In this chapter, we provide RD Sharma Class 7 ex 17.4 Solutions Chapter 17 Constructions for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 17.4 Solutions Chapter 17 Constructions Maths pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 7 |

Subject | Maths |

Chapter | Chapter 17 |

Chapter Name | Constructions |

Exercise | Ex 17.4 |

Table of Contents

**RD Sharma Solutions for Class 7 Chapter 17 Constructions** **Ex 17.4 Download PDF**

**Constructions Exercise 17.4**

**Question: 1**

Construct ∆ABC in which BC = 4 cm, ∠B = 50° and ∠C = 70°.

**Solution:**

**Steps of construction:**

- Draw a line segment BC of length 4 cm.
- Draw ∠CBX such that ∠CBX=50°.
- Draw ∠BCY with Y on the same side of BC as X such that ∠BCY=70°.
- Let CY and BX intersect at A.
- ABC is the required triangle.

**Question: 2**

Draw ∆ABC in which BC = 8 cm, ∠B = 50° and ∠A = 50°.

∠ABC + ∠BCA + ∠CAB = 180°

∠BCA = 180° − ∠CAB − ∠ABC

∠BCA = 180°− 100° = 80°

**Solution:**

**Steps of construction:**

- Draw a line segment BC of length 8 cm.
- Draw ∠CBX such that ∠CBX = 50°.
- Draw ∠BCY with Y on the same side of BC as X such that ∠BCY = 80°.
- Let CY and BX intersect at A.

**Question: 3**

Draw ∆ABC in which ∠Q = 80°, ∠R = 55° and QR = 4.5 cm. Draw the perpendicular bisector of side QR.

**Solution:**

**Steps of construction:**

- Draw a line segment QR = 4.5 cm.
- Draw ∠RQX = 80° and ∠QRY = 55°.
- Let QX and RY intersect at P so that PQR is the required triangle.
- With Q as centre and radius more that 2.25 cm, draw arcs on either sides of QR.
- With R as centre and radius more than 2.25 cm, draw arcs intersecting the previous arcs at M and N.
- Join MN

MN is the required perpendicular bisector of QR.

**Question: 4**

Construct ∆ABC in which AB = 6.4 cm, ∠A = 45° and ∠B = 60°

**Solution:**

**Steps of construction:**

Draw a line segment AB = 6.4 cm.

Draw ∠BAX = 45°.

Draw ∠ABY with Y on the same side of AB as X such that ∠ABY = 60°.

Let AX and BY intersect at C.

ABC is the required triangle.

**Question: 5**

Draw ∆ABC in which AC = 6 cm, ∠A = 90° and ∠B = 60°.

∠A + ∠B + ∠C = 180°

Therefore ∠C = 180°− 60°− 90°= 30°

**Solution:**

**Steps of construction:**

- Draw a line segment AC = 6 cm.
- Draw ∠ACX = 30°.
- Draw ∠CAY with Y on the same side of AC as X such that ∠CAY = 90°.
- Join CX and AY. Let these intersect at B.

ABC is the required triangle where angle ∠ABC = 60°.

**All Chapter RD Sharma Solutions For Class 7 Maths**

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