In this chapter, we provide RD Sharma Solutions for Chapter 16 Circles Ex 16.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 16 Circles Ex 16.3 Maths pdf, free RD Sharma Solutions for Chapter 16 Circles Ex 16.3 Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 16 |

Chapter Name | Circles |

Exercise | Ex 16.3 |

**RD Sharma Solutions for Class 9 Chapter 16 Circles Ex 16.3 Download PDF**

Question 1.

Construct a ∆ABC in which BC = 3.6 cm, AB + AC = 4.8 cm and ∠B = 60°.

Solution:

Steps of construction :

(i) Draw a line segment BC = 3.6 cm.

(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 4.8 cm.

(iii) Join EC.

(iv) Draw perpendicular bisector of CE which intersects BE at A.

(v) Join AC.

∆ABC is the required triangle.

Question 2.

Construct a ∆ABC in which AB + AC = 5.6 cm, BC = 4.5 cm and ∠B = 45°.

Solution:

Steps of construction :

(i) Draw a line segment BC = 4.5 cm.

(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 5.6 cm and join CE.

(iii) Draw the perpendicular bisector of CE which intersects BE at A.

(iv) Join AC.

∆ABC is the required triangle.

Question 3.

Construct a AABC in which BC = 3.4 cm, AB – AC = 1.5 cm and ∠B = 45°.

Solution:

Steps of construction :

(i) Draw a line segment BC = 3.4 cm.

(ii) At B, draw a ray BX making an angle of 45° and cut off BE = 1.5 cm.

(iii) Join EC.

(iv) Draw the perpendicular bisector of CE which intersects BE produced at A.

(v) Join AC.

∆ABC is the required triangle.

Question 4.

Using ruler and compasses only, construct a ∆ABC, given base BC = 7 cm, ∠ABC = 60° and AB + AC = 12 cm.

Solution:

Steps of construction :

(i) Draw a line segment BC = 7 cm!

(ii) At B, draw a ray BX making an angle of 60° and cut off BE = 12 cm.

(iii) Join EC.

(iv) Draw the perpendicular bisector of EC which intersects BE at A.

(v) Join AC.

∆ABC is the required triangle.

Question 5.

Construct a right-angled triangle whose perimeter is equal to 10 cm and one acute angle equal to 60°.

Solution:

Steps of construction :

(i) Draw a line segment PQ = 10 cm.

(ii) At P, draw a ray PX making an angle of 90° and at Q, QY making an angle of 60°.

(iii) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.

(iv) Draw the perpendicular bisectors of AP and AQ intersecting PQ at B and C respectively,

(v) Join AB and AC.

∆ABC is the required triangle.

Question 6.

Construct a triangle ABC such that BC = 6cm, AB = 6 cm and median AD = 4 cm.

Solution:

Steps of construction :

(i) Draw a line segment BC = 6 cm and bisect it at D.

(ii) With centre B and radius 6 cm and with centre D and radius 4 cm, draw arcs intersecting each other at A.

(iii) Join AD and AB and AC.

Then ∆ABC is the required triangle.

Question 7.

Construct a right triangle ABC whose base BC is 6 cm and the sum of hypotenuse AC and other side AB is 10 cm.

Solution:

Steps of construction :

(i) Draw a line segment BC = 6 cm

(ii) At B, draw a ray BX making an angle of 90° and cut off BE = 10 cm.

(iii) Join EC and draw the perpendicular bisector of CE which intersects BE at A.

(iv) Join AC.

∆ABC is the required triangle.

Question 8.

Construct a triangle whose perimeter is 6.4 cm, and angles at the base are 60° and 45°.

Solution:

Steps of construction :

(i) Draw a line segment PQ = 6.4 cm.

(ii) At P draw a ray PX making an angle of 60° and at Q, a ray QY making an angle of 45°.

(iii) Draw the bisector of ∠P and ∠Q meeting each other at A.

(iv) Draw the perpendicular bisectors of PA and QA intersecting PQ at B and C respectively.

(v) Join AB and AC.

∆ABC is the required triangle.

Question 9.

Using ruler and compasses only, construct a ∆ABC, from the following data:

AB + BC + CA = 12 cm, ∠B = 45° and ∠C = 60°.

Solution:

Steps of construction :

(i) Draw a line segment PQ = 12 cm.

(ii) Draw ray PX at P making are angle of 45° and at Q, QY making an angle of 60°.

(Hi) Draw the angle bisectors of ∠P and ∠Q meeting each other at A.

(v) Draw the perpendicular bisector of AP and AQ intersecting PQ at B and C respectively.

(v) Join AB and AC.

∆ABC is the required triangle.

Question 10.

Construct a triangle XYZ in which ∠Y = 30° , ∠Z = 90° XY +YZ + ZX = 11

Solution:

Steps of construction :

(i) Draw a line segment PQ =11 cm.

(ii) At P, draw a ray PL making an angle of 30° and Q, draw another ray QM making an angle of 90°.

(iii) Draw the angle bisector of ∠P and ∠Q intersecting each other at X.

(iv) Draw the perpendicular bisector of XP and XQ. Which intersect PQ at Y and Z respectively.

(v) Join XY and XZ.

Then ∆XYZ is the required triangle

**All Chapter RD Sharma Solutions For Class 9 Maths**

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**All Subject NCERT Solutions For Class 9**

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