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

Construction of Point Angle at 120 Degree

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

Ray AS with the initial point A, an arc BCD, cutting AS at B

To construct an angle at point A

Steps of construction:

1. Draw a ray AS.

2. With its initial point A as centre and any radius, draw an arc BCD, cutting AS at B.

3. With same radius and center at B, cut the arc intersecting the circle with center A at C

4. With centre C and same radius, draw an arc, cutting the circle with center A at D.

5. With D and C as centres, and any convenient radius ( ) draw two arcs intersecting at P

6.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

Ray OA, O is the center of the 45° angle

Steps of Construction:

1. Draw a ray OA

2. With O as centre and any suitable radius draw an arc cutting OA at B.

3. 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.

4. With C as centre and radius more than half CD draw an arc.

5. With D as centre and same radius draw another arc to cut the previous arc at E

6. 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

1. With B as center and radius more than half of BF draws an arc.

2. With F as center and radius more than half of BF draws another arc intersecting previous arc at G.

3. Now OG is the bisector of angle . Since is the bisector of . So,

Justification:

1. We already proved that constructed this way would be .

2. Now, in and ,

OG is common

(by construction)

(radius of the same circle at O)

3. Therefore, by SSS

4. Therefore,

5. Also,

6. Therefore,

Q-3 Construct the angle of the following measurements:

1.

2.

3.

Solution:

i) Steps of Construction:

∠QAB=30°

∠QAB=30° ray AQ, initial point A cut AQ at P

1. Draw a ray AQ.

2. With its initial point A as centre and any radius, draw an arc, cutting AQ at P.

3. With centre P and same radius. Draw an arc, cutting the arc of step 2 in D

4. With P and D as centres, and any convenient radius ( ), draw two arcs intersecting at B.

5. Join OB. Then

ii) Step of construction:

POB=90°

POB=90and∠POD=22.5°

1. Draw an angle

2. Draw the bisector OC of , then

3. Bisect , such that

4. Thus,

iii) Steps of Construction:

Ray AB at ∠CAB=15°

Ray AB at ∠CAB=15° ,∠BAC=15°

1. Construct an

2. Bisect so that

3. Bisect ,so that

4.