Triangles: Important Questions

Very Short Answer Questions [1 Marks]

Ques 1. Find the measure of each exterior angle of an equilateral triangle.

Ans. We know that each interior angle of an equilateral triangle is 60°.

∴ Each exterior angle = 180° – 60° = 120°

Ques 2. If in ∆ABC, ∠A = ∠B + ∠C, then write the shape of the given triangle.
Ans. Here, ∠A = ∠B + ∠C
And in ∆ABC, by angle sum property, we have
∠A + ∠B + C = 180°
⇒ ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
Hence, the given triangle is a right triangle.

Read More: Triangles Important Question

Ques 3. In ∆PQR, PQ = QR and ∠R = 50°, then find the measure of ∠Q.
Ans. Here, in ∆PQR, PQ = QR
⇒ ∠R = ∠P = 50° (given)
Now, ∠P + ∠Q + ∠R = 180°
50° + ∠Q + 50° = 180°
⇒ ∠Q = 180° – 50° – 50°
= 80°

Ques 4. If ∆SKY ≅ ∆MON by SSS congruence rule, then write three equalities of corresponding angles.
Ans. Since ∆SKY ≅ ∆MON by SSS congruence rule, then three equalities of corresponding angles are ∠S = ∠M, ∠K = ∠O and ∠Y = ∠N.

Ques 5. Is ∆ABC possible, if AB = 6 cm, BC = 4 cm and AC = 1.5 cm ?
Ans. Since 4 + 1.5 = 5.5 ≠ 6
Thus, the triangle is not possible.

Ques 6. In ∆MNO, if ∠N = 90°, then write the longest side.
Ans. We know that, side opposite to the largest angle is longest.
∴ Longest side = MO.

Ques 7. In ∆ABC, if AB = AC and ∠B = 70°, find ∠A.
Ans. Here, in ∆ABC AB = AC ∠C = ∠B [∠s opp. to equal sides of a ∆)
Now, ∠A + ∠B + ∠C = 180°
⇒ ∠A + 70° + 70° = 180° [∵ ∠B = 70°]
⇒ ∠A = 180° – 70° – 70° = 40°

Ques 8. In ∆ABC, if AD is a median, then show that AB + AC > 2AD.
Ans. 

Produce AD to E, such that AD = DE.
In ∆ADB and ∆EDC, we have
BD = CD, AD = DE and ∠1 = ∠2
∆ADB ≅ ∆EDC

AB = CE
Now, in ∆AEC, we have
AC + CE > AE
AC + AB > AD + DE
AB + AC > 2AD [∵ AD = DE]

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Short Answer Questions [2 Marks]

Ques 1. In the figure given below, AD = BC and BD = AC, prove that ∠DAB = ∠CBA.

Ans. In ∆DAB and ∆CBA, we have

AD = BC [given]

BD = AC [given]

AB = AB [common]

∴ ∆DAB ≅ ∆CBA [by SSS congruence axiom]

Thus, ∠DAB =∠CBA [c.p.c.t.] 

Ques 2. In the figure given below , ∆ABD and ABCD are isosceles triangles on the same base BD. Prove that ∠ABC = ∠ADC.

Ans. In ∆ABD, we have

AB = AD (given)

∠ABD = ∠ADB [angles opposite to equal sides are equal] …(i)

In ∆BCD, we have

CB = CD

⇒ ∠CBD = ∠CDB [angles opposite to equal sides are equal] … (ii)

Adding (i) and (ii), we have

∠ABD + ∠CBD = ∠ADB + ∠CDB

⇒ ∠ABC = ∠ADC

Also read: Questions on Triangles

Ques 3: In the given figure, if ∠1 = ∠2 and ∠3 = ∠4, then prove that BC = CD.

Ans. 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

Ques 4. In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD < BC

Ans. 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

Ques 5. In the given figure, AC > AB and D is a point on AC such that AB = AD. Show that BC > CD.

Ans. 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]

Ques 6. In a triangle ABC, D is the midpoint of side AC such that BD =½ AC. Show that ∠ABC is a right angle.
Ans.

Here, in ∆ABC, D is the midpoint of AC.
⇒ AD = CD = ½ AC …(i)
Also, BD = ½ AC… (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°

Ques 7. ABC is an isosceles triangle with AB = AC. P and Q are points on AB and AC respectively such that AP = AQ. Prove that CP = BQ.
Ans. 

In ∆ABQ and ∆ACP, we have
AB = AC (given)
∠BAQ = ∠CAP [common]
AQ = AP (given)
∴ By SAS congruence criteria, we have
∆ABQ ≅ ∆ACP
CP = BQ

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Long Answer Questions [3 Marks]

Ques 1. In the given figure, ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC, AD is extended to intersect BC at P. Show that : (i) ∆ABD ≅ ∆ACD (ii) ∆ABP ≅ ∆ACP

Ans.

(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

Read Also: Altitude and Median of Triangle

Ques 2. In the given figure, it is given that AE = AD and BD = CE. Prove that ∆AEB ≅ ∆ADC. 

Ans. 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

Ques 3. In the given figure, in ∆ABC, ∠B = 30°, ∠C = 65° and the bisector of ∠A meets BC in X. Arrange AX, BX and CX in ascending order of magnitude.

Ans. Here, AX bisects ∠BAC.

∴ ∠BAX = ∠CAX = x (say)

Now, ∠A + ∠B + C = 180° [angle sum property of a triangle]

⇒ 2x + 30° + 65° = 180°

⇒ 2x + 95 = 180°

⇒ 2x = 180° – 95°

⇒ 2x = 85°

⇒ x = 85∘/2 = 42.59

In ∆ABX, we have x > 30°

BAX > ∠ABX

⇒ BX > AX (side opp. to larger angle is greater)

⇒ AX < BX

Also, in ∆ACX, we have 65° > x

⇒ ∠ACX > ∠CAX

⇒ AX > CX [side opp. to larger angle is greater]

⇒ CX > AX … (ii)

Hence, from (i) and (ii), we have

CX < AX < BX

Ques 4. In figure, ‘S’ is any point on the side QR of APQR. Prove that PQ + QR + RP > 2PS. 

Ans. In ∆PQS, we have

PQ + QS > PS …(i)

[∵ sum of any two sides of a triangle is greater than the third side]

In ∆PRS, we have

RP + RS > PS …(ii)

Adding (i) and (ii), we have

PQ + (QS + RS) + RP > 2PS

Hence, PQ + QR + RP > 2PS. [∵ QS + RS = QR]

Ques 5. If two isosceles triangles have a common base, prove that the line joining their vertices bisects them at right angles.

Ans. Here, two triangles ABC and BDC having the common

base BC, such that AB = AC and DB = DC.

Now, in ∆ABD and ∆ACD

AB = AC [given]

BD = CD [given]

AD = AD [common]

∴ ΔABD ≅ ΔΑCD [by SSS congruence axiom]

⇒ ∠1 = ∠2 [c.p.c.t.]

Again, in ∆ABE and ∆ACE, we have

AB = AC [given]

∠1 = ∠2 [proved above]

AE = AE [common]

∆ABE = ∆ACE [by SAS congruence axiom]

BE = CE [c.p.c.t.]

and ∠3 = ∠4 [c.p.c.t.]

But ∠3 + ∠4 = 180° [a linear pair]

⇒ ∠3 = ∠4 = 90°

Hence, AD bisects BC at right angles.

Read Also: Angle Formula

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Very Long Answer Questions [5 Marks]

Ques 1. In the given figure, AP and DP are bisectors of two adjacent angles A and D of quadrilateral ABCD. Prove that 2 ∠APD = ∠B + 2C.

Ans. Here, AP and DP are angle bisectors of ∠A and ∠D

∴ ∠DAP = ½ ∠DAB and ∠ADP = ½ ∠ADC ……(i)

In ∆APD, ∠APD + ∠DAP + ∠ADP = 180°

⇒ ∠APD + ½ ∠DAB + ½ ∠ADC = 180°

⇒ ∠APD = 180° – ½ (∠DAB + ∠ADC)

⇒ 2∠APD = 360° – (∠DAB + ∠ADC) ……(ii)

Also, ∠A + ∠B + C + ∠D = 360°

∠B + 2C = 360° – (∠A + ∠D)

∠B + C = 360° – (∠DAB + ∠ADC) ……(iii)

From (ii) and (iii), we obtain

2∠APD = ∠B + ∠C

Ques 2. In figure, ABCD is a square and EF is parallel to diagonal BD and EM = FM. Prove that

(i) DF = BE (i) AM bisects ∠BAD.

Ans. 
(i) EF || BD = ∠1 = ∠2 and ∠3 = ∠4 [corresponding ∠s]

Also, ∠2 = ∠4

⇒ ∠1 = ∠3

⇒ CE = CF (sides opp. to equals ∠s of a ∆]

∴ DF = BE [∵ BC – CE = CD – CF)

(ii) In ∆ADF and ∆ABE, we have

AD = AB [sides of a square]

DF = BE [proved above]

∠D = ∠B = 90°

⇒ ∆ADF ≅ ∆ABE [by SAS congruence axiom]

⇒ AF = AE and ∠5 = ∠6 … (i) [c.p.c.t.]

In ∆AMF and ∆AME

AF = AE [proved above]

AM = AM [common]

FM = EM (given)

∴ ∆AMF ≅ ∆AME [by SSS congruence axiom]

∴ ∠7 = ∠8 …(ii) [c.p.c.t.]

Adding (i) and (ii), we have

∠5 + ∠7 = ∠6 + ∠8

∠DAM = ∠BAM

∴ AM bisects ∠BAD.

Ques 3. In figure, ABC is an isosceles triangle with AB = AC. D is a point in the interior of ∆ABC such that ∠BCD = ∠CBD. Prove that AD bisects ∠BAC of ∆ABC.

Ans. 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.

Ques 4. Show that the difference of any two sides of a triangle is less than the third side.

Ans.

Consider a triangle ABC

To Prove :

(i) AC – AB < BC

(ii) BC – AC < AB

(iii) BC – AB < AC

Construction : Take a point D on AC

such that AD = AB.

Join BD.

Proof : In ∆ABD, we have ∠3 > ∠1 …(i)

[∵ exterior ∠ is greater than each of interior opposite angle in a ∆]

Similarly, in ∆BCD, we have

∠2 > ∠4 …..(ii) [∵ ext. ∠ is greater than interior opp. angle in a ∆]

In ∆ABD, we have

AD = AB [by construction]

∠1 = ∠2 …(iii) [angles opp. to equal sides are equal in a triangle]

From (i), (ii) and (iii), we have

⇒ ∠3 > ∠4 =

⇒ BC > CD

⇒ CD < BC

AC – AD < BC

AC – AB < BC [∵ AD = AB]

Hence, AC – AB < BC

Similarly, we can prove

BC – AC < AB

and BC – AB < AC