NCERT Class 9 Solutions: Triangles (Chapter 7) Exercise 7.1 – Part 3

Angle Angle Side congruence condition

Give Angle Angle Side congruence condition of ΔENR and ΔRNV

Give Angle Angle Side Congruence Condition

Give Angle Angle Side congruence condition of ΔENR and ΔRNV

Q-5. Line l is the bisector of an angle Equation and B is any point on l. BP and BQ are perpendiculars from B to the arms of Equation (see Fig). Show that:

  1. Equation

  2. Equation or B is equidistant from the arms of Equation

Angle β Angle β: Angle between A, E, A' Angle α Angle α: Angle between F, G, F'_1 Segment f Segment f: Segment [E, F] Segment g Segment g: Segment [F, G] Vector u Vector u: Vector[A, B] Vector u Vector u: Vector[A, B] Vector v Vector v: Vector[A, C] Vector v Vector v: Vector[A, C] Vector w Vector w: Vector[A, D] Vector w Vector w: Vector[A, D] Vector a Vector a: Vector[H, I] Vector a Vector a: Vector[H, I] A text1 = "A" B text2 = "B" P text3 = "P" Q text4 = "Q" l text5 = "l"

Angle a and B Is Any Point on L

Angle a and B is any point on l that line is bisector. BP and BQ are perpendicular from B tohe arms of angle A

Solution:

Give, l is the bisector of an angle Equation

BP and BQ are perpendiculars.

  1. In Equation and Equation

    Equation (Right angles)

    Equation (l is bisector)

    Equation (Common line)

    Therefore, Equation (by Angle-Angle-Side congruence condition).

  2. Equation by Corresponding Part of Congruent Triangles.

    Therefore, B is equidistant from the arms of Equation .

Q-6 In the Figure. , Equation . Show that Equation

Angle α Angle α: Angle between E, A, B Segment f Segment f: Segment [A, B] Segment f Segment f: Segment [A, B] Segment g Segment g: Segment [B, C] Segment h Segment h: Segment [A, D] Segment h Segment h: Segment [A, D] Segment i Segment i: Segment [A, C] Segment i Segment i: Segment [A, C] Segment i Segment i: Segment [A, C] Segment j Segment j: Segment [A, E] Segment j Segment j: Segment [A, E] Segment j Segment j: Segment [A, E] Segment k Segment k: Segment [E, C] Segment l Segment l: Segment [F, G] A text1 = "A" B text2 = "B" C text3 = "C" D text4 = "D" E text5 = "E"

Point of a,B,C,D,E ,∠BAD = ∠EAC

point of A,B,C,D,E ,∠BAD = ∠EAC also AC=AE,AB=AD

Solution:

Given,

Equation

Equation

To show,

Equation

Proof,

Equation (Adding ∠DAC both sides)

Equation

Equation

In Equation and Equation

Equation (Given)

Equation

Equation (Given)

Therefore, Equation (by Side-Angle-Side congruence condition).

Equation (By Corresponding Part of Congruent Triangles)

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