• Skip to main content
  • Skip to secondary menu
  • Skip to primary sidebar
  • Skip to footer
  • RBSE Model Papers
    • RBSE Class 12th Board Model Papers 2022
    • RBSE Class 10th Board Model Papers 2022
    • RBSE Class 8th Board Model Papers 2022
    • RBSE Class 5th Board Model Papers 2022
  • RBSE Books
  • RBSE Solutions for Class 10
    • RBSE Solutions for Class 10 Maths
    • RBSE Solutions for Class 10 Science
    • RBSE Solutions for Class 10 Social Science
    • RBSE Solutions for Class 10 English First Flight & Footprints without Feet
    • RBSE Solutions for Class 10 Hindi
    • RBSE Solutions for Class 10 Sanskrit
    • RBSE Solutions for Class 10 Rajasthan Adhyayan
    • RBSE Solutions for Class 10 Physical Education
  • RBSE Solutions for Class 9
    • RBSE Solutions for Class 9 Maths
    • RBSE Solutions for Class 9 Science
    • RBSE Solutions for Class 9 Social Science
    • RBSE Solutions for Class 9 English
    • RBSE Solutions for Class 9 Hindi
    • RBSE Solutions for Class 9 Sanskrit
    • RBSE Solutions for Class 9 Rajasthan Adhyayan
    • RBSE Solutions for Class 9 Physical Education
    • RBSE Solutions for Class 9 Information Technology
  • RBSE Solutions for Class 8
    • RBSE Solutions for Class 8 Maths
    • RBSE Solutions for Class 8 Science
    • RBSE Solutions for Class 8 Social Science
    • RBSE Solutions for Class 8 English
    • RBSE Solutions for Class 8 Hindi
    • RBSE Solutions for Class 8 Sanskrit
    • RBSE Solutions

RBSE Solutions

Rajasthan Board Textbook Solutions for Class 5, 6, 7, 8, 9, 10, 11 and 12

  • RBSE Solutions for Class 7
    • RBSE Solutions for Class 7 Maths
    • RBSE Solutions for Class 7 Science
    • RBSE Solutions for Class 7 Social Science
    • RBSE Solutions for Class 7 English
    • RBSE Solutions for Class 7 Hindi
    • RBSE Solutions for Class 7 Sanskrit
  • RBSE Solutions for Class 6
    • RBSE Solutions for Class 6 Maths
    • RBSE Solutions for Class 6 Science
    • RBSE Solutions for Class 6 Social Science
    • RBSE Solutions for Class 6 English
    • RBSE Solutions for Class 6 Hindi
    • RBSE Solutions for Class 6 Sanskrit
  • RBSE Solutions for Class 5
    • RBSE Solutions for Class 5 Maths
    • RBSE Solutions for Class 5 Environmental Studies
    • RBSE Solutions for Class 5 English
    • RBSE Solutions for Class 5 Hindi
  • RBSE Solutions Class 12
    • RBSE Solutions for Class 12 Maths
    • RBSE Solutions for Class 12 Physics
    • RBSE Solutions for Class 12 Chemistry
    • RBSE Solutions for Class 12 Biology
    • RBSE Solutions for Class 12 English
    • RBSE Solutions for Class 12 Hindi
    • RBSE Solutions for Class 12 Sanskrit
  • RBSE Class 11

RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2

April 15, 2019 by Fazal Leave a Comment

RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Exercise 11.2.

Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Ex 11.2

Question 1.
D and E are to points on the (RBSESolutions.com) sides AB and AC of ΔABC, respectively. Such that DE || BC, then
(i) In AD = 6 cm, DB = 9 cm and AE = 8 cm then find AC.
(ii) If \(\frac { AD }{ DB }\) = \(\frac { 4 }{ 13 }\) and AC = 20.4 cm. Then find EC.
(iii) \(\frac { AD }{ DB }\) = \(\frac { 7 }{ 4 }\) and AE = 6.3 cm then find AC.
(iv) In AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3 then find x.
Solution :
(i) In ∆ABC, DE || BC
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 1
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 2
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 3

RBSE Solutions

Question 2.
Points D and E lie on sides AB and AC of ΔABC, respectively. In (RBSESolutions.com) the following questions check whether DE || BC or not.
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 9.0 cm and AE = 1.8 cm
(iii) AD = 10.5 cm, BD = 4.5 cm, AC = 4.8 cm and AE = 2.8 cm
(iv) AD = 5.7 cm, BD = 9.5 cm, AE = 3.3 cm and EC = 5.5 cm
Solution :
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 4
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 5

Question 3.
In given figure, L, M and N are (RBSESolutions.com) points on OA, OB and OC such that LM || AM and MN || BC then show that LN || AC.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 6
Solution :
Given : In ∆ABC, point L,M, N respectively lie OA, OB and OC such that LM || AB and MN || BC.
To prove : LN || AC
Proof : In ∆ABO
LM || AB (given)
\(\frac { OL }{ LA }\) = \(\frac { OM }{ MB }\) …..(ii) (By Basic prop. theorem)
Again, In ∆BCO
MN || BC
\(\frac { OM }{ MB }\) = \(\frac { ON }{ NC }\) ….(ii) (By Basc prop. theorem)
From equation (i) and (ii),
⇒ \(\frac { OL }{ LA }\) = \(\frac { OM }{ MB }\) = \(\frac { ON }{ NC }\)
\(\frac { OL }{ LA }\) = \(\frac { ON }{ NC }\) (By Converse of B. P. theorem)
⇒ LN || AC

Question 4.
In ∆ABC, D and E points lie on sides AB and AC such (RBSESolutions.com) that BD = CE. If ∠B = ∠C then show that DE || BC.
Solution :
In ∆ABC, D and E lie on sides BD = CE and ∠B = ∠C
To prove : DE || BC
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 7
Proof : In ∆ABC,
∠B = ∠C
∴ AB = AC (opposite sides of equals angles are same)
⇒ AD + DB = AE + CE [∵ DB = CE given]
⇒ AD = AE
∵ ∠D = ∠E [Opposite angles of equal sides are same]
⇒ ∆ADE is an isosceles triangle
So, ∆ABC and ∆ADE are isosceles triangle in (RBSESolutions.com) which vertex angle ∠A is common in both triangles.
∴ ∠B = ∠C = ∠D = ∠E
∠B = ∠D [corresponding angle]
⇒ DE || BC

RBSE Solutions

Question 5.
In fig. DE || BC and CD || EF then prove that : AD2 = AB × AF
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 8
Solution :
Given : ∆ABC in (RBSESolutions.com) which DE || BC and CD || EF
To prove : AD2 = AB × AF
Proof : In ∆ABC
DE || BC
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 9
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 10

Question 6.
In fig., EF || DC || AB Then (RBSESolutions.com) prove that : \(\frac { AE }{ ED }\) = \(\frac { BF }{ FC }\)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 11
Solution :
Given : EF || DC || AB
To prove : \(\frac { AE }{ ED }\) = \(\frac { BF }{ FC }\)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 12
Construction : Join AC
Proof : In ∆ADC
GE || DC
∴ \(\frac { AE }{ ED }\) = \(\frac { AG }{ GC }\) ….(i) (By Basic proportionality theorem)
Now In ∆ABC
AB || GF
∴ \(\frac { CF }{ BF }\) = \(\frac { CG }{ AG }\) (By Basic Prop. Theorem)
or \(\frac { BF }{ CF }\) = \(\frac { AG }{ GC }\) ….(ii)
From equation (i) and (ii)
\(\frac { AE }{ ED }\) = \(\frac { BF }{ FC }\)

Question 7.
ABCD is a parallelogram. P is any (RBSESolutions.com) point on side BC of ||gm. If DP and AB produced, meet at L then prove that:
(i) \(\frac { DP }{ PL }\) = \(\frac { DC }{ BL }\)
(ii) \(\frac { DL }{ DP }\) = \(\frac { AL }{ DC }\)
Solution :
Given : ABCD ¡s a parallelogram. Point P lies on side BC. Produced DP and AB meet at point L.
(i) To prove : \(\frac { DP }{ PL }\) = \(\frac { DC }{ BL }\)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 13
Proof : In ∆LDA
BP || AD
\(\frac { LP }{ DP }\) = \(\frac { LB }{ AB }\) (By B.P. Theorem)
or \(\frac { DP }{ PL }\) = \(\frac { DC }{ BL }\) [∵ AB = CD sides of ||gm]
⇒ \(\frac { DP }{ PL }\) = \(\frac { DC }{ BL }\)
(ii) To prove
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 14

RBSE Solutions

Question 8.
Point D and E lie on side AB of ∆ABC such (RBSESolutions.com) that AD = BE. If DP || BC and EQ || AC then prove that PQ || AB.
Solution :
Given : ∆ABC in which
AD = BE,
DP || BC and EQ || AC
To prove PQ || AB
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 15
Proof : In ∆ABC DP || BC
\(\frac { AD }{ BD }\) = \(\frac { AP }{ CP }\) …..(i) (By B.P. Theorem)
and QE || AC
\(\frac { BE }{ AE }\) = \(\frac { BQ }{ QC }\) …..(ii)
∵ AD = BE (given) ……(iii)
AD + DE = BE + DE
AE = BD …..(iv)
putting values from (RBSESolutions.com) equation (iii) and (iv) in (ii)
\(\frac { BE }{ AE }\) = \(\frac { AD }{ BD }\) = \(\frac { BQ }{ QC }\)
From equation (i) and (v)
\(\frac { AP }{ DB }\) = \(\frac { AP }{ CP }\) = \(\frac { BQ }{ QC }\)
⇒ \(\frac { AP }{ CP }\) = \(\frac { BQ }{ QC }\)
⇒ \(\frac { CP }{ AP }\) = \(\frac { QC }{ BQ }\)
⇒ PQ || AB

Question 9.
ABCD is a Trapezium in which AB || DC and its diagonals intersects each other at point O. Show that : \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\) (Higher Secondary Board Raj 2012)
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 16
Solution :
Given : Trapezium ABCD in which AC and BC are (RBSESolutions.com) two diagonals which intersects each other at point O.
To prove : \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\)
Construction : From O, Draw OE || CD.
Proof : In ∆ADC
OE || DC
\(\frac { AE }{ ED }\) = \(\frac { AO }{ CO }\) ……(i) (By Basic prop. Theorem)
In Trapezium ABCD
AB || CD
OE || CD
OE || AB
Now, In ∆ABD
OE || AB
\(\frac { ED }{ AE }\) = \(\frac { DO }{ BO }\)
⇒ \(\frac { AE }{ ED }\) = \(\frac { BO }{ DO }\)
From equation (i) and (ii)
\(\frac { AO }{ CO }\) = \(\frac { BO }{ DO }\)
⇒ AO × DO = BO × CO
⇒ \(\frac { AO }{ BO }\) = \(\frac { CO }{ DO }\)

Question 10.
Points D and E lie on sides AB and AC of ∆ABC such (RBSESolutions.com) that BD = CE then show that ∆ABC Is isosceles triangles.
Solution :
Given : In ∆ABC, D and E lie on sides AB and AC such that BD = CE.
To prove : ∆ABC is an isosceles triangle.
RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 17
Construction : Join DC and BE.
Proof : In ∆BCD and ∆BEC
BD = CE (given)
∠BDC = ∠BEC [∵ angles of same base an some height]
BC = BC (common)
By SAS rule ∆BCD = ∆BEC
∴ ∠ABC = ∠ACE (by CPCT)
⇒ AB = AC (Sides opposite to (RBSESolutions.com) equal angles are same)
Hence ∆ABC is an isosceles triangle.

RBSE Solutions

We hope the given RBSE Solutions for Class 10 Maths Chapter 11 Similarity Ex 11.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 11 Similarity Exercise 11.2, drop a comment below and we will get back to you at the earliest.

Share this:

  • Click to share on WhatsApp (Opens in new window)
  • Click to share on Twitter (Opens in new window)
  • Click to share on Facebook (Opens in new window)

Related

Filed Under: Class 10

Reader Interactions

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Primary Sidebar

Rajasthan Board Questions and Answers

Recent Posts

  • RBSE Class 10 Science Notes Chapter 11 Work, Energy and Power
  • RBSE Solutions for Class 11 Physics Chapter 4 Laws of Motion
  • RBSE Solutions for Class 11 Physics Chapter 13 Thermodynamics
  • RBSE Class 10 Maths Important Questions Chapter 3 Pair of Linear Equations in Two Variables
  • RBSE Solutions for Class 10 Maths Chapter 2 Real Numbers Ex 2.2
  • RBSE Solutions for Class 7 Maths Chapter 15 Comparison of Quantities Additional Questions
  • RBSE Solutions for Class 11 Chemistry Chapter 7 Equilibrium
  • RBSE Solutions for Class 11 Chemistry Chapter 1 Basic Concepts of Chemistry
  • RBSE Solutions for Class 11 Chemistry Chapter 5 States of Matter: Gas and Liquid
  • RBSE Solutions for Class 12 Physics Chapter 15 Nuclear Physics
  • RBSE Class 9 Science Important Questions in Hindi & English Medium

Footer

RBSE Solutions for Class 12
RBSE Solutions for Class 11
RBSE Solutions for Class 10
RBSE Solutions for Class 9
RBSE Solutions for Class 8
RBSE Solutions for Class 7
RBSE Solutions for Class 6
RBSE Solutions for Class 5
RBSE Solutions for Class 12 Maths
RBSE Solutions for Class 11 Maths
RBSE Solutions for Class 10 Maths
RBSE Solutions for Class 9 Maths
RBSE Solutions for Class 8 Maths
RBSE Solutions for Class 7 Maths
RBSE Solutions for Class 6 Maths
RBSE Solutions for Class 5 Maths
Target Batch
RBSE Class 11 Political Science Notes
RBSE Class 11 Geography Notes
RBSE Class 11 History Notes

Copyright © 2022 RBSE Solutions

 

Loading Comments...