NCERT MATHS SOLUTIONS (9TH STANDARD) by Dr.Prabhat Tiwari

Author:Dr.Prabhat Tiwari [Tiwari, Dr.Prabhat]
Language: eng
Format: epub
Publisher: UNKNOWN
Published: 2020-11-14T00:00:00+00:00

DON

BOM [By RHS congruency]

DN = BM [By CPCT]

Also ar (

DON) = ar (

BOM) ……….(i)

Again, In

DCN and

ABM,

CD = AB [Given]

DNC =

BMA =

[By construction]

DN = BM [Prove above]

DCN

BAM [By RHS congruency]

ar (

DCN) = ar (

BAM) ……….(ii)

Adding eq. (i) and (ii),

ar (

DON) ar (

DCN) = ar (

BOM) ar ( BAM)

ar (

DOC) = ar (

AOB)

(ii) Since ar (

DOC) = ar (

AOB)

Adding ar

BOC both sides,

ar (

DOC) ar

BOC = ar (

AOB) ar BOC

ar (

DCB) = ar (

ACB)

(iii) Since ar (

DCB) = ar (

ACB)

Therefore, these two triangles in addition to be on the same base CB lie between two same parallels CB and DA.

DA

CB

Now AB = CD and DA CB

Therefore, ABCD is a parallelogram.

7. D and E are points on sides AB and AC respectively of

ABC such that ar (DBC) = ar (EBC). Prove that DE

BC.

Ans. Given: ar (DBC) = ar (EBC)

To Prove:- DE II BC.

Proof:- Since two triangles of equal area have common base BC.

Therefore, DE II BC [

Two triangles having same base (or equal bases) and equal areas lie between the same parallel]

8. XY is a line parallel to side BC of triangle ABC. If BE

AC and CF

AB meet XY at E and F respectively, show that ar (ABE) = ar (ACF).

Ans. Given: XY is a line parallel to side BC of a triangle ABC. If BE || AC and CF || AB meet XY at E and E respectively.

To Prove:- ar(ABE) = ar(ACF)

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