NCERT Class 9 Solutions: Constructions (Chapter 11) Exercise 11.1 – Part 1

Download PDF of This Page (Size: 221K) ↧

Construction of point angle at

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:

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