Class 9 Maths Chapter 6 Previous Year Questions - Triangles

Sol. In ΔABC, ∠ B = 60° and ∠ C = 63°

⇒ ∠ A = 180º - (60° + 63°) = 57°⇒ The greatest side is opp. to the greatest angle, i.e., 63°∴ Side AB is the greatest.

Sol. BC = AB ⇒ ∠A = ∠C

∵ ∠B = 80°
∴ ∠A + ∠C = 180° - 80° = 100°
⇒ ∠A = ∠C = 50º

Sol. SSA is not the criterion for congruency.

Sol. ∵ (30 - a)º and (125 + 2a)º are supplement to each other.
∴ (30 - a + 125 + 2a)º = 180º
⇒ a = 180º - 125º - 30º = 25º
⇒ Value of a = 25°

Sol. Let the measure of each of the equal acute angle of the Δ be x
∴ We have: x + x + 90° = 180°
⇒ x + x = 180° - 90° = 90°
⇒ x= (90^o/2)= 45°

Sol:

In ∆ABC and ACDA, we have

∠1 = ∠2 (given)

AC = AC [common]

∠3 = ∠4 [given]

So, by using ASA congruence axiom

∆ABC ≅ ∆CDA

Since corresponding parts of congruent triangles are equal

∴ BC = CD

Sol:

Here, in ∆ABD, AB = AD

∠ABD = ∠ADB

[∠s opp. to equal sides of a ∆]

In ∆BAD

ext. ∠BDC = ∠BAD + ∠ABD

⇒ ∠BDC > ∠ABD ….(ii)

Also, in ∆BDC .

ext. ∠ADB > ∠CBD …(iii)

From (ii) and (iii), we have

∠BDC > CD [∵ sides opp. to greater angle is larger]

Sol:
Here, ∠B < ∠A
⇒ AO < BO …..(i)
and ∠C < ∠D
⇒ OD < CO …..(ii)
[∴ side opposite to greater angle is longer]
Adding (i) and (ii), we obtain
AO + OD < BO + CO
AD < BC

Sol:

Here, in ∆ABC, D is the mid-point of AC.
⇒ AD = CD = 1/2AC …(i)
Also, BD = 1/2AC… (ii) [given]
From (i) and (ii), we obtain
AD = BD and CD = BD
⇒ ∠2 = ∠4 and ∠1 = ∠3 …..(iii)
In ∆ABC, we have
∠ABC + ∠ACB + ∠CAB = 180°
⇒ ∠1 + ∠2 + ∠3 + ∠4 = 180°
⇒ ∠1 + ∠2 + ∠1 + ∠2 = 180° [using (iii)]
⇒ 2(∠1 + ∠2) = 180°
⇒ ∠1 + ∠2 = 90°
Hence, ∠ABC = 90° 

Sol:
We have AE = AD … (i)
and CE = BD … (ii)
On adding (i) and (ii),
we have AE + CE = AD + BD
⇒ AC = AB
Now, in ∆AEB and ∆ADC,
we have AE = AD [given]
AB = AC [proved above]
∠A = ∠A [common]
∴ By SAS congruence axiom, we have
∆AEB = ∆ADC

Sol:
(i) In ∆ABD and ∆ACD
AB = AC [given]
BD = CD [given]
AD = AD [common)]
∴ By SSS congruence axiom, we have
∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP
AB = AC [given]
∠BAP = ∠CAP [c.p.cit. as ∆ABD ≅ ∆ACD]
AP = AP [common]
∴ By SAS congruence axiom, we have
∆ABP ≅ ∆ACP 

Sol:
Given : ∆ACB in which 4C = 90° and M is the mid-point of AB.
To Prove :
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC = 90°
(iii) ∆DBC ≅ ∆ACB
(iv) CM = 1/2AB
Proof : Consider ∆AMC and ∆BMD,
we have AM = BM [given]
CM = DM [by construction]
∠AMC = ∠BMD [vertically opposite angles]
∴ ∆AMC ≅ ∆BMD [by SAS congruence axiom]
⇒ AC = DB …(i) [by c.p.c.t.]
and ∠1 = ∠2 [by c.p.c.t.]
But ∠1 and ∠2 are alternate angles.

⇒ BD || CA
Now, BD || CA and BC is transversal.
∴ ∠ACB + ∠CBD = 180°
⇒ 90° + CBD = 180°
⇒ ∠CBD = 90°
In ∆DBC and ∆ACB,
we have CB = BC [common]
DB = AC [using (i)]
∠CBD = ∠BCA
∴ ∆DBC ≅ ∆ACB
⇒ DC = AB
⇒ 1/2AB = 1/2DC
⇒ 1/2AB = CM or CM = 1/2AB (∵ CM = 1/2DC)

Sol: 
In ∆BDC, we have ∠DBC = ∠DCB (given).
⇒ CD = BD (sides opp. to equal ∠s of ∆DBC)
Now, in ∆ABD and ∆ACD,
we have AB = AC [given]
BD = CD [proved above]
AD = AD [common]
∴ By using SSS congruence axiom, we obtain
∆ABD ≅ ∆ACD
⇒ ∠BAD = ∠CAD [c.p.ç.t.]
Hence, AD bisects ∠BAC of ∆ABC.

Sol:

Given : Two As ABC and DEF in which
∠B = ∠E,
∠C = ∠F and BC = EF
To Prove : ∆ABC = ∆DEF
Proof : We have three possibilities
Case I. If AB = DE,
we have AB = DE,
∠B = ∠E and BC = EF.
So, by SAS congruence axiom, we have ∆ABC ≅ ∆DEF

Case II. If AB < ED, then take a point Mon ED
such that EM = AB.
Join MF.
Now, in ∆ABC and ∆MEF,
we have
AB = ME, ∠B = ∠E and BC = EF.
So, by SAS congruence axiom,
we have ΔΑΒC ≅ ΔΜEF
⇒ ∠ACB = ∠MFE
But ∠ACB = ∠DFE
∴ ∠MFE = ∠DFE

Which is possible only when FM coincides with B FD i.e., M coincides with D.
Thus, AB = DE
∴ In ∆ABC and ∆DEF, we have
AB = DE,
∠B = ∠E and BC = EF
So, by SAS congruence axiom, we have
∆ABC ≅ ∆DEF
Case III. When AB > ED
Take a point M on ED produced
such that EM = AB.
Join MF
Proceeding as in Case II, we can prove that
∆ABC = ∆DEF
Hence, in all cases, we have
∆ABC = ∆DEF.