Explore topic-wise InterviewSolutions in .

We have: 

∠AFD = ∠EFB (vertically opposite angles) 

∵ DA || BC ∴ ∠DAF = ∠BEF (alternative angles) 

∆ DAF ~ ∆ BEF (AA similarity theorem) 

⇒ AF/EF = FD/FB

Or, AF × FB = FD × EF 

This completes the proof.

Correct option is (C) CE/ED

In \(\triangle BCD,\) EF || DB

\(\therefore\) \(\triangle CEF\sim\triangle CDB\)    (Corresponding angles are equal as EF ||DB)

\(\therefore\) \(\frac{CE}{CD}=\frac{CF}{CB}\)                 (By properties of similar triangles)

\(\Rightarrow\) \(\frac{CE}{CD-CE}=\frac{CF}{CB-CF}\)    \((If\,\frac ab=\frac cd\,then\,\frac a{b-a}=\frac c{d-c})\)

\(\Rightarrow\) \(\frac{CE}{ED}=\frac{CF}{FB}\)

Hence, \(\frac{CF}{FB}\) = \(\frac{CE}{ED}\)

Correct option is: (C) \(\frac{CE}{ED}\)

Given,

AB ‖ CD

And, EF cuts them

So,

30° + 65° + ∠PQR = 180°

95° + ∠PQR = 180°

∠PQR = 85°

∠APQ + ∠PQC = 180°(Co. interior angle)

25° + y + 85° + 30° = 180°

y = 40°

In ΔPQR

∠PQR + ∠PRQ + ∠QPR = 180°

85° + x + y = 180°

x = 55°

Thus,

x = 55° and y = 40°

B) (1, 2, 3)

The correct answer is B.

We have to check all the options by using the Pythagoras theorem.

ex: in the first option 

3²+4²=5²

this condition satisfies all the options except 2 and option.

(d) ∠ 1 + ∠ 2 = ∠ 3 + ∠ 4

As we know that, exterior angle is equal to the sum of opposite interior angles

Consider, Δ ABC

As BC is extended

∠A + ∠B = ∠3 + ∠4

Therefore, ∠1 + ∠2 = ∠3 + ∠4

Correct answer is (c) 40°

Given,

AD perpendicular to BC

∠A = 100°

In ΔADB

∠ADB + ∠B + ∠DAC = 180°

90° + ∠B + \(\frac{1}{2}\)∠A = 180°

∠B + \(\frac{1}{2}\) x 100° = 180° – 90°

∠B + 50° = 90°

∠B = 40°

We know that AB || CD and AE is a transversal

From the figure we know that ∠BAE and ∠DOE are corresponding angles

So we get

∠BAE = ∠DOE = 65^o

From the figure we know that ∠DOE and ∠COE form a linear pair of angles

So we get

∠DOE + ∠COE = 180^o

By substituting the values

65^o + ∠COE = 180^o

On further calculation

∠COE = 180^o – 65^o

By subtraction

∠COE = 115^o

We know that the sum of all the angles in triangle OCE is 180^o.

∠OEC + ∠ECO + ∠COE = 180^o

By substituting the values we get

20^o + ∠ECO + 115^o = 180^o

On further calculation

∠ECO = 180^o – 20^o – 115^o

By subtraction

∠ECO = 45^o

Therefore, ∠ECO = 45^o

∠A = 70° , ∠B = 40° [Given] 

∠ACD is an exterior angle of ∆ABC. [Given] 

∴ ∠ACD = ∠A + ∠B 

= 70° + 40° 

∴ ∠ACD = 110°

Given,

∠ACD = 105°

∠EAF = 45°

∠EAF = ∠BAC (Vertically opposite angle)

∠BAC = 45°

∠ACD + ∠ACB = 180°(Linear pair)

105° + ∠ACB = 180°

∠ACB = 180° – 105°

= 75°

In ΔABC

∠BAC + ∠ABC + ∠ACB = 180°

45° + ∠ABC + 75° = 180°

∠ABC = 180° – 120°

= 60°

Thus, all three angles of a triangle are 45°, 60° and 75°

Since,

AB ‖ DE

∠ABC = ∠CDE (Alternate angles)

∠ABC = 40°

In ΔABC

∠A + ∠B + ∠ACB = 180°

30° + 40° + ∠ACB = 180°

∠ACB = 180° – 70°

= 110° (i)

Now,

∠ACD + ∠ACB = 180°(Linear pair)

∠ACD + 110° = 180°[From (i)]

∠ACD = 180° – 110°

= 70°

Hence, ∠ACD = 70°.

Correct option is  D) 8 cm

i. ∠ACB = 50° [Given]

 In ∆ABC, seg AC ≅ seg AB [Given] 

∴ ∠ABC ≅ ∠ACB [Isosceles triangle theorem] 

∴ x = 50°

ii. ∠DBC = 60° [Given] 

In ABDC, seg BD ≅ seg DC [Given] 

∴ ∠DCB ≅ ∠DBC [Isosceles triangle theorem] 

∴ y = 60°

iii. ∠ABD = ∠ABC + ∠DBC [Angle addition property] 

= 50° + 60° 

∴ ∠ABD = 110° 

iv. ∠ACD = ∠ACB + ∠DCB [Angle addition property] 

= 50° + 60° 

∴ ∠ACD = 110° 

∴ x = 50°, y = 60°,

 ∠ABD = 110°, ∠ACD = 110°

Data: ABC and DBC are two isosceles triangles on the same base BC. 

To Prove: 

∠ABD = ∠ACD 

Proof: 

In ∆ABC, AB = AC, 

∴ Opposite angle 

∠ABC = ∠ACB …………. (i) 

Similarly in ∆BDC, BD = DC. 

Opposite angle 

∠DBC = ∠DCB ………….. (ii) 

From (i) and (ii), 

∠ABC = ∠ACB 

Adding ∠DBC and ∠DCB on both sides, 

∠ABC + ∠DBC = ∠ABD 

∠ACB + ∠DCB = ∠ACD 

Equals are added to equal angles. 

∴ ∠ABD = ∠ACD.

(a) In ∆NOM and ∆MLN

LM = ON    [given]

MN = MN   [common side]

LN = OM    [given]

(b) Yes, by SSS congruence criterion,

∆NOM ≅ ∆MLN

Correct answer is 

(i) Yes, (ASA)

(ii) Yes, CPCT

(iii) Yes, CPCT

Correct answer is 

(a) LM = ON, LN = OM, MN = NM

(b) Yes, (SSS)

(b) Isosceles 

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

(b) Isosceles 

In an isosceles triangle, the perpendicular from the vertex to the base bisects the base.

CORRECT OPTION :-

(b) Isosceles 

By using angle bisector in ∆ABC, we have 

AB/AC = BD/DC 

⇒ 10/14 = 6 − x/x

⇒ 10x = 84 – 14x 

⇒ 24x = 84 

⇒ x = 3.5 

Hence, the correct answer is option (b).

(d) 7.5 cm 

It is given that AD bisects angle A 

Therefore, applying angle bisector theorem, we get: 

BD/DC = AB/AC 

⇒ 4/5 = 6/x 

⇒ x = 5 × 6/4 = 7.5 

Hence, AC = 7.5 cm

AX bisects ∠DAC

∠CAD = 2 x ∠DAC

∠CAD = 2 x 70°

= 140°

By exterior angle theorem,

∠CAD = ∠B + ∠C

140° = ∠C + ∠C (Therefore, ∠B = ∠C)

140° = 2∠C

∠C = 70°

Therefore,

∠C = ∠ACB = 70°

Correct answer is  (b) 45°

(b) 45^o

From the given figure,

In triangle PQS, ∠PSQ + ∠QPS = 110^o … [from exterior angle property of a triangle]

We know that, sum of all angle of the triangle is equal to 180^o.

So, ∠PSQ + ∠QPS + ∠PQS = 180^o

110^o + ∠PQS = 180^o – 110^o

∠PQS = 70^o

Now, consider the triangle PRS,

∠PSQ = x + 25^o … [from the exterior angle property of a triangle]

x = 70^o – 25^o

x = 45^o

AB = 6√3 cm

⇒ AB² = 108 cm² 

AC = 12 cm 

⇒ AC² = 144 cm²

BC = 6 cm 

⇒ BC² = 36 cm 

∴ AC² = AB² + BC² 

Since, the square of the longest side is equal to the sum of two sides, so ∆ABC is a right angled triangle. 

∴ The angle opposite to ∠90° 

Hence, the correct answer is option (c)

a)  From the figure, 

∠A = 35°, ∠B = x° and exterior angle at C = 70° 

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

Exterior angle at C = ∠A + ∠B = 70° 

⇒ 35° + x= 70° 

⇒ 35° + x – 35° 

= 70° – 35° 

∴ x = 35°

b) From the figure, ∠P = 4x, ∠Q = 3x and exterior angle at R = 119° 

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

Exterior angle at R = ∠P + ∠Q = 119° 

⇒ 4x + 3x = 119° 

⇒ 7x = 119°

⇒ 7x/7 = 119°/7

∴ x = 17°

a) From the figure 

∠A = 60°, ∠B = 73° 

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

Exterior angle at C = ∠A + ∠B 

= 600 + 730

= 133°

b) From the figure, 

∠D = 90°, ∠E = 30° 

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

Exterior angle at 

F = ∠D + ∠E 

= 90° +.30° 

= 120°

Exterior angles of ∆XYZ are ∠PXY, ∠ZYQ and ∠XZR. 

Exterior angle of X is ∠PXY. 

Exterior angle of Y is ∠QYZ. 

Exterior angle of Z is ∠XZR.

C) ∠A = ∠B + ∠C

Given the ratio of the angles of a ’ triangle are 2 : 4 : 3. 

2x : 4x : 3x 

Sum of the angles of a triangle is 180°. 

⇒ 2x + 4x + 3x = 180° 

⇒ 9x = 180°

⇒ 9x/9 = 180°/9

∴ x = 20° 

Angles are 

⇒ 2x : 4x :,3x 

2(20°) : 4(20°): 3(20°) 

40°: 80° : 60°

∴ Angles of a triangle are 40°, 80°, 60°.

(i) FALSE. 

In triangle sum of three angles is 180°. In triangle, if two angles are two right angles (90° + 90° = 180°). Then, sum of three angles is greater than 180°.

(ii) TRUE.

(iii) FALSE. 

In triangle sum of three angles is 180°. In triangle, if two angles are two obtuse angles, then sum of three angles is greater than 180°.

Since the perimeters and two sides are proportional
Therefore, the third side is proportional to the corresponding third side.
i.e., The two triangles will be similar by SSS criterion.

 We observe that,

1. ∆ABC is an equilateral triangle, 

2. ∆PQR is an isosceles triangle, and 

3. ∆XYZ is a scalene triangle.

The ratio of their areas will be 1 : 4.

Here, two diagrams PQRS and ABCD are rhombus and square whose sides are in proportional. But the corresponding angle are not equal. Hence these are not similar.

The names of other two sides are: seg BC and seg AC 

The names of other angles are: ∠BCA and ∠CAB

Given the three angles of a triangular signboard are 2x°, (x – 10)° and (x +30)°. 

We know that in a triangle, 

2x + (x – 10) + (x + 30) = 180°

⇒ 2x + x- 10° + x + 30 = 180° 

⇒ 4x + 20 = 180° 

⇒ 4x + 20 – 20 = 180° – 20° 

⇒ 4x = 160°

⇒ 4x/4 = 160°/4

∴ x = 40° 

Angles are 2x°, (x- 10)°, (x + 30)° 2(40°), 

40° – 10, 40 + 30 

Angles of signboard are : 80°, 30°, 70°.

Given in ∆SKV, ∠K = 60°, ∠V = 70° 

We know that in ∆SKV, 

∠S + ∠K + ∠V = 180° 

⇒ ∠S + 60° + 70° = 180° 

⇒ ∠S + 130° = 180° 

⇒∠S + 130°- 130° = 180°- 130° 

⇒ ∠S = 50° 

∴ ∠S = 50°

False

If all angles of a triangle are less than 60°, then their sum will be less than 180°. But in a triangle sum of all angles is 180°.

∴ Triangle is not possible.

In ∆CUT, 

∠C = 64°, ∠U = 46° and exterior angle 

∠CTE = x =? ∠C + ∠U + ∠T= 180° 

⇒ 64° + 46° + ∠T = 180° 

⇒ 110° + ∠T= 180° 

⇒ 110° + ∠T – 110° = 180°- 110° 

∠UTC = ∠T = 70° 

∠UTC + ∠CTE = 180° (linear pair of angles) 

⇒ 70° + x° = 180° 

⇒ 70° + x° – 70° = 180°-70° 

∴ x = 110°

The three sides \(\overline{AB}, \overline{BC}, \overline{CA}\) and the three angles ∠A, ∠B and ∠C are the six elements of ∆ABC.