Chapter 5 Plane Geometry and Line and Angle Ex 5.3

Rajasthan Board RBSE Class 9 Maths Solutions Chapter 5 Plane Geometry and Line and Angle Ex 5.3

Question 1.
Draw a line segment AB = 10 cm. Bisect this line segment and verify by measuring the lengths of the two segments.
Solution.
Construction:
1. Draw a line segment AB = 10 cm
2. With A and B as centres and a radius more than half of AB, draw arcs on both side of the line segment AB, cutting each other at M and N respectively.

3. Join MN which intersect AB at L. Hence, L bisect AB.
By measuring we can say AL = BL = 5 cm.

Question 2.
Construct an angle of 120° . Bisect the angle and verify by measuring the two angles.
Solution.
Steps of Construction:
1. Draw a line segment of any measure named as PQ.
2. With P as centre draw an angle of 120° with the help of protractor.

3. With P as centre again draw an arc which cuts line segments PQ and PS at M and N.
4. With M and N as centres and with radius little more than half the arc, draw arcs cutting each other at T. Join PT which is the required bisector of ∠SPQ = 120°.
∴ ∠TPQ = ∠TPS = 60° (each)

Question 3.
Draw an angle of 40° using protractor. Now, with the help of ruler and compass, construct an angle equal to this angle.
Solution.
Steps of Construction:
1. Draw ∠BAC = 40° with the help of protractor. By taking A as centre draw an arc which cuts AB and AC at P & Q.

2. Now take O as centre on another line segment EF and taking the same radius PQ draw an arc which cuts OF at R. Now again take R as centre and draw the arc which cut at S.
3. Join O to S which is the required ∠ROS = 40°.

Question 4.
Draw a line segment of length 6 cm. From a point P outside this line, draw a perpendicular on this line.
Solution.
Steps of construction:
1. First of all draw a line segment AB = 6 cm

2. Take a point P (outside the line) and draw an arc in such a way that its cuts AB into two points C and D. Again by taking C & D as centre take suitable arc and draw two arc which intersect at Q. Join PQ.
Hence, PQ is the required perpendicular on AB from P.

Question 5.
Construct ∠ABC = 120°. Through A draw a line parallel to BC.
Solution.
Steps of construction:
1. Draw a line segment BC of any measure.
2. By taking B as centre, draw an angle of 120° with the help of compass and ruler.
3. Now at A, we will draw a line parallel to BC by making 60° at A.

(Reason: Lines are parallel if sum of the interior angles on the same side of transversal is 180°)

Question 6.
Draw a line segment of length 9 cm using ruler and compass. Divide this line segment into three equal parts.
Solution.
Steps of construction:
1. Draw a line segment AB = 9 cm.
2. Make any acute angle at A.
3. Starting from A as centre and any convenient radius set off three equal distances.
4. Join the third arc to B.
5. Then we will draw line segments parallel to EB.

Hence AC = CD = DB = 3 cm each.

Question 7.
Draw a line segment of length 10 cm. Using ruler and compass divide this segment in the ratio 3 : 2. Measure the length of these segments.
Solution.
Steps of construction:
1. Draw any line segment say AB = 10 cm
2. At A, draw a ray AY making an acute angle YAB.
3. Using compass mark five points on AY such that
AP = PQ = QR = RS = ST.
4. Join TB.
5. Through point R draw a line parallel to TB i.e RL which divides AB in the ratio 3 : 2.

On measuring AB we find that AL = 6 cm and LB = 4 cm.

Question 8.
Draw a line segment AB of length 6 cm, using ruler and compass divide this in the ratio 1:2:3.
Solution.
Steps of construction:
1. Draw a line segment AB = 6 cm.
2. Make any acute angle at A. i.e. ∠CAB.
3. Starting from A as centre and any convenient radius set off 6 equal distances.
4. Join 6th arc to B.
5. Then we will draw line segment PM and QN parallel to OB.

Hence AN : NM : MB =1:2:3.

Question 9.
Using ruler and compass, construct the following angles.
(i) 150°
(ii) 105°
(iii) 75°
Solution.
Steps of Construction
(i) 150° = 189° –  x 60 = 180° – 30°
First draw ∠AOC = 60° then draw the bisector of ∠AOC.
Hence ∠BOP = 150° is the required angle.

(ii) Let OB be a line segment and at O construct an angle of 90°. With this vertical line construct an angle of 15° which gives required ∠BOP = 105°.

(iii) Let OB be a straight line and at O draw an angle of 60° and 90° then draw bisector of these two angles, which is the required ∠BOP = 75°.