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Sum of angles in a triangle is 180°. If each angle is less than 60°, then the sum of angles of a triangle is < 180°. So, triangle cannot form. 

So, we cannot find a triangle in which each angle is less than 60°.

a) In ∆IJK, IJ ≠ KJ ≠ IK 

So, ∆IJK is a scalene triangle.

b) In ∆LMN, ∠M = 90 

So, ∆LMN is a right triangle.

c) In ∆TUV, TU = UV = TV = 4.5 cm 

So, ∆TUV is an equilateral triangle.

In ∆TOP given ∠T = 6O, ∠O = y° 

∠OPT = x° and ∠RPQ = 68° 

∠OFT = ∠RPQ (Vertically, opposite angles are equal) 

x = 68° We know that in ∆TOP

∠T + ∠O + ∠P= 1800 

⇒ 60° + y° + x° = 180° 

⇒ 60° + y° + 68° = 180° 

⇒ 128° + y – 128° = 180°- 128° 

∴ y = 52° ‘ 

∴ x = 68°and y = 52°

Equilateral triangle (viii) 

Isosceles triangle (iv), (ix), (x) 

Scalene triangle (i), (ii), (iii), (v), (vi), (vii).

i) In ∆ABC, 

AB ≠ BC ≠ AC 

So, ∆ABC is scalene triangle.

ii) In ∆DEF, 

DE = EF = FD = 2 cm 

So, ∆DEF is an equilateral triangle.

iii) In ∆XYZ, 

XZ = YZ = 3.5 cm 

So, ∆XYZ is an isosceles triangle.

iv) Isosceles triangle (∵ Two sides are equal)

v)  Equilateral triangle (∵ Three sides are equal)

vi)  Scalene triangle (∵ NO two sides are equal).

a) In ∆TUV, ∠T = 90° 

So, ∆TUV is Right-angled triangle.

b) In ∆QRS, ∠Q = 108°

So, ∆QRS is an Obtuse angled triangle.

c) In ∆IJK, all are acute angles. 

So, ∆IJK is an acute angled triangle.

a) From the figure, ∠A = 40°, ∠B = 60°, ∠D = 45° 

Exterior angle at C = x° 

Exterior angle at E = y° 

Exterior angle of a triangle is equal to the sum of its opposite interior angles. 

Exterior angle at C = ∠A + ∠B 

x = 40° + 60° 

∴ x = 100°

Exterior angle at E = ∠C + ∠D 

y = x + 45° 

y = 100° + 45° 

∴ y = 145°

b) From the figure, 

∠M = y, ∠N = 70° 

Exterior angle at M = x° 

Exterior angle at L = 120° 

Exterior angle of a triangle is equal to the sum of its opposite interior angles. 

Exterior angle at L = ∠M + ∠N = 120° 

⇒ y + 70° = 120° 

⇒ y + 70 – 70 = 120 – 70

∴ y = 50° 

∠QLN + ∠MLN = 180° (linear pair of angles) 

120° + ∠MLN = 180° 

120° + ∠MLN – 120° = 180°- 120° 

∠MLN = 60° 

∴ ∠L = 60° 

Exterior angle at M = ∠L + ∠N 

x = 60° + 70° = 130° 

∴ x = 130° 

∴ x = 130° and y = 50°

In the figure, 

x° + 65° = 180° (linear pair of angles) 

∴ x°= 180°- 65°= 115° 

Also y° + 30° = 65° (exterior angle property) 

y° = 65° – 30° = 35°

In ΔABC, 

107° = x° + 57° (exterior angle property) 

∴ x° = 107° – 57° =50° In ΔADC,

∠CAD + ∠ADC + ∠ACD = 180° (angle – sum property) 

40° + 107°+ x°= 180° 

∴ x° = 180° – 147° = 33° 

Also in ΔABC, 

∠A +∠B +∠C = 180° 

40° + ∠B + (33° + 65°) = 180° 

∠B + 138° = 180° 

B = 180° – 138° = 42° 

Now y° = ∠A + ∠B for ΔABC (exterior angle property) 

=40° + 42° = 82°

In ΔBCD, ∠DBC + ∠BCD = ∠BDA (exterior angle property)

 ∠DBC + 65° -104°

 ∴ ∠DBC = 104° – 65° = 39°

 Also ∠BDA + ∠BDC = 180° (linear pair of angles) 

104° + ∠BDC = 180° 

∠CDB or ∠BDC = 180° – 104° = 76° 

Now in ΔABD, 

∠BAD + ∠ADB + ∠DHA = 180°

3∠DBA + ∠DBA + 104° = 180° (given) 

4∠DBA + 104° – 180° 

∴ 4∠DBA = 180° – 104°= 76° 

∴∠DBA =  76°/4 = 19° 

Now ∠ABC = ∠DBA + ∠DBC = 19° + 39° = 58°

Given : AB and CD are two chords 

To Prove: 

(a) ∆ PAC - ∆ PDB 

(b) PA. PB = PC.PD 

Proof: ∠ABD + ∠ACD = 180° …(1) (Opposite angles of a cyclic quadrilateral are supplementary) 

∠PCA + ∠ACD = 180° …(2) (Linear Pair Angles ) 

Using (1) and (2), we get 

∠ABD = ∠PCA 

∠A = ∠A (Common)

By AA similarity-criterion ∆ PAC - ∆ PDB 

When two triangles are similar, then the rations of the lengths of their corresponding sides are proportional. 

∴ PA/PD = PC/PB 

⇒ PA.PB = PC.PD

Data : 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. 

To Prove : 

(i) ∆DAP ≅ ∆EBP 

(ii) AD = BE 

Proof : (i) In ∆DAP and ∆EBP, 

AP = BP (∵ P is the mid-point of AB) 

∠BAD = ∠ABE (Data) 

∠APD = ∠BPE 

∵ (∠EPA = ∠DPB 

Adding ∠EPD to both sides) 

∠EPA + ∠EPD 

= ∠DPB + ∠EPD 

∴ ∠APD = -BPE. 

Now, Angle, Side, Angle postulate. 

∴ ∆DAP ≅ ∆EBP . 

(ii) As it is ∆DAP ≅ ∆EBP , 

Three sides and three angles are equal to each other. 

∴ AD = BE.

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on the both sides of the perpendicular are similar to the whole triangle and also to each other. 

(a) Now using the same property in In ∆BDC, we get 

∆CQD ~ ∆DQB 

CQ/DQ = DQ/QB 

⇒ DQ² = QB. CQ 

Now. Since all the angles in quadrilateral BQDP are right angles. 

Hence, BQDP is a rectangle. 

So, QB = DP and DQ = PB 

∴ DQ² = DP.CQ 

(b) Similarly, ∆APD ~ ∆DPB 

AP/DP = PD/PB 

⇒ DP² = AP.PB

⇒ DP² = AP.DQ [∵ DQ = PB]

(i) In ΔAMC and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMC = ∠BMD (Vertically opposite angles)

CM = DM (Given)

∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)

∴ AC = BD (By CPCT)

And, ∠ACM = ∠BDM (By CPCT)

(ii) ∠ACM = ∠BDM

However, ∠ACM and ∠BDM are alternate interior angles. Since

alternate angles are equal, It can be said that DB || AC

∠DBC + ∠ACB = 180º (Co-interior angles)

∠DBC + 90º = 180º

∠DBC = 90º

(iii) In ΔDBC and ΔACB, 

DB = AC (Already proved)

∠DBC = ∠ACB (Each 90 )

BC = CB (Common)

∴ ΔDBC ≅ ΔACB (SAS congruence rule)

(iv) ΔDBC ≅ ΔACB 

AB = DC (By CPCT) 

AB = 2 CM

∴ CM =1/2 AB