NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 11 |

Chapter Name |
Constructions |

Exercise |
Ex 11.1 |

Number of Questions Solved |
7 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1

**In each of the following, give the justification of the construction also:**

**Question 1.**

**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 another line segment (or ray) AX at any convenient acute angle with AB.

3. Along AX, locate 13 (5 + 8) points A

_{1}, A

_{2}, A

_{3}, …… , A

_{13}such that AA

_{1}= A

_{1}A

_{2}= ……… = A

_{12}A

_{13 }4. Join BA

_{13}.

5. At point, A

_{5}draw a straight line A

_{5}C parallel to A

_{13}B making an angle equal to ∠AA

_{13}B

**.**

Thus, the point C divides the line segment AB in the ratio 5: 8.

**Justification:**

**Question 2.
**

**Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are of the corresponding sides of the first triangle.**

**Solution:**

**Steps of Construction:**

1. Draw AC – 6 cm (ii) With A and Cas centers and radii 4 cm.

2. 5 cm respectively draw two ares intersecting each other at B. Join BA and BC.

3. Draw a ray AY making an acute angle with AC.

4. Locate three points and Ron AY, such that AP – POQR.

5. Join CR.

6. Through o. draw a line QC’ parallel to RC (by making an angle equal to ∠ARC) meeting the line segment AC at C’.

7. Similarly, through C, draw a line B’C parallel to CB

Thus, ABC is the required triangle, which is similar to ∆ABC with scale factor

**Justification:**

By construction, we have:

**Question 3.
**

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

**Solution:**

**Steps of Construction:**

1. Construct AABC such that AB = 5 cm, BC = 6 cm and CA = 7 cm.

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

3. Locate 7 points on BX such that BD = DE = EF = FG – GH – IJ.

4. Join HC.

5. Through J, draw a line JC parallel to HC, meeting produced line BC at C’.

6. Through C’, draw a line 7 parallel to CA meeting the produced line BA at A’.

Thus A∆ABC is a required triangle

**Justification:**

By construction, we have:

**Question 4.**

**Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1 times the corresponding sides of the isosceles triangle.**

**Solution:**

**Steps of Construction:
**1. Construct an isosceles triangle ABC in which BC – 8 cm and altitude AD is 4 cm.

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

3. Locate 3 points on BX, such that BP – PQ = QR.

4. Join QC.

5. Through R, draw a line RC parallel to QC, meeting produced line BC at C’.

6. Through C, draw a line CA parallel to CA, meeting the produced line BA at A’.

Thus, ∆A’BC’ is the required isosceles triangle

**Justification:**

**Question 5.**

**Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are of the corresponding sides of the triangle ABC.**

**Solution:**

**Steps of Construction:
**1. Construct a ABC in which BC = 6 cm, AB = 5 cm and ∠ABC = 60°.

2. Draw a ray BX making any acute angle with BC.

3. Locate 4 points B

_{1}, B2, B

_{3}and B

_{4}on BX, such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}=B

_{3}B

_{4}

4. Join B

_{4}C.

5. Through B

_{3}draw a line B

_{3}C’ parallel to B

_{4}C, intersecting the line segment BC at C’.

6. Through C, draw a line parallel to CA, intersecting the line segment BA at A’.

Thus, ∆A’BC’ is the required triangle.

**Justification:**

**Question 6.**

**Draw a triangle ABC with side BC = 7 cm, ∠B = 45°, ∠A = 105°. Then, construct a triangle whose sides are times the corresponding sides of ∆ABC.**

**Solution:**

**Steps of Construction:
**1. Draw a line segment BC – 7 cm.

2. Draw ∠ABC = 45° and ∠ACB = 30°, i.e., ∠BAC = 105°.

3. We get ∆ABC

4. Draw a ray BX making an acute angle with BC

5. Mark four points B

_{1}, B

_{2}, B

_{3}and B

_{4}on BX, such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}= B

_{3}B

_{4.}

6. Join B

_{3}C.

7. Through B

_{4}draw a line B

_{4}C’ parallel to B

_{3}C, intersecting the extended line segment BC at C’.

8. Through C’, draw a line A’C’ parallel to CA, intersecting the extended line segment BA at A’.

Thus, A’BC’ is the required triangle.

**Justification:**

**Question 7.
**

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

**Solution:**

**Steps of Construction:**

1. Construct a ∆ABC, such that BC = 4 cm, CA = 3 cm and ∠BCA = 90°

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

3. Mark five points B

_{1}, B

_{2}, B

_{3}, B

_{4}and B

_{5}on BX, such that BB

_{1}= B

_{1}B

_{2}= B

_{2}B

_{3}= B

_{3}B

_{4}= B

_{4}B

_{5.}

4. Join B

_{3}C.

5. Through B

_{5}, draw B

_{5}C’ parallel to B

_{3}C intersecting BC produced at C’.

6. Through C’, draw C’A’ parallel to CA intersecting AB produced at A’.

Thus, ∆A’BC’ is the required right triangle.

**Justification:**

We hope the NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1 help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 11 Constructions Ex 11.1, drop a comment below and we will get back to you at the earliest.

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