Rs Aggrawal 2020 2021 Solutions for Class 6 Maths Chapter 14 Constructions Using Ruler And A Pair Of Compasses are provided here with simple step-by-step explanations. These solutions for Constructions Using Ruler And A Pair Of Compasses are extremely popular among Class 6 students for Maths Constructions Using Ruler And A Pair Of Compasses Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggrawal 2020 2021 Book of Class 6 Maths Chapter 14 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Rs Aggrawal 2020 2021 Solutions. All Rs Aggrawal 2020 2021 Solutions for class Class 6 Maths are prepared by experts and are 100% accurate.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line segment PQ, which is equal 6.2 cm.

2. With P as the centre and radius more than half of PQ, draw arcs, one on each side of PQ.

3. With Q as the centre and the same radius as before, draw arcs cutting the previously drawn arcs at A and B, respectively.

4. Draw AB, meeting PQ at R.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line segment AB = 5.6 cm.

2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.

3. With B as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at P and Q, respectively.

4. Draw PQ, meeting AB at R.

#### Page No 186:

#### Answer:

Here $\angle $AOB is given.

Steps for construction:

1. Draw a ray QP.

2. With O as the centre and any suitable radius, draw an arc cutting OA and OB at C and E, respectively..

3. With Q as the centre and the same radius as in step (2), draw an arc cutting QP at D.

4. With D as the centre and radius equal to CE, cut the arc through D at F.

5. Draw QF and produce it to point R.

$\therefore $ $\angle $PQR = $\angle $AOB

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw $\angle $BAC = 50$\xb0$ with the help of protractor.

2. With A as the centre and any convenient radius, draw an arc cutting AB and AC at Q and P, respectively.

3. With P as the centre and radius more than half of PQ, draw an arc.

4. With Q as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point S.

5. Draw SA and produce it to point R.

Then, ray AR bisects $\angle $BAC.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw $\angle $AOB = 85$\xb0$ with the help of a protractor.

2. With O as the centre and any convenient radius, draw an arc cutting OA and OB at P and Q, respectively.

3. With P as the centre and radius more than half of PQ, draw an arc.

4. With Q as the centre and the same radius as before, draw another arc cutting the previously drawn arc at a point R.

5. Draw RO and produce it to point X.

Then, ray OX bisects $\angle $AOB.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line AB.

2. Take a point P on line AB.

3. With P as the centre, draw an arc of any radius, which intersects line AB at M and N, respectively.

4. With M as the centre and radius more than half of MN, draw an arc.

5. With N as the centre and the same radius as in step (4), draw an arc that cuts the previously drawn arc at R.

6. Draw PR.

PR is the required line, which is perpendicular to AB.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line AB.

2. Take a point P outside AB.

3. With P as the centre and a convenient radius, draw an arc intersecting AB at M and N, respectively.

4. With M as the centre and radius more than half of MN, draw an arc.

5. With N as the centre and the same radius, draw an arc cutting the previously drawn arc at Q.

6. Draw PQ meeting AB at S.

PQ is the required line that passes through P and is perpendicular to AB.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line AB.

2. Take a point P outside AB and another point O on AB.

3. Draw PO.

4. Draw $\angle $FPO such that $\angle $FPO is equal to AOP.

5. Extend FP to E.

Then, the line EF passes through the point P and EF||AB.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line BX and take a point A, such that AB is equal to 4.5 cm.

2. Draw $\angle $ABP = 60$\xb0$ with the help of protractor.

3. With A as the centre and a radius of 5 cm, draw an arc cutting PB at C.

4. Draw AC.

5. Now, draw $\angle $BCY = 60$\xb0$.

6. Then, draw $\angle $ABW, such that $\angle $ABW is equal to$\angle $CAX, which cut the ray CY at D.

7. Draw BD.

When we measure BD and CD, we have:

BD = 5 cm and CD = 4.5 cm

#### Page No 186:

#### Answer:

Steps of constructions

1. Draw a line segment AB, which is equal to 6 cm.

2. Take a point C on AB such that AC is equal to 2.5 cm.

3. With C as the centre, draw an arc cutting AB at M and N.

4. With M as the centre and radius more than half of MN, draw an arc.

5. With N as the centre and the same radius as before, draw another arc cutting the perviously drawn arc at S.

6. Draw SC and produce it to D.

#### Page No 186:

#### Answer:

Steps for construction:

1. Draw a line segment AB, which is equal to 5.6 cm.

2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.

3. With B as the centre and the same radius as before, draw arcs cutting the perviously drawn arcs at M and N, respectively.

4. Draw MN, meeting AB at R.

#### Page No 188:

#### Answer:

(i)

Steps for construction:

1. Draw a ray QP.

2. With Q as the centre and any convenient radius, draw an arc cutting QP at N.

3. With N as the centre and the same radius as before, draw another arc to cut the previous arc at M.

4. Draw QM and produce it to R.

^{$\angle $} PQR is the required angle of 60^{o}.

(ii)

Steps for construction:

1. Draw a ray QP.

2. With Q as the centre and any convenient radius, draw an arc cutting QP at N.

3. With N as the centre and the same radius, cut the arc at A. Again, with A as the centre and the same radius, cut the arc at M.

4. Draw QM and produce it to R.

^{$\angle \mathrm{PQR}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{angle}\mathrm{of}120\xb0.$}

(iii)

Steps for construction:

1. Draw a line PX.

2. Take a point Q on AC. With Q as the centre and any convenient radius, draw an arc cutting AX at M and N.

3. With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at W.

5. Draw QW produce it to R.

$\angle \mathrm{PQR}\mathrm{is}\mathrm{required}\mathrm{angle}\mathrm{of}90\xb0.$

#### Page No 188:

#### Answer:

Constructions steps:

1. Draw a ray QP.

2. Wth Q as the centre and any convenient radius,draw an arc cutting QP at N.

3. With N as the centre and radius same as before, draw another arc to cut the previous arc at M.

4. Draw QM and produce it to R.

$\angle \mathrm{PQR}\mathrm{is}\mathrm{an}\mathrm{a}\mathrm{ngle}\mathrm{of}60\xb0.$

5. With M as the centre and radius more than half of MN, draw an arc.

6. With N as the centre and radius same as in step (5), draw another arc, cutting the previously drawn arc at point X.

7. Draw QX and produce it to point S.

Ray QS is the bisector of $\angle PQR$.

#### Page No 188:

#### Answer:

Construction steps:

1. Draw a line PR.

2. Take a point Q on PR. With Q as the centre and any convenient radius, draw an arc cutting AC at M and N.

3.With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at X.

5. Draw QX, meeting the arc at Z. Produce it to W.

6. With Z as the centre and radius more than half of ZN, draw an arc.

7. With N as the centre and the same radius as in step (6), draw another arc, cutting the previously drawn arc at a point Y.

8. Draw QY and produce it to point S.

$\angle \mathrm{PQS}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{angle}\mathrm{of}45\xb0.$

#### Page No 188:

#### Answer:

(i)

Steps for construction:

1. Draw a line XY and take a point O.

2. With O as the centre and any suitable radius, draw an arc cutting XY at M and N.

3.With N as the centre and the same radius,draw an arc cutting MN at R.

4.With R as the centre and the same radius as before, draw another arc cutting MN at Q .

5. With Q as the centre and radius less than MQ draw an arc.

6. With M as the centre and the same radius draw another arc cutting the previously drawn arc at P

5. Join PO.

$\therefore $ $\angle $XOP = 150$\xb0$

(ii)

Steps for construction:

1. Draw a ray XY.

2. With X as the centre and any convenient radius, draw an arc cutting XY at M.

3. With M as the centre and the same radius, draw an arc cutting the previously drawn arc at N.

4. Draw YN and produce it to B.

4. Draw the bisector AY of $\angle $XYB.

5. Again, draw the bisector YZ of $\angle $XYA.

$\therefore $ $\angle $XYZ = 15$\xb0$

(iii)

Steps for construction:

1. Draw a line XY and take a point A.

2. With A as the centre and any convenient radius, draw an arc cutting XY at M and N.

3. With N as the centre and the same radius, draw an arc.

4. With M as the centre and the same radius as before, draw another arc cutting the previously drawn arc at R.

5. Draw RA.

6. Draw draw the bisector ZA of $\angle $YAR.

$\therefore $ $\angle $XAZ = 135$\xb0$

(iv)

Steps for construction:

1. Draw a line XY.

2. Take a point A on XY. With A as the centre and any convenient radius, draw an arc cutting XY at M and N.

3. With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at P.

5. Draw PA meeting the arc at C. Produce it to E.

6. With C as the centre and radius more than half of CN, draw an arc.

7. With N as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point Q.

8. Draw AQ and produce it to point F.

9. Draw the bisector ZA of $\angle $XAF.

$\therefore $ $\angle $XAZ = 22.5$\xb0$

(v)

Steps for construction:

1. Draw a line XY.

2. Take a point O on XY. With O as the centre and any convenient radius, draw an arc cutting XY at M and N. Draw arcs with the same radius cutting MN at P and Q.

3. With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at B.

5. Draw BO meeting the arc at E.

6. With Q as the centre and radius more than half of PE, draw an arc.

7. With E as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point A.

8. Draw AO and produce it to point Z.

$\therefore $ $\angle $XOZ = 105$\xb0$

(vi)

Steps for construction:

1. Draw a line XY.

2. Take a point O on XY. With O as the centre and any convenient radius, draw an arc cutting XY at M and N. Draw arcs with the same radius cutting MN at P.

3. With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at R.

5. Draw RO meeting the arc at E. Produce it to A.

6. With P as the centre and radius more than half of PE, draw an arc.

7. With E as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point S.

8. Draw OS and produce it to point Z.

$\therefore $ $\angle $XOZ = 75$\xb0$

(vii)

Steps for construction:

1. Draw a line XY and take a point O.

2. With O as the centre and any convinient radius, draw an arc cutting XY at M and N.

3. With N as the centre and the same radius, draw an arc.

4. With M as the centre and the same radius as before, draw another arc cutting the previously drawn arc at Q.

5. Draw QO.

6. Draw PO bisector of $\angle $YOA.

7. Draw ZO bisector of $\angle $POX.

$\therefore $ $\angle $XAZ = 67.5$\xb0$

(viii)

Steps for construction:

1. Draw a line PR.

2. Take a point Q on PR. With Q as the centre and any convenient radius, draw an arc cutting AC at M and N.

3. With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at X.

5. Draw QX, meeting the arc at Z. Produce it to W.

6. With Z as the centre and radius more than half of ZN, draw an arc.

7. With N as the centre and the same radius as in step (6), draw another arc cutting the previously drawn arc at a point Y.

8. Draw QY and produce it to point S.

#### Page No 188:

#### Answer:

Construction steps:

1. Draw a ray AX.

2. With A as the centre, cut the ray AX at B such that AB is equal to 5 cm.

3. With B as the centre and any convenient radius, draw an arc cutting AX at M and N.

4. With N as the centre and radius more than half of MN, draw an arc.

5. With M as the centre and the same radius as before, draw another arc to cut the previous arc at Y.

6. Draw BY and produce it to W.

7. With B as the centre and a radius of 3.5 cm, cut ray BW at point C.

8. With C as the centre and a radius of 5 cm, draw an arc on the right side of BC.

9. With A as the centre and a radius of 3.5 cm, draw an arc cutting the previous arc at D.

10. Join CD and AD.

ABCD is the required rectangle.

#### Page No 188:

#### Answer:

Construction steps:

1. Draw a ray AX.

2. With A as centre cut the ray AX at B such that AB=5 cm

3. With B as centre and any convenient radius,draw an arc cutting AX at M and N.

4. With N as centre and radius more than half of MN draw an arc.

5. With M as centre and the same radius as before,draw another arc to cut the previous arc at Y.

6. Join BY and produced it to W.

7. With B as centre and radius 5 cm cut ray BW at point C.

8.With C as centre and radius 5 cm draw an arc on right side of BC.

9. With A as centre and radius 5 cm draw an arc cutting the previous arc at D.

10.Join CD and AD.

ABCD is required square.

#### Page No 189:

#### Answer:

(i) We can draw infinite number of lines passing through a given point.

(ii) Only one line can be drawn with two given points.

(iii) We can draw one line with three given points if all the three point are collinear.

But, if the points are not collinear, then we cannot draw any line passing through the points.

#### Page No 189:

#### Answer:

(i) It is an acute angle because it is more than 0$\xb0$ and less than 90$\xb0$.

(ii) It is an obtuse angle because it is more than 90$\xb0$ and less than 180$\xb0$.

(iii) It is a reflex angle because it is more than 180$\xb0$ and less than 360$\xb0$.

(iv) It is a right angle because it is 90$\xb0$.

(v) It is an straight angle because it is 180$\xb0$.

#### Page No 189:

#### Answer:

Steps for construction:

1. Draw a line segment AB, which is equal to 6 cm.

2. With A as the centre and radius more than half of AB, draw arcs, one on each side of AB.

3. With B as the centre and radius same as before, draw arcs, cutting the perviously drawn arcs at M and N, respectively.

4. Draw MN meeting AB at D.

MN is the required perpendicular bisector of AB.

#### Page No 189:

#### Answer:

Steps of construction:

1. Draw a ray QP.

2.With Q as the centre and any convenient radius, draw an arc cutting QP at N.

3.With N as the centre and the same radius, cut the arc at A. Again, with A as the centre and the same radius, cut the arc at M.

4. Draw QM and produce it to R.

$\angle \mathrm{PQR}\mathrm{is}120\xb0.$

5. With M as the centre and radius more than half of MN, draw an arc.

6. With N as the centre and the same radius mentioned in step(5), draw another arc, cutting the previously drawn arc at point X.

7. Draw QX and produce it to point S.

$\mathrm{Ray}\mathrm{QS}\mathrm{is}\mathrm{a}\mathrm{bisector}\mathrm{of}\angle \mathrm{PQR}.$

#### Page No 189:

#### Answer:

Construction steps:

1. Draw a line OA.

2. Take a point B on OA. With B as the centre and any convenient radius, draw an arc cutting OA at M and N.

3. With N as the centre and radius more than half of MN, draw an arc.

4. With M as the centre and the same radius as before, draw another arc to cut the previous arc at W.

5. Draw WB, meeting the arc at S. Produce it to C.

$\angle \mathrm{ABC}\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{angle}\mathrm{of}90\xb0$.

6. With S as the centre and radius more than half of SN, draw an arc.

7. With N as centre and the same radius as in step (6), draw another arc, cutting the previously drawn arc at point X.

8. Draw BX and produce it to point D.

$\mathrm{Ray}\mathrm{BD}\mathrm{is}\mathrm{the}\mathrm{angle}\mathrm{bisctor}\mathrm{of}\angle \mathrm{ABC}.$ $\mathrm{Ray}\mathrm{BD}i\mathrm{s}\mathrm{the}\mathrm{angle}\mathrm{bisctor}\mathrm{of}\angle \mathrm{ABC}.$

#### Page No 189:

#### Answer:

Steps of construction:

1. Draw a ray AX.

2. With A as the centre, cut the ray XA at B, such that AB is equal to 3.5 cm.

3. With B as the centre and with any convenient radius, draw an arc cutting AX at M and N.

4. With N as the centre and with radius more than half of MN, draw an arc.

5. With M as the centre and with the radius same as before, draw another arc to cut the previous arc at Y.

6. Draw BY and produced it to W.

7. With B as the centre and a radius of 5.4 cm, cut ray BW at point C.

8. With C as the centre and a radius 3.5 cm, draw an arc on the right side of BC.

9. With A as the centre and a radius 5.4 cm, draw an arc cutting the previous arc at D.

10. Join CD and AD.

ABCD is the required rectangle.

#### Page No 189:

#### Answer:

(c) A line

A line has no end points. We can produce it infinitely in both directions.

#### Page No 189:

#### Answer:

(b) A ray

A ray has one end point. We can produce a ray infinitely in one direction.

#### Page No 189:

#### Answer:

(a) A line segment

A line segment has two end points and both of them are fixed. Thus, a line segment has a fixed length.

#### Page No 189:

#### Answer:

(b) in a line

When the common points of two planes intersect, they form a line.

#### Page No 189:

#### Answer:

(b) 135$\mathit{\xb0}$

$\frac{3}{2}\mathrm{right}\mathrm{angles}=\frac{3}{2}\times 90\xb0=135\xb0$

#### Page No 189:

#### Answer:

(c) on the angle

#### Page No 189:

#### Answer:

(d) a reflex angle

This is because it is more than 180$\xb0$ and less than 360$\xb0$.

#### Page No 189:

#### Answer:

(i) A line has __no__ end point.

(ii) A ray has __one__ end point

(iii) A line __cannot__ be drawn on a paper.

(iv) 0° < acute angle < 90° < obtuse angle < 180°.

(v) The standard unit of measuring an angle is __degree(°)__.

#### Page No 189:

#### Answer:

(i) False

If two line segments do not intersect, they may or may not be parallel.

(ii)False

If two rays do not intersect, they may or may not be parallel.

(iii) True

(iv)True

(v)False

We can produce a ray in one direction.

(vi)False

$\overrightarrow{AB}$ means A is fixed and B is not fixed. In other words, we can produce AB towards B.

$\overrightarrow{BA}$ means B is fixed and A is not fixed. In other words, we can produce B towards A.

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