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Given ∠DCB = 1000 and ∠DBA = 128° 

In the given figure ∠CBD + ∠DBA = 180° 

∠CBD + 128° = 180° 

∠CBD = 52° 

Now exterior angle x = Sum of interior opposite angles. 

x = ∠DCB + ∠CBD 

= 100° + 52° 

= 152° 

x = 152°

∠BAD = ∠BAE + ∠EAD 

= 40°+ 30° = 70°.

And ∠CDA = 70° 

∠BAD = ∠CDA 

But they form a pair of alternate angles 

⇒ AB || CD 

Also ∠BAE + ∠AEF = 40° + 140° = 180°

But they form a pair of interior opposite angles. 

⇒ AB || EF 

From (1) and (2), we get 

AB || CD || EF 

⇒ CD || EF

(i) ∠ABG and ∠GBC are adjacent angles. 

(ii) ∠BCF and ∠FCD are adjacent angles. 

(iii) ∠BCF and ∠FCE are adjacent angles. 

(iv) ∠FCE and ∠ECD are adjacent angles.

∠LIK and ∠KIJ are adjacent angles. 

∴ ∠JIL = ∠LIK + ∠KIJ

= 38° + 27° 

= 65° 

∴ ∠JIL = 65°

∠HEF = ∠HEG + ∠GEF 

120° = ∠HEG + 34° 

120° - 34° = ∠GEH + 34° – 34° 

∠GEH = 86°

If the angle are linear pair, then their sum is 180°. 

Given one angle is right angle ie 90°. 

∴ The other angle = 180° - 90° = 90° 

∴ The other angle also a right angle

The correct answer is (C) N

Lines l and m intersect at a point and making a pair of vertically opposite angles x° and 150°. 

We know that vertically opposite angles are equal. 

x = 150°

The correct answer is (D) 85°

In ΔABC, ∠C = ∠B then AB = AC.

In ΔABC, ∠C = ∠B then AB = AC.

Given polynomials are f(x) = x² + x – 6 

and g(x) = x² + 3x – 18 

f(x) and g(x) having common factors 

(x + a), So 

x + a = 0 ⇒ x = -a 

f(-a) = (-a)² + (-a) – 6 

= a² – a – 6 

g(-a) = (-a)² + 3(-a) – 18 

= a² – 3a – 18 

f(-a) = g(-a) 

a² – a – 6 = a² – 3a – 18 

a² – a² – a + 3a = -18 + 6 

2a = -12 

a = -6.

i) There is a unique line that passes through the given two distinct points. 

ii) We can draw a circle with any centre and radius.

Given a + b = 5 and 

a² + b² = 11

 (a + b)² = (5)² 

a² + b² + 2ab = 25 

11 + 2ab = 25 

2ab = 14 

ab = 7 

a³ + b³ = (a + b) (a² + b² – ab) 

= 5(11 – 7) 

= 5 × 4 = 20 

∴ a³ + b³ = 20.

If a > b so, 

Let a = 3 and b = 2 

a² = 3² and b² = 2² 

9 > 4 so, 

∴ If a > b, then a² > b².

i) Things which are equal to the same things are equal to one another. 

ii) Things which coincide with one another are equal to one another. 

iii) The whole is greater than the part.

False.

The angle is not affected, if the arms of an angle on the paper are increased or decreased.

The correct option is (A) 0.

The number of diagonals of a triangle is 0.

True.

From the question, line PQ || line m.

Then, parts of those lines are also parallel.

Therefore, line segment PQ || m.

A polygon of six sides is called a hexagon.

False.

Infinite number of line can pass through a given point.

True

All equilateral triangles are isosceles also.

A triangle with all its sides of unequal lengths is called a Scalene triangle.

False 

They will intersect, when they are produced.

False.

only one line can pass through two given points.

False.

Because, two line segments are intersecting at only one point.

False.

Parallel lines never meet each other.

True.

From the figure, both ∠ABC and ∠CBA contains common angle B.

(A) If B is the image of A in line l and D is the image of C in line l, then AC = BD.

(B) The number of scales in a protractor for measuring the angles is Two

The correct answer is True.

The correct answer is False.

The number of triangles in Fig. is 5. Their names are ΔAOC, ΔCOD, ΔAOB, ΔACB, ΔACD.

From the figure,

(a) ∠AOD is a/an right angle.

Is anyone angle in the triangle is equal to 90^o then the triangle is called right triangle.

∠AOD = ∠AOB + ∠BOC + ∠COD

= 30^o + 20^o + 40^o

= 90^o

(b) ∠COA is a/an acute angle

An acute angle is an angle formed between 0^o to 90^o.

∠COA = ∠COB + ∠BOA

= 20° + 30°

= 50^o

(c) ∠AOE is a/an obtuse angle

An obtuse angle is an angle formed between 90^o to 180^o.

∠AOE = ∠EOD + ∠DOC + ∠COB + ∠BOA

= 40° + 40° + 20° + 30°

= 130°

In Fig., points lying in the interior of the triangle PQR are right, that in the exterior are acute and that on the triangle itself are obtuse.