NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.2 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.2.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 12 |

Chapter Name |
Constructions |

Exercise |
Ex 12.2 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 12 Constructions Ex 12.2

**Question 1.**

**Construct a ∆ ABC in which BC = 7 cm, ∠B = 75° and AB + AC = 13 cm.**

**Solution:**

Given that, in ∆ ABC, BC = 7 cm, ∠B = 75° and AS + AC = 13 cm

Steps of construction

- Draw the base BC = 7 cm

- At the point 6 make an ∠XBC = 75°.
- Cut a line segment BD equal to AB + AC = 13 cm from the ray BX.
- Join DC.
- Make an ∠DCY = ∠BDC.
- Let CY intersect BX at A.

Then, ABC is the required triangle.

**Question 2.**

**Construct a ∆ ABC in which BC = 8 cm, ∠B = 45° and AB – AC = 35 cm.**

**Solution:**

Given that, in ∆ ABC,

BC = 8 cm, ∠B = 45°and AB – AC = 3.5 cm

**Steps of construction**

- Draw the base BC = 8 cm
- At the point B make an ∠XBC = 45°.
- Cut the line segment BD equal to AB – AC = 3.5 cm from the ray BX.
- Join DC.
- Draw the perpendicular bisector, say PQ of DC.
- Let it intersect BX at a point A
- Join AC.

**Question 3.**

**Construct a ∆ ABC in which QR = 6 cm, ∠Q = 60° and PR – PQ = 2 cm.**

**Solution:**

Given that, in ∆ ABC, QR = 6 crn ∠Q = 60° and PR – PQ = 2 cm

**Steps of construction**

- Draw the base QR = 6 cm
- At the point Q make an ∠XQR = 60°.
- Cut line segment QS = PR- PQ (= 2 cm) from the line QX extended on opposite side of line segment QR.
- Join SR.
- Draw the perpendicular bisector LM of SR.
- Let LM intersect QX at P.
- Join PR.

**Question 4.**

**Construct a ∆ XYZ in which ∠Y = 30°, ∠Y = 90° and XY + YZ + ZX = 11 cm.**

**Solution:**

Given that, in ∆XYZ ∠Y = 30°, ∠Z = 90° and XY + YZ + ZX = 11cm

**Steps of construction**

- Draw a line segment BC = XY + YZ + ZX = 11 cm
- Make ∠LBC = ∠Y = 30° and ∠MCB = ∠Z = 90°.
- Bisect ∠LBC and ∠MCB. Let these bisectors meet at a point X.
- Draw perpendicular bisectors DE of XB and FG of XC.
- Let DE intersect BC at Y and FC intersect BC at Z.
- Join XY and XZ.

Then, XYZ is the required triangle.

**Question 5.**

**Construct a right triangle whose base is 12 cm and sum of its hypotenuse and other side is 18 cm.**

**Solution:**

Given that, in A ABC, base BC = 12 cm, ∠B = 90° and AB + BC= 18 cm.

**Steps of construction**

- Draw the base BC = 12 cm
- At the point 6, make an ∠XBC = 90°.
- Cut a line segment BD = AB+ AC = 18 cm from the ray BX.
- Join DC.
- Draw the perpendicular bisector PQ of CD to intersect SD at a point A

Join AC.

Then, ABC is the required right triangle.