RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2 is part of RBSE Solutions for Class 9 Maths. Here we have given Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.2.

Board |
RBSE |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Congruence and Inequalities of Triangles |

Exercise |
Ex 7.2 |

Number of Questions Solved |
11 |

Category |
RBSE Solutions |

## Rajasthan Board RBSE Class 9 Maths Solutions Chapter 7 Congruence and Inequalities of Triangles Ex 7.2

Question 1.

In figure, AB = AC and ∠B = 58°, then find (RBSESolutions.com) the value of ∠A.

Solution.

∠A = 180° – (58° + 58°) = 180° – 116° = 64°

Question 2.

In figure, AD = BD and ∠C = ∠E. Prove that BC = AE.

Solution.

In ∆AED and ∆BCD

AD = BD (given)

∠ADE = ∠BDC

(vertically opposite angles)

∠E = ∠C

∴ ∆AED ≅ ∆BCD

(by AAS congruency property)

=> AE = BC (by c.p.c.t)

Question 3.

If AD be a median of (RBSESolutions.com) an isosceles ABC and ∠A = 120° and AB = AC then find ∠ADB.

Solution.

In ∆ABC

AB = AC (given)

∠B = ∠C = x (say)

x + x + 120° = 180°

=> 2x + 120 = 180°

=> 2x = 60° => x = 30°

∠ADB = 180° – (60° + 30°)

=> ∠ADB = 90°

Question 4.

If the bisector of an angle of a triangle (RBSESolutions.com) also bisects the opposite side, show that the triangle is isosceles.

Solution.

Given: In ∆ABC, the bisector of ∠BAC meets BC at D such that AD ⊥ BC

To prove: AB = AC or ∆ABC is an isosceles triangle.

Proof: In ∆’s ABD and ACD

∠BAD = ∠D AC (given)

∠ADB = ∠ADC

(each is a right angle)

and AD = AD (common)

∆ABD ≅ ∆ACD (by AAS congruency property)

=> AB = AC (c.p.c.t)

=> ABC is an isosceles ∆.

Hence proved.

Question 5.

In figure, AB = AC and BE = CD. Prove that AD = AE.

Solution.

In ∆ABD and ∆AEC

BE = DC (given) …(i)

Subtracting DE from (RBSESolutions.com) both side of (i), we get

BE – DE = DC – DE

=> BD = EC

AB = AC (given)

=> ∠B = ∠C

(angle opposite to equal sides are equal)

=> ∆ABD ≅ ∆AEC

=> AD = AE (by c.p.c.t)

Hence proved

Question 6.

Two points E and F of the sides AD and BC respectively of (RBSESolutions.com) a square ABCD such that AF = BE then prove that.

(i) ∠BAF = ∠ABE

(ii) BF = AE

Solution.

∵ ABCD is a square

=> ∠A = ∠B = 90°

Now in ∆AEB and ∆AFB

AF = BE

∠BAE = ∠ABF = 90° (given)

AB = AB (common side)

∴ ∆EAB ≅ ∆FBA

=> ∠FAB = ∠EBA (by c.p.c.t)

and also BF = AE (by c.p.c.t)

Hence proved.

Question 7.

AD and BC are equal perpendiculars to (RBSESolutions.com) a line segment AB (see figure). Show that CD bisects AB.

Solution.

In ∆OBC and ∆OAD

∵ ∠B = ∠A = 90°

(as AD and BC are perpendiculars)

and OB = OA (given)

and ∠BOC = ∠AOD

(vertically opposite angles)

∴ ∆OBC ≅ ∠OAD

(by AAS congruence rule)

=> OA = OB (by c.p.c.t)

=> CD bisects AB

Hence proved.

Question 8.

In an isosceles triangle (RBSESolutions.com) with AB = AC, the bisectors of ∠B and ∠C meet at O. Produce BO upto M then prove that ∠MOC = ∠ABC.

Solution.

In ∆ABC with AB = AC

∠B = ∠C (given)

In ∆BOC

∠MOC = ∠OBC + ∠OCB

Question 9.

Line l is the bisector of an angle ∠A and ∠B is any point on l. BP and BQ are (RBSESolutions.com) perpendicular from B to the arms of ∠A (see figure). Show that

(i) ∆APB = ∆AQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Solution.

(i) In ∆APB and ∆AQB

∠BQA = ∠BPA = 90° (given)

AB = AB (common side)

and ∠QAB = ∠PAB

(∵ l is the bisector of ∠A)

∴ ∆APB ≅ ∆AQB

(by AAS congruence rule)

(ii) ∵ ∆APB ≅ ∆AQB

=> BP = BQ (by c.p.c.t)

i.e. B is equidistant from the arms of ∠A.

Question 10.

In figure, AC = AE, AB = AD and ∠BAD = ∠EAC, show that BC = DE.

Solution.

We are given that ∠BAD = ∠EAC

Adding ∠DAC to both sides, we get

∠BAD + ∠DAC = ∠EAC + ∠DAC

=> ∠BAC = ∠DAE …(i)

Now in ∆BAC and ∆EAD, we have

∠BAC = ∠DAE [using (i)]

AB = AD (given)

AC = AE (given)

∆BAC ≅ ∆EAD

(by SAS congruence rule)

=> BC = DE (by c.p.c.t)

Hence proved.

Question 11.

In right triangle ABC, right angled at C, M is the mid-point (RBSESolutions.com) of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure).

Show that:

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle

(iii) ∆DBC ≅ ∆ACB

(iv) CM = [latex]\frac { 1 }{ 2 }[/latex] AB

Solution.

(i) In ∆AMC and ∆BMD

DM = CM (given)

∠3 = ∠4

(vertically opposite angles)

AM = MB

(as M is the mid-point of AB)

=> ∆AMC ≅ ∆BMD

(by SAS congruency rule)

(ii) ∵ ∆AMC ≅ ∆BMD

=> ∠1 = ∠2 (by c.p.c.t)

But they are alternate angles,

so AC || BD

=> ∠ACB + ∠DBC = 180

(sum of the interior angles on (RBSESolutions.com) the same side of a transversal is equal to 180°)

But ∠ACB = 90° (given)

=> ∠DBC = 180° = 90°

(iii) In ∆DBC and ∆ACB

∵ ∠DBC = ∠ACB = 90°

BC = BC (common side)

BD = CA

(∵ ∆AMC ≅ ∆BMD)

∴ ∆DBC ≅ ∆ACB

(by SAS congruency rule)

(iv) ∵ ∆DBC ≅ ∆ACB

=> DC = AB

But M mid-point of DC

=> 2CM = CD

=> 2CM = AB [∵ DC = AB]

=> CM = [latex]\frac { 1 }{ 2 }[/latex] AB.

Hence proved.

We hope the given RBSE Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2 will help you. If you have any query regarding RBSE Rajasthan Board Solutions for Class 9 Maths Chapter 7 Congruence and Inequalities of Triangles Ex 7.2, drop a comment below and we will get back to you at the earliest.

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