# RD Sharma Solutions Class 9 Constructions Exercise 17.2

## RD Sharma Solutions Class 9 Chapter 17 Exercise 17.2

Q1. Draw an angle and label it as $\angle BAC$. Construct another angle, equal to $\angle BAC$

Steps of construction:

1. Draw an angle ABC and a line segment QR.
2. With center A and any radius, draw an arc which intersects $\angle BAC$at E and D.
3. With Q as a centre and same radius draw an arc which intersects QR at S.
4. With S as center and radius equal to DE, draw an arc which intersects the previous arc at T.
5. Draw a line segment joining Q and T.

Therefore $\angle PQR$= $\angle BAC$

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

Steps of construction:

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

Therefore $\angle ABR$=  $\angle RBC$=  $60^{0}$

Q3. Using your protractor, draw an angle of $108^{0}$. With this given angle as given, draw an angle of $54^{0}$.

Steps of construction:

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

Therefore $\angle RBC$=  $54^{0}$

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

Steps of construction:

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

Therefore $\angle RBC$=  $45^{0}$

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:

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

Therefore $\angle SCT$=  $90^{0}$.

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:

1. Draw a pair of vertically opposite angle $\angle AOC$ and $\angle DOB$.
2. 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.
3. Keeping P and Q as center and radius more than half of PQ draw two arcs which intersect each other at T.
4. Join TO.
5. Keeping R and S as center and radius more than half of RS draw two arcs which intersect each other at U.
6. Join OU.

Therefore TOU is a straight line

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

Steps of construction:

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

Therefore $\angle FAB$=  $90^{0}$

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

Steps of construction:

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

Therefore $\angle HAB$=  $135^{0}$

Q9. Using a protractor, draw an angle of measure $72^{0}$. With this angle as given draw angles of measure $36^{0}$ and $54^{0}$.

Steps of construction:

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

Therefore $\angle HBC$=  $54^{0}$

$\angle FBC$=  $36^{0}$

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

1. $45^{0}$
2. $90^{0}$

1. $45^{0}$

Steps of construction:

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

Therefore $\angle HAB$=  $45^{0}$

1. $90^{0}$

Steps of construction

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

Therefore $\angle FAB$=  $90^{0}$

Q11. Construct the angles of the following measurements:

1. $30^{0}$
2. $75^{0}$
3. $105^{0}$
4. $135^{0}$
5. $15^{0}$
6. $22\frac{1}{2}^{0}$

1. $30^{0}$

Steps of construction:

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

Therefore $\angle EAB$=  $30^{0}$

1. $75^{0}$

Steps of construction:

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

Therefore $\angle HAB$=  $75^{0}$

1. $105^{0}$

Steps of construction:

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

Therefore $\angle HAB$=  $105^{0}$

1. $135^{0}$

Steps of construction:

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

Therefore $\angle HAB$=  $135^{0}$

1. $15^{0}$

Steps of construction:

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

Therefore $\angle GAB$=  $15^{0}$

1. $22\frac{1}{2}^{0}$

Steps of construction:

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

Therefore $\angle JAB$=  $22\frac{1}{2}^{0}$