Question

solution of NCERT class 9 maths chapter 7 exercise 7.1

Answer 1

Solution:

In △PQR and △PQS , we have

PR = PS

∠RPQ=∠SPQ (PQ bisects ∠P)

PQ = PQ (common)

△PQR=△PQS (By SAS congruence)

Hence Proved.

Therefore, QR = QS (CPCT)

 

Question 2:

ABCD is a quadrilateral in which AD = BC and ∠DAB= ∠CBA (see Fig.). Prove that

(i) △ABD=△BAC

(ii) BD = AC

(iii) ∠ABD=∠BAC



Solution:

In the given figure, ABCD is a quadrilateral in which AD = BC and ∠DAB= ∠CBA.

In △DAB and △BAC, we have

AD = BC [Given]

∠DAB=∠CBA [Given]

AB = AB [Common]

△ABD=△BAC [By SAS congruence]

BD = AC [CPCT]

and ∠ABD=∠BAC [CPCT]

Hence Proved.

 

Question 3:

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



Solution:

In △AOD=△BOC, we have,

∠AOD=∠BOC [Vertically opposite angles)

∠CBO=∠DAO(Each=90o)

AD = BC [Given]

△AOD=△BOC [By AAS congruence]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

 

Question 4:

L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that △ABC+△CDA .



Solution:

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In △ABCand△CDA, we have,

∠BAC=∠DCA (Alternate angles)

∠BCA=∠DAC (Alternate angles)

AC = AC (Common)

△ABC=△CDA (By SAS congruence)

Hence Proved.

 

Question 5:

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of ∠A (See fig.). Show that:

(i)  △APB and △AQB

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



Solution:

In △APB and △AQB , we have

∠PAB=∠AQB (l is the bisector of ∠A)

∠APB=∠AQB (Each=90o)

AB = AB

△APB=△AQB (By AAS congruence)

Also, BP = BQ (By CPCT)

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

Hence proved.

 

Question 6:

In the fig, AC = AE, AB = AD and △BAD=△EAC . Show that BC = DE.



Solution:

∠BAD=△EAC (Given)

∠BAD+∠DAC=∠EAC+∠DAC(Adding∠DAC to both sides)

∠BAC=∠EAC

Now, in △ABCand△ADE , We have

AB = AD

AC = AE

∠BAC=∠DAE

△ABC=△ADE (By SAS congruence)

BC = DE (CPCT)

Hence Proved.

 

Question 7:

AB is a line segment and p is its mid-point. D and E are points on the same side of AB such that ∠BAD=∠ABE and ∠EPA=∠DPB(see fig.). Show that

            (i) △DAP=△EBP

            (ii) AD = BE



Solution:

In △DAP and △EBP, we have

AP = BP (P is the midpoint of the line segment AB)

∠BAD=∠ABE (Given)

∠EPB=∠DPA

(∠EPA=∠DPB⇒∠EPA+∠DPE=∠DPB+∠DPE)

△DPA=△EPB

AD = BE (ASA)

AD = BE

Hence Proved.

 

Question 8:

In right triangle ABC, right angles at C, M is the mid-point 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 fig.). Show that :

            (i) △AMC=△BMD

            (II) ∠DBC is a right angle.

            (iii) △DBC=△ACB

            (iv) CM=12AB



Solution:

In △BMBand△DMC , we have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

∠DMB=∠AMC (Vertically opposite angles)

△AMC=△BMD (By SAS)

Hence Proved.

(ii) AC || BD (∠DBM and ∠CAM are .alternate angles)

⇒∠DBC+∠ACB=180o (Sum of co-interior angles)

⇒∠DBC=90o

Hence proved.

(iii) In △DBC amd △ACB, We have

DB = AC (CPCT)

BC = BC (Common)

∠DBC=∠ACB(Each=90o)

△DBC=△ACB (By SAS)

Hence proved.

(iv) AB = CD

⇒12AB=12CD (CPCT)

Hence, 12AB=CM Proved.

Answer 2

Solution is in the above attachment.

Regrets for handwriting _/\_

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