TS 6th Class Maths Solutions Chapter 13 Practical Geometry Ex 13.5

Students can practice Telangana SCERT Class 6 Maths Solutions Chapter 13 Practical Geometry Ex 13.5 to get the best methods of solving problems.

TS 6th Class Maths Solutions Chapter 13 Practical Geometry Exercise 13.5

Question 1.
Construct ∠ABC = 60° without using protractor.
Answer:

Steps of construction:

1. Draw a line l and choose a point B on it.
2. Place the pointer of the compasses at B and draw an arc of convenient radius which cuts the line / at a point say C.
3. Take C as centre and with the same radius (as in step 2) in draw an arc.
4. Now take B as centre and with the same radius (as in step 2), draw another arc cutting the previous arc (drawn in step 2) at A.
5. Join AB, we get ∠ABC whose measure is 60°.

Question 2.
Construct an angle of 120° with using protractor and compasses.
Answer:
With using compasses:

Steps of construction:

1.  Draw any ray OA.
2. Place the pointer of the compasses at O with O as centre and any convenient radius draw an arc cutting OA at M.
3. With M as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at P.
4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
5. Join OQ. Then ∠AOQ is the required angle.

With Using protractor:

Steps of construction :

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at
   Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark at a point P at 120°.
4. Join QP. ∠PQR is the required angle.

Question 3.
Construct the following angles using ruler and compasses. Write the steps of construction in each case.
(i) 75°
Answer:

Steps of construction:

1. Draw any ray OA.
2. Place the pointer of the compasses at O. With ‘O’ as centre and any convenient radius draw an arc cutting OA at M.
3. With M as centre and without altering radius (as in step 2) draw an arc which, cuts the first arc at P.
4. With P as centre and without altering the radius (as in step 2) draw an arc which cuts the first arc at Q.
5. Join OQ. ∠AOQ = 120° (∵∠AOP + ∠POQ = 60°)
   ∠AOQ = ∠AOP + ∠POQ = 60° + 60° = 120°
6. Draw OB the perpendicular bisector of ∠POQ.
   Now ∠POB = ∠QOB = 30°
   ∴∠AOB = 90°
7. Draw OC the perpendicular bisector of ∠POB.
   Now ∠POC = ∠BOC = 15°
8. ∠AOC = ∠AOP + ∠OC
   = 60° + 15°
   = 75°
   (or)
   ∠AOC = ∠AOB – ∠BOC
   = 90° – 15°
   = 75°
   ∠AOC is the required angle whose measure is 75°.

(ii) 15°
Answer:

Steps of construction:

1. Draw a line l and mark a point O on it.
2. Place the pointer of the compasses at O and draw an arc of convenient radius which cuts the line / at a point say A.
3. With the pointer at A (as centre) and the same radius as in the step – 2, now draw an arc that passes through O.
4. Let the two arcs intersect at B. Join OB. We get ∠BOA whose measure is 60°.
5. Draw the bisector OC of ∠AOB. Now ∠AOC = 30°.
   [∠AOC = ∠BOC = 30° ]
6. Draw the bisector OD of ∠AOC. Now ∠AOD = 15°.
   [∠AOD = ∠COD = 15°]
7. ∠AOD = 15° is the required angle.

(iii) 105°
Answer:

Steps of construction:

1. Draw any ray OM.
2. Place the pointer of the compasses at O. With O as centre and any convenient radius draw an arc cutting OM at A.
3. With A as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
4. With B as centre and without altering radius (as in step 2) draw an arc which cuts the first arc at B.
5. Join QC. ∠AOC = 120°.
6. We know that ∠AOB = ∠BOC = \(\frac{1}{2}\) ∠AOC = \(\frac{1}{2}\) × 120° = 60°
7. Draw OD the bisector of ∠BOC.
   ∴ ∠BOD = ∠DOC = \(\frac{1}{2}\) ∠BOC = \(\frac{1}{2}\) × 60° = 30°
8. Draw OE the bisector of ∠DOC.
   ∴ ∠DOE = ∠EOC = \(\frac{1}{2}\) ∠DOC = \(\frac{1}{2}\) × 30° = 15°
9. Now ∠AOE = ∠AOD + ∠DOE = 90° + 15° = 105°
10. ∠AOE = 105° is the required angle.

Question 4.
Draw the angles given in Q. 3 using a protractor.
Answer:
The angles given in Q. 3 are 75°, 105° and 15°.

(i) 75°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark a point P at 75°.
4. Join QP. ∠RQP is the required angle.

(ii) 105°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark a point P at 105°
4. Join QP. ∠RQP is the required angle.

(iii) 15°

Steps of construction:

1. Draw a ray QR of any length.
2. Place the centre point of the protractor at Q and the line aligned with the \(\overline{\mathrm{QR}}\).
3. Mark a point P at 15°
4. Join QP. ∠RQP is the required angle.

Question 5.
Construct ∠ABC = 50° and then draw another angle ∠XYZ equal to ∠ABC without using a protactor.
Answer:

Steps of construction:

1. Make an angle ∠ABC = 50° v
2. Draw a line l and choose a point Y on it.
3. Use the same compasses setting to draw an arc with Y as centre, cutting l in Z.
4. Set your compasses to the length MN.
5. With the same radius place the compasses pointer at Z and draw an arc to cut the arc drawn earlier at X.
6. Join XY. This gives us ∠XYZ. It has the same measure as ∠ABC i.e,, 50°.

Question 6.
Construct ∠DEF = 60°. Bisect it, measure each half by using a protractor.
Answer:

Steps of construction :

1. Draw a line P and mark a point ‘E’ on it.
2. Place the pointer of the compasses at E and draw an arc of convenient radius which cuts the line \(\overleftrightarrow{\mathrm{PQ}}\) at a point say F.
3. With the pointer at F (as centre), now draw an arc that passes through E.
4. Let the two arcs intersect at D. Join ED. We get ∠DEF whose measure is 60°.
5. Draw OC the bisector of ∠DEF.
6. Now ∠FEC = ∠CED = 30°.