D and C as centres, and any convenient radius () draw two arcs intersecting at P
Join AP. Then
Justification:
AB and AC are the radius and AB = BC by construction, therefore,
Therefore is an equilateral triangle. So,
Similarly,, therefore is also an equilateral triangle. So
AP bisects , so
Now
Q-2 Construct an angle of at the initial point of a given ray and justify the construction.
Solution:
Ray OA, O is the Centre of the 45° Angle
Steps of Construction:
Draw a ray OA
With O as centre and any suitable radius draw an arc cutting OA at B.
With B as centre and same radius cut the previously drawn arc at C and then with C as centre and same radius cut the arc in step 1 at D.
With C as centre and radius more than half CD draw an arc.
With D as centre and same radius draw another arc to cut the previous arc at E
Join OE. Then (we proved this above). Let OE cut the circle with center O at F
Now we will draw the bisector OG of Then
With B as center and radius more than half of BF draws an arc.
With F as center and radius more than half of BF draws another arc intersecting previous arc at G.
Now OG is the bisector of angle. Since is the bisector of. So,
Justification:
We already proved that constructed this way would be .
Now, in and,
OG is common
(by construction)
(radius of the same circle at O)
Therefore, by SSS
Therefore,
Also,
Therefore,
Q-3 Construct the angle of the following measurements:
1.
2.
3.
Solution:
i) Steps of Construction:
∠QAB=30°
Draw a ray AQ.
With its initial point A as centre and any radius, draw an arc, cutting AQ at P.
With centre P and same radius. Draw an arc, cutting the arc of step 2 in D
With P and D as centres, and any convenient radius (), draw two arcs intersecting at B.
Join OB. Then
ii) Step of construction:
POB=90°
Draw an angle
Draw the bisector OC of , then
Bisect , such that
Thus,
iii) Steps of Construction:
Ray AB at ∠CAB=15°
Construct an
Bisect so that
Bisect ,so that