AP State Syllabus AP Board 6th Class Maths Solutions Chapter 10 Practical Geometry Ex 10.4 Textbook Questions and Answers.

## AP State Syllabus 6th Class Maths Solutions 10th Lesson Practical Geometry Ex 10.4

Question 1.

Construct ∠ABC = 60° without using protractor.

Solution:

Steps of construction :

- Draw a line l and mark a point B on it.
- Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line l at a point C.
- With the pointer at C as centre and with the same radius as above. Now draw an arc that passes through B.
- Cut intersecting point of two arcs is A. Join BA. ∠ABC = 60° is formed.

Hence the required angle ∠ABC = 60° is constructed.

Question 2.

Construct an angle of 120° with using protractor and compasses.

Solution:

Steps of construction :

- Draw a ray \(\overline{\mathrm{OA}}\) of any length.
- Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OA at M.
- With M as centre and without altering radius draw an arc which cuts the previous arc at P.
- With P as centre and without changing radius draw an arc which cuts the first arc at Q.
- Join OQ. Then ∠AOQ is the required angle.

Hence the required angle ∠AOQ = 120° is constructed.

Question 3.

Construct the following angles using ruler and compasses. Write the steps of construction in each case.

i) 75°

ii) 15°

iii) 105°

Solution:

i) 75°

Steps of construction:

- Construct ∠RAM 900
- Construct ∠PAM = 600 AM as common arm
- Now, ∠PAR = ∠RAM – ∠PAM = 90° – 60° = 30°
- Draw bisector to ∠PAR which is \(\overrightarrow{\mathrm{AQ}}\).
- Now, ∠MAQ = ∠PAM + ∠PAQ = 60° + \(\frac{30^{\circ}}{2}\)

= 60° + 15° = 75° (QED)

ii) 15°

Steps of construction:

- Construct ∠BQR = 60°
- Draw the bisector to ∠BQR which is \(\overrightarrow{\mathrm{QC}}\)
- Now,∠CQR = \(\frac{60^{\circ}}{2}\) = 30°
- Draw the bisector \(\overrightarrow{\mathrm{QP}}\) to∠CQR.
- ∠PQR = \(\frac{1}{2}\) x 30° = 15° (Q.E.D)

iii) 105°

Steps of construction :

- Construct ∠IDS = 90°
- Construct ∠ADS = \(\frac{\angle \mathrm{RDS}}{2}=\frac{30^{\circ}}{2}\) = 15°
- Now, ∠ADI = ∠IDS + ∠ADS = 90° + 15°

∴ ∠ADI = 105° (Q.E.D)