R S Aggarwal Solutions for Class 10 Maths Chapter 9 Constructions, contains solutions for exercise 9A questions. This exercise is all about division of a line segment in a given ratio using different methods and construction of a triangle similar to a given triangle as per given scale factor. Students can download the R S Aggarwal Solutions of Class 10 and clear their doubts on any Maths topic.

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## 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 P_{5}Q

8. Draw a line P_{4}Q’ which is parallel to P_{5}Q

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 R_{5}C

8. Draw a line R_{7}Câ€™ which is parallel to R_{5}C

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.

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

Class 10 Maths Chapter 9 Constructions Exercise 9A is based on the topics:

- Division of a line segment
- Construction of triangles