The document RD Sharma Solutions -Ex-17.2, Constructions, Class 9, Maths Class 9 Notes | EduRev is a part of the Class 9 Course RD Sharma Solutions for Class 9 Mathematics.

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**Q1. Draw an angle and label it as ∠BAC. Construct another angle, equal to ∠BAC**

** **

__Steps of construction:__

- Draw an angle ABC and a line segment QR.
- With center A and any radius, draw an arc which intersects ∠BACat E and D.
- With Q as a centre and same radius draw an arc which intersects QR at S.
- With S as center and radius equal to DE, draw an arc which intersects the previous arc at T.
- Draw a line segment joining Q and T.

Therefore ∠PQR= ∠BAC

**Q2. Draw an obtuse angle. Bisect it. Measure each of the angles so formed.**

__Steps of construction:__

- Draw an angle ∠ABC of 120
^{0}. - With B as a centre and any radius, draw an arc which intersects AB at P and BC at Q.
- With P as center and radius more than half of PQ draw an arc.
- With Q as a center and same radius draw an arc which cuts the previous arc at R.
- Join BR.

Therefore ∠ABR= ∠RBC= 60^{0}

** **** **

**Q3. Using your protractor, draw an angle of 1080. With this given angle as given, draw an angle of 54 ^{0}.**

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__Steps of construction:__

- Draw an angle ABC of 108
^{0}. - With B as the center and any radius draw an arc which intersects AB at P and BC at Q.
- With P as center and radius more than half of PQ draw an arc.
- With Q as the centre and same radius draw an arc which intersects the previous arc at R.
- Join BR.

Therefore ∠RBC= 54^{0}

** **

**Q4. Using the protractor, draw a right angle. Bisect it to get an angle of measure 45 ^{0}.**

** **

__Steps of construction:__

- Draw an angle ABC of 90
^{0}. - With B as the centre and any radius draw an arc which intersects AB at P and BC at Q.
- With P as center and radius more than half of PQ draw an arc.
- With Q as center and same radius draw an arc which intersects the previous arc at R.
- Join RB.

Therefore ∠RBC= 45^{0}

**Q5. Draw a linear pair of angles. Bisect each of the two angles. Verify that the two bisecting rays are perpendicular to each other.**

__Steps of construction:__

- Draw two angles DCA and DCB forming linear pair
- With center C and any radius draw an arc which intersects AC at P and CD at Q and CB at R
- With center P and Q and any radius draw two arcs which intersect each other at S
- Join SC
- With Q and R as center and any radius draw two arcs which intersect each other at T
- Join TC

Therefore ∠SCT= 90^{0}.

** **

**Q6. Draw a pair of vertically opposite angles. Bisect each of the two angles. Verify that the bisecting rays are in the same line.**

__Steps of Construction:__

- Draw a pair of vertically opposite angle ∠AOC and ∠DOB.
- Keeping O as the center and any radius draw two arcs which intersect OA at P, OC at Q, OB at S and OD at R.
- Keeping P and Q as center and radius more than half of PQ draw two arcs which intersect each other at T.
- Join TO.
- Keeping R and S as center and radius more than half of RS draw two arcs which intersect each other at U.
- Join OU.

Therefore TOU is a straight line

** **** **

**Q7. Using rulers and compasses only, draw a right angle.**

__Steps of construction:__

- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the center and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
- Join FA.

Therefore ∠FAB= 90^{0}

** **

**Q8.Using rulers and compasses only, draw an angle of measure 135 ^{0}.**

__Steps of construction:__

- Draw a line segment AB and produce BA to C.
- Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E.
- Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
- Join FA which intersects the arc in (2) at G.
- Keeping G and D as center and radius more than half of GD draw arcs which intersect each other at H.
- Join HA.

Therefore ∠HAB= 135^{0}

** **

**Q9. Using a protractor, draw an angle of measure 72 ^{0}. With this angle as given draw angles of measure 360and 540.**

__Steps of construction:__

- Draw an ∠ABC of 72
^{0}with the help of a protractor. - Keeping B as center and any radius draw an arc which intersects AB at D and BC at E.
- Keeping D and E as center and radius more than half of DE draw two arcs which intersect each other at F.
- Join FB which intersects the arc in (2) at G.
- Keeping D and G as center and radius more than half of DG draw two arcs which intersect each other at H
- Join HB

Therefore ∠HBC= 54^{0}

∠FBC= 36^{0}

**Q10. Construct the following angles at the initial point of a given ray and justify the construction:**

**45**^{0}**90**^{0}

__Answers__

**45**^{0}

__Steps of construction:__

- Draw a line segment AB and produce BA to C.
- Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E.
- Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
- Join FA which intersects the arc in (2) at G.
- Keeping G and E as center and radius more than half of GE draw arcs which intersect each other at H.
- Join HA.

Therefore ∠HAB= 45^{0}** **

**2. 90 ^{0}**

__Steps of construction__

- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the center and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
- Join FA.

Therefore ∠FAB= 90^{0}

**Q11. Construct the angles of the following measurements:**

**30**^{0}**75**^{0}**105**^{0}**135**^{0}**15**^{0}

**ANSWERS: **

**30**^{0}

__Steps of construction:__

- Draw a line segment AB.
- Keeping A as the centre and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D and C as center and radius more than half of DC draw arcs which intersect each other at E.
- Join EA.

Therefore ∠EAB= 30^{0}

**2. 75 ^{0}**

__Steps of construction:__

- Draw a line segment AB
- Keeping A as center and any radius draw an arc which intersects AB at C
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D
- Keeping D as center and same radius draw an arc which intersects arc in (2) at E
- Keeping E and D as center and radius more than half of ED, draw arcs intersecting each other at F
- Join FA which intersects arc in (2) at G
- Keeping G and D as center and radius more than half of GD draw arcs intersecting each other at H
- Join HA

Therefore ∠HAB= 75^{0}

**3.105 ^{0}**

__Steps of construction:__

- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the centre and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each other at F.
- Join FA which intersects arc in (2) at G.
- Keeping E and G as center and radius more than half of EG draw arcs which intersect each other at H.
- Join HA.

Therefore ∠HAB= 105^{0}** **** **

**4. 135 ^{0}**

__Steps of construction:__

- Draw a line segment AB and produce BA to C.
- Keeping A as the center and any radius draw an arc which intersects AC at D and AB at E.
- Keeping D and E as center and radius more than half of DE draw arcs which intersect each other at F.
- Join FA which intersects the arc in (2) at G.
- Keeping G and D as center and radius more than half of GD draw arcs which intersect each other at H
- Join HA.

Therefore ∠HAB= 135^{0}** **

**5. 15 ^{0}**

__Steps of construction:__

- Draw a line segment AB.
- Keeping A as the centre and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D and C as center and radius more than half of DC draw arcs which intersect each other at E.
- Join EA which intersects arc in (2) at F.
- Keeping F and C as center and radius more than half of FC draw arcs which intersect each other at G.
- Join GA.

Therefore ∠GAB= 15^{0}

**6. **

__Steps of construction:__

- Draw a line segment AB.
- Keeping A as the center and any radius draw an arc which intersects AB at C.
- Keeping C as center and the same radius draw an arc which intersects the previous arc at D.
- Keeping D as the centre and same radius draw an arc which intersects arc in (2) at E.
- Keeping E and D as center and radius more than half of ED draw arcs which intersect each at F.
- Join FA which intersects arc in (2) at G.
- Keeping G and C as center and radius more than half of GC draw arcs intersecting each other at point H.
- Join HA which intersects the arc in (2) at a point I.
- Keeping I and C as center and radius more than half of IC draw arcs intersecting each other at point J.
- Join JA.

Therefore ∠JAB=

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