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Chapter 7 Triangles Ex-7.1 Interview Questions Answers

Question 1 :
In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that ∆ABC ≅ ∆ABD. What can you say about BC and BD?

Answer 1 :

In quadrilateral ACBD, we have AC = AD and AB being the bisector of ∠A.
Now, In ∆ABC and ∆ABD,
AC = AD (Given)
∠ CAB = ∠ DAB ( AB bisects ∠ CAB)
and AB = AB (Common)
∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)
∴ BC = BD (By CPCT)

Question 2 :
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that

(i) ∆ABD ∆BAC
(ii) BD = AC
(iii) ABD = BAC

Answer 2 :

In quadrilateral ACBD, we have AD = BC and DAB = CBA

(i) In ∆ ABC and ∆ BAC,
AD = BC (Given)
DAB = CBA (Given)
AB = AB (Common)
∆ ABD ∆BAC (By SAS congruence)

(ii) Since ∆ABD ∆BAC
BD =AC [By C.P.C.T.]

(iii) Since ∆ABD ∆BAC
ABD = BAC [By C.P.C.T.]

Question 3 :
AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Answer 3 :

In ∆BOC and ∆AOD, we have
BOC = AOD
BC = AD [Given]
BOC = AOD [Vertically opposite angles]
∆OBC ∆OAD [By AAS congruency]
OB =OA [By C.P.C.T.]
i.e., O is the mid-point of AB.
Thus, CD bisects AB.

Question 4 :

l and m are two parallellines intersected by another pair of parallel lines p and q (see figure). Showthat ∆ABC = ∆CDA.

Answer 4 :

p || qand AC is a transversal,
BAC = DCA …(1) [Alternate interior angles]
Also l || m and AC is a transversal,
BCA = DAC …(2)
[Alternate interior angles]
Now, in ∆ABC and ∆CDA, we have
BAC = DCA [From (1)]
CA = AC [Common]
BCA = DAC [From (2)]
∆ABC ∆CDA [By ASA congruency]

Question 5 : Line l is the bisector of an A and B is any point on l. BP and BQ are perpendiculars from B to the arms ofLA (see figure). Show that
(i) ∆APB ∆AQB

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

Answer 5 :

We have, l is the bisector of QAP.
QAB = PAB
Q = P [Each 90°]
ABQ = ABP
[By angle sum property of A]
Now, in ∆APB and ∆AQB, we have
ABP = ABQ [Proved above]
AB = BA [Common]
PAB = QAB [Given]
∆APB ∆AQB [By ASA congruency]
Since ∆APB
∆AQB
BP =BQ [By C.P.C.T.]
i. e., [Perpendicular distance of B from AP]
= [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of
A.

Question 6 : In figure, AC = AE, AB = AD and BAD = EAC. Show that BC = DE.

Answer 6 :

We have, BAD = EAC
Adding
DAC onboth sides, we have
BAD + DAC = EAC + DAC
BAC = DAE
Now, in ∆ABC and ∆ADE. we have
BAC = DAE [Proved above]
AB = AD [Given]
AC = AE [Given]
∆ABC ∆ADE [By SAS congruency]
BC =DE [By C.P.C.T.]

Question 7 : AS is a line segment and P is its mid-point. D and Eare points on the same side of AB such that BAD = ABE and EPA = DPB. (see figure). Showthat
(i) ∆DAP
∆EBP
(ii) AD = BE

Answer 7 :

We have, P is the mid-point of AB.
AP = BP
EPA = DPB [Given]
Adding
EPD onboth sides, we get
EPA + EPD = DPB + EPD
APD = BPE

(i) Now, in ∆DAP and ∆EBP, we have
PAD = PBE [ ∵∠BAD = ABE]
AP = BP [Proved above]
DPA = EPB [Proved above]
∆DAP ∆EBP [By ASA congruency]

(ii) Since, ∆ DAP ∆ EBP
AD =BE [By C.P.C.T.]

Question 8 : In right triangle ABC, right angled at C, M isthe mid-point of hypotenuse AB. C is joined to M and produced to a point D suchthat 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

Answer 8 : Since M is the mid – point of AB.
BM = AM


(i) In ∆AMC and ∆BMD, we have
CM = DM [Given]

(ii) Since ∆AMC ∆BMD
MAC = MBD [By C.P.C.T.]
But they form a pair of alternate interior angles.
AC ||DB
Now, BC is a transversal which intersects parallel lines AC and DB,
BCA + DBC = 180° [Co-interior angles]
But
BCA =90° [∆ABC is right angled at C]
90° + DBC = 180°
DBC = 90°

(iii) Again, ∆AMC ∆BMD [Proved above]
AC =BD [By C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
DBC = ACB [Each 90°]
BC = CB [Common]
∆DBC ∆ACB [By SAS congruency]

(iv) As∆DBC ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
CM = DC= AB

CM = AB


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Chapter 7 Triangles Ex-7.1 Contributors

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