Chapter 11: Constructions

According to the question,

An angle ABC = 110^o.

To draw the bisector of ∠ABC

Steps of construction:

1. With B as the centre and a convenient radius, draw an arc to intersect the rays BA at P and BC at Q, respectively.

2. With centre P and a radius greater than half of PQ, draw an arc.

3. With centre Q and the same radius (as in step 2), draw another arc to cut the previous arc at R.

4. Draw ray BR.

The ray BR is the required bisector of ∠ABC.

∠ABD = ∠CBD = 55⁰

∠ABD = ∠CBD = 1/2 ∠ABC = 1/2 X 110⁰ = 55⁰

According to the question,

A line segment AB of length 4cm.

To draw a perpendicular to AB through A and B, respectively.

Steps of construction:

1. Draw AB = 4 cm.

2. With A as the centre, draw an arc, cutting AB at P.

3. With P as the centre and the same radius, draw an arc cutting the arc drawn in step 2 at Q.

4. With Q as the centre and the same radius, draw an arc, cutting the arc drawn in step 3 at R.

5. With R as the centre and the same radius, draw an arc, cutting the arc drawn in step 5 at X.

6. Draw OX and produce it to C and D.

7. Now, repeat the steps from 2 to 7 to draw the line EF perpendicular through B.

Yes, these lines are parallel because the sum of the interior angles on the same side of the transversal is 180^o.

According to the question:

We have to draw an angle of 80° with the help of a protractor. Then construct angles of

(i) 40°

(ii)160°

(iii) 120°

Steps of construction:

1. Draw a ray OA.

2. With the help of a protractor, construction ∠BOA = 80^o

3. Taking O as the centre and any suitable radius, draw an arc to intersect rays OA and OB at points P and Q, respectively.

4. Bisect ∠BOA as done in Q1. Let ray OC be the bisector of ∠BOA, then

∠ROA= ½ ∠BOA = ½ × 80^o = 40^o

5. With Q as the centre and radius equal to PQ, draw an arc to cut the extended arc PQ at R.

Join OR and produce it to form ray OD, then ∠DOA = 2∠BOA = 2 × 80 = 40

6. Bisect ∠DOB as in Q1. Let OE be the bisector of ∠DOB is, then

∠EOA = ∠EOB +∠BOA = ½ ∠DOB +∠BOA= ½ (80) + 80 = 40 + 80 = 120^o

According to the question,

We have to construct a triangle whose sides are 3.6 cm, 3.0 cm and 4.8 cm.

And to bisect the smallest angle and measure each part.

Steps of construction:

Step 1: Draw a line AB = 4.8 cm.

Step 2: Now, take a radius of 3 cm and centre ‘A’ and draw an arc. And take a radius of 3.6 cm and

centre ‘B’, draw an arc that intersects our previous arc at ‘C’.

Step 3: Join CA and CB we get the required triangle ABC.