Get R S Aggarwal solutions for class 10 Chapter 9 Constructions here. We provide detailed explanations to all the exercise questions enlisted in the textbook. Subject experts have used simple language to explain the concepts and other technical jargons in detail. RS Aggarwal Maths Class 10 Solutions are the best guide for the students appearing for the board examinations.

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### Access Solutions of Maths RS Aggarwal Chapter 9 – Constructions

Get detailed solutions for all the questions listed under the below exercises:

## Exercise 9A

**Question 1: Draw a line segment AB of length 7 cm. Using ruler and compasses, find a point P on AB such that AP/AB = 3/5. **

**Solution:**

Steps of Construction:

1. Draw a line segment AB = 7cm.

2. Draw a line from A making an acute angle with line segment AB.

3. Taking A as center draw an arc cutting at A_{1} on the line.

And with the same radius consider A_{1} as a center and draw another arc cutting line at A_{2}.

Repeat the same procedure and divide the line AX from A into 8 equal parts:

AA_{1}, A_{1}A_{2}, A_{2}A_{3}, A_{3}A_{4}, A_{4}A_{5}, A_{5}A_{6}, A_{6}A_{7} and A_{7}A_{8 }

4. Join A_{8} and B by drawing a line.

5. Draw a parallel line to A_{8}B from A_{3} which divides line segment AB at point P.

6. P is the required point such that AP/AB = 3/5.

**Question 2:**

**(i) Draw a line segment of length 8 cm and divide it internally in the ratio 4:5.**

**Solution**:

Steps of construction:

1. Draw a line segment AB = 8 cm.

2. Draw a ray AX making an acute angle at A with AB.

3. Draw another ray BY parallel to AX making an acute angle. Make sure angle must be same as considered in step 2.

4.Taking A as center draw an arc cutting at A_{1} on the line.

Taking same radius consider A_{1} as a center and draw another arc cutting line at A_{2}.

Repeat the same procedure and divide the line AX into 4 points A_{1}, A_{2}, A_{3}, A_{4}.

In such a way, AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}

5. Similar to step 4,

Taking B as center draw an arc cutting at B_{1} on the line.

Taking same radius (set in step 4) consider B_{1} as a center and draw another arc cutting line at B_{2}.

Repeat the same procedure and divide the line BY into 5 points in such a way that BB_{1} = B_{1}B_{2}= B_{2}B_{3}= B_{3}B_{4} = B_{4}B_{5}

6. Join A_{4}B_{5}

7.Line A_{4}B_{5} intersect AB at a point P.

Therefore, P is the point dividing the line segment AB internally in the ratio of 4 : 5.

**(ii)Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts.**

**Solution:**

Steps of construction:

1. Draw a line segment AB = 7.6 cm.

2. Draw a ray AX making an acute angle at A with AB.

3. Draw another ray BY parallel to AX making an acute angle. Make sure angle must be same as considered in step 2.

4. Taking A as center draw an arc cutting at A_{1} on the line.

Taking same radius consider A_{1} as a center and draw another arc cutting line at A_{2}.

Repeat the same procedure and divide the line AX into 5 points A_{1}, A_{2}, A_{3}, A_{4} and A_{5}

In such a way, AA_{1} = A_{1}A_{2} = A_{2}A_{3} = A_{3}A_{4} = A_{4}A_{5}

5. Similar to step 4,

Taking B as center draw an arc cutting at B_{1} on the line.

Taking same radius (set in step 4) consider B_{1} as a center and draw another arc cutting line at B_{2}.

Repeat the same procedure and divide the line BY into 8 points in such a way that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5} = B_{5}B_{6} = B_{6}B_{7} = B_{7}B_{8}

6. Join A_{5}B_{8}

7. Line A_{5}B_{8} intersect AB at a point P in the ratio 5:8

8. Measurement: PB = 4.7 cm and AP = 2.9 cm

**Question 3: Construct a âˆ†PQR, in which PQ = 6 cm, QR = 7 cm and PR = 8 cm. Then, construct another triangle whose sides are 4/5 times the corresponding sides of âˆ†PQR.**

**Solution**:

Steps of construction:

1. Draw a line segment PQ = 6 cm.

2. Draw an arc, using P as a center and radius = 8 cm

3. Draw another arc, using Q as a center and radius = 7 cm

4. Now, join PR and QR to get â–³PQR

5. Draw a ray PX by making an acute angle, angle QPX

6. Divide PX into 4 equal parts

7. Join P5Q

8. Draw a line P4Q’ which is parallel to P5Q

9. Similar to step 8, draw a line Qâ€™Râ€™ which is parallel to QR

Therefore, â–³PQâ€™Râ€™ is the required triangle.

**Question 4: Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of first triangle.**

**Solution:**

Steps of construction:

1. Draw a line segment BC = 5 cm.

2. Draw an arc, using B as a center and radius = 6 cm

3. Draw another arc, using C as a center and radius = 7 cm

4. Now, join AC and AB to get â–³ABC

5. Draw a ray BX by making an acute angle, angle CBX

6.

7. Join R5C

8. Draw a line R7Câ€™ which is parallel to R5C

9. Similar to step 8, draw a line Câ€™Aâ€™ which is parallel to CA

Therefore,â–³Aâ€™BCâ€™ is the required triangle.

**Question 5: Construct a âˆ†ABC, with BC = 7 cm, âˆ B = 60^o and AB = 6 cm. Construct another triangle whose sides are times the corresponding sides of âˆ†ABC.**

**Solution: **

Steps of Construction:

1. Draw a triangle ABC with BC = 7 cm, âˆ B = 60^0 and AB = 6 cm.

2. Draw a ray BX making an acute angle with the line BC.

3. Divide BX into 4 equal arcs starting from B till B_{4}. As shown in the below image.

4. Join B_{4}C. Also draw a line from B_{3} parallel to B_{4}C passing BC at Câ€™.

5.Draw another line from Câ€™ parallel to CA passing AB at Aâ€™.

Thus, â–³Aâ€™BCâ€™ is required triangle.

**Question 6: Construct a âˆ†ABC in which AB = 6 cm, âˆ A = 30^o and âˆ B = 60^o. Construct another âˆ†ABâ€™Câ€™ similar to âˆ†ABC with base ABâ€™ = 8 cm. **

**Solution:**

Construct a âˆ†ABC in which AB = 6 cm, âˆ A = 30^o and âˆ B = 60^o. Construct another âˆ†ABâ€™Câ€™ similar to âˆ†ABC with base ABâ€™ = 8 cm.

1. Draw a line segment AB = 6cm and extend it to Bâ€™ such that ABâ€™ = 8 cm.

2. Make an angle of 30Â° from A and angle of 60Â° from B. The point where these extended rays meet is C.

3. Draw an angle of 60Â° from Bâ€™ (Similar triangles)

4. Extend line AC along C such that ray passing through point B’ Cut AX at C’.

5. AB’C’ is the required triangle.

**Question 7: Construct a âˆ†ABC in which BC = 8 cm, âˆ B = 45^0 and âˆ C = 60^o. Construct another triangle similar to âˆ†ABC such that its sides are 3/5 of the corresponding sides of âˆ†ABC. **

**Solution**:

Steps of Construction:

1. Draw a line segment BC = 8 cm.

2. Construct a triangle ABC based on given instructions.

3. Draw an arc below BC, by making an acute angle, angle CBZ

4. Divide BZ into 5 equal parts.

5. Join B_{5}C

6. Draw a line B_{3}C’ which is parallel to B_{5}C and join C’A’ which is parallel to CA.

Therefore, â–³Aâ€™BCâ€™ is the required triangle.

**Question 8: To construct a triangle similar to âˆ†ABC in which BC = 4.5 cm, âˆ B = 45^o and âˆ C = 60^o, using a scale factor of 3/7, BC will be divided in the ratio.**

**(a) 3 : 4 (b) 4 : 7 (c) 3 : 10 (d) 3 : 7**

**Solution:**

BC will be divided in the ratio 3 : 4. Option (a) is correct.

**Question 9: Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1_1/2 (or 3/2) times the corresponding sides of the isosceles triangle.**

**Solution**:

Steps of Construction:

1. Draw a line segment BC = 8 cm.

2. Draw a perpendicular bisector PQ of BC dividing BC at point M.

3. From QP cut off a distance MA = 4cm

4. Join AC and AB

5. Draw an arc below BC, by making an acute angle, angle CBX

6.

7. Join B_{2}C

8. Join B_{3}C’ which is parallel to B_{2}C and Join A’C’ which is parallel to AC

Thus, â–³Aâ€™BCâ€™ is required triangle.

**Question 10: Draw a right triangle in which sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then, construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. **

**Solution:**

Steps to construction:

1. Draw a line segment BC = 4 cm.

2. Construct, angle B = 90 degrees

3. Cut BA at the radius of 3 cm from BQ

4. Join AC (we have right triangle ABC)

5. Draw a ray BX and make an acute below BC, i.e. angle CBX

6.

7. Join B_{3}C

8. Join B_{5}C’ which is parallel to B_{3}C and C’A’ which is parallel to AC.

## Exercise 9B

**Question 1: Draw a circle of radius 3 cm. From a point P, 7 cm away from the centre of the circle, draw two tangents to the circle. Also, measure the lengths of the tangents. **

**Solution:**

Steps of Construction:

1. Draw a circle of radius 3 cm from center O.

2. Set a point P which is 7 cm far from point O. Join OP.

3. Draw a perpendicular bisector of OP which cuts OP at point Q.

4. Now, considering Q as a center and equal radius (OQ = PQ). Draw a circle.

5. Both circles intersect at points A and B.

6. Join PA and PB.

Therefore, AP and BP are the required tangents.

Measurements: AP = BP = 6.1 cm

**Question 2: Draw two tangents to a circle of radius 3.5 cm from a point P at a distance of 6.2 cm from its centre.**

**Solution:**

**Question 3: Draw a circle of radius 3 cm. Take two point P and Q on one of its diameters extended on both sides, each at a distance of 7 cm on opposite sides of its centre. Draw tangents to the circle from each of these two points P and Q.**

**Solution:**

**Question 4: Draw a circle with centre O and radius 4 cm. Draw any diameter AB of this circle. Construct tangents to the circle at each of the two end points of the diameter AB.**

**Solution:**

**Question 5: Draw a circle with the help of a bangle. Take any point P outside the circle. Construct the pair of tangents from the point P to the circle.**

**Solution:**

Steps of Construction:

1. Draw a circle with the help of a bangle.

2. Draw a point P outside the circle and draw two chords ST and QR.

3. Draw a perpendicular bisector form ST and QR.

4. Join PO.

5. Draw a circle taking U as a center and radius OU (letâ€™s say U is the midpoint of OP)

6. Both the circles are intersecting at W and V. Join PW and PV.

Therefore, PW and PV are required tangents.

**Question 6: Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.**

**Solution: **

Thus, AS, AR, BP and BQ are the required tangents.

**Question 7: Draw a circle of radius 4.2 cm. Draw a pair of tangents to this circle inclined to each other at an angle of 45Â°.**

**Solution: **

Both the lines intersect each other at point P. So PC and PA are tangents.

**Question 8: Write the steps of construction for drawing a pair of tangents to a circle of radius 3 cm, which are inclined to each other at an angle of 60^o.**

**Solution**:

**Question 9: Draw a circle of radius 3 cm. Draw a tangent to the circle making an angle of 30Â° with a line passing through the centre.**

**Solution:**

**Question 10: Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.**

**Solution: **

## RS Aggarwal Solutions for Class 10 Maths Chapter 9 Constructions

In this chapter students will study important concepts on constructions as listed below:

- Division of a line segment.
- Construct similar triangles.
- Construction of tangents to a circle.
- Calculation of the length of a tangent

### Key Features of RS Aggarwal Solutions for Class 10 Maths Chapter 9 Constructions

1. Step by step problem solving approach used to solve problems which helps students to clear their basics.

2. Easy for quick revision.

3. All questions are solved by subject experts at BYJUâ€™S.

4. Helpful for the other competitive exams as well.