MCQ: Geometry - 1 - SSC CGL MCQ

# MCQ: Geometry - 1 - SSC CGL MCQ

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## 15 Questions MCQ Test Quantitative Aptitude for SSC CGL - MCQ: Geometry - 1

MCQ: Geometry - 1 for SSC CGL 2024 is part of Quantitative Aptitude for SSC CGL preparation. The MCQ: Geometry - 1 questions and answers have been prepared according to the SSC CGL exam syllabus.The MCQ: Geometry - 1 MCQs are made for SSC CGL 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MCQ: Geometry - 1 below.
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MCQ: Geometry - 1 - Question 1

### In a polygon ABCDE, BC is perpendicular to AB, CD to AC and DE  to AD, then AB2 + BC2 + CD2 + DE2 is

Detailed Solution for MCQ: Geometry - 1 - Question 1

From the given question, by using Pythagoras theorem, we get,

We need to find
⇒AB+ BC+ CD+ DE2
By putting the value from the equation (1), we get,
⇒ AC2 + CD+ DE2
⇒ By putting the value from the equation (2), we get,
By putting the value from the equation (3), we get,
⇒ AE2
Hence, the correct answer is AE2.

MCQ: Geometry - 1 - Question 2

### In a triangle ABC, incentre is O and ∠BOC = 110o, then the measure of ∠BAC

Detailed Solution for MCQ: Geometry - 1 - Question 2

As given,
∠BOC = 110o
And, we know incentre of a triangle is a point where angle bisector of triangles meet.
So, which means ∠OCB =∠OBC
∠BOC +∠OCB+∠OBC = 180o
⇒∠OCB + ∠OBC = 180o − 110o
⇒∠OCB +∠OBC = 70o

Since, ∠ABC = 2∠OBC
∠ACB = 2∠OCB
So, ∠ABC+∠ACB = 2 × 70o
⇒∠ABC + ∠ACB = 140o
Now, in triangle ABC.
∠ABC +∠BAC +∠ACB = 180o
⇒∠BAC  +140o = 180o
⇒∠BAC = 180o − 140o
⇒∠BAC = 40o
Hence, the correct answer is 40o.

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MCQ: Geometry - 1 - Question 3

### The three equal circles touches each other externally. If the centres of these circles be A, B, C then ΔABC is ________

Detailed Solution for MCQ: Geometry - 1 - Question 3

The figure is as follows:

Assuming, the radius of circle be r.
Then,
AB = BC = CA = 2r
So, all the three sides are equal of the triangle.
△ABC is an equilateral triangle.
Hence, the required answer is equilateral triangleequilateral triangle.

MCQ: Geometry - 1 - Question 4

Two angles are supplementary and the ratio of the angels is 1:4. what is the value of smaller angle?

Detailed Solution for MCQ: Geometry - 1 - Question 4

As we know that the angles are supplementary so sum of angles will be 180 degree.
Let us assume that the ratio factor is r.
According to question,
Angles are supplementary and have a ratio of 1:4.
r + 4r = 180
⇒ 5r = 180
⇒ r = 180/5
⇒ r = 36

MCQ: Geometry - 1 - Question 5

In the given, AB || CD. Then X is equal to:

Detailed Solution for MCQ: Geometry - 1 - Question 5

Through O, draw a line l parallel to both AB and CD. Then
∠1 = 45° (alt. ∠S)
and ∠2 = 30° (alt. ∠S)
∴ ∠BOC = ∠1 + ∠2 = 45° + 30° = 75°
So, X = 360° – ∠BOC = 360° – 75° = 285°
Hence X = 285°.

MCQ: Geometry - 1 - Question 6

In the adjoining figure, AB || CD, t is the traversal, EG and FG are the bisectors of ∠BEF and ∠DFE respectively, then ∠EGF is equal to:

Detailed Solution for MCQ: Geometry - 1 - Question 6

AB || CD and t transversal intersects them at E and F
∠BEF + ∠EFD = 180° (co-interior angles)

⇒ ∠FEG + ∠EFG = 90°
In Δ GEF
∠EGF + ∠FEG + ∠EFG = 180°
∴ ∠EGF + 90° = 180°
∴ ∠EGF = 90°.
The above result can be restated as :
If two parallel lines are cut by a traversal, then the bisectors of the interior angles on the same side of the traversal intersect each other at right angles.

MCQ: Geometry - 1 - Question 7

The complement of 30°20′ is:

Detailed Solution for MCQ: Geometry - 1 - Question 7

Complement of 30°20′ = 90° – ( 30°20′ ) = 90° – ( 30° + 20′ )
= (89° – 30°) + (1° – 20′)
= 59° + 60′ – 20′ [ ∴ 1° = 60°′]
= 59° + 40′ = 59°40′.

MCQ: Geometry - 1 - Question 8

An angle is equal to one-third of its supplement. Its measure is equal to :

Detailed Solution for MCQ: Geometry - 1 - Question 8

Let the measured of the required angle be P degree.
Then, its supplement = 180 – P
Now use the formula,

3P + P180°
⇒ P = 45°

MCQ: Geometry - 1 - Question 9

Find the measure of an angle, if six times its complement is 12° less than twice its supplement :

Detailed Solution for MCQ: Geometry - 1 - Question 9

Let, the measure of the required angled be A°.
Then, measure of its complement = ( 90 – A )° measure of its supplement = (180 – A)°
According to question,
6(90° – A) = 2(180° – A) –12°
⇒ 540° – 6A = 360° – 2A – 12°
⇒ 4A = 192°
⇒ A = 48°.

MCQ: Geometry - 1 - Question 10

In the following figure, ∠B : ∠C = 2 : 3, find ∠B + ∠C.

Detailed Solution for MCQ: Geometry - 1 - Question 10

∠DAC = ∠B + ∠C
(Exterior angle prop. of a Δ ABC)
According to question,
130° = 2A + 3A
5A = 130°
A = 26°
∴ ∠B = 52°; ∠C = 78°

MCQ: Geometry - 1 - Question 11

A, B, C are the three angles of a Δ. If A − B = 15° and B − C = 30°. Then ∠A is equal to :

Detailed Solution for MCQ: Geometry - 1 - Question 11

Since A, B and C are the angles of a Δ,
∴ A + B + C = 180° .................... (1)
According to question,
A – B = 15° ;
⇒ A = B + 15°...................(2)
B – C = 30°;
⇒ B = C + 30°;....................(3)
Put the value of B from equation (2) in Equation (1), we will get
∴ A = B + 15°
A = C + 30° + 15°
A = C + 45° ......................(4)
From a equation,
∴ A + B + C = 180°
⇒ (C + 45°) + (C + 30°) + C = 180°
⇒ 3C + 45° + 30° = 180°
⇒ 3C = 180° – 75° = 105°
⇒ C = 35° ...........................(5)
From equation (4)
A = C + 45°
Put the value of C from equation (5) , we will get
∴ ∠A = 35° + 45° = 80°.

MCQ: Geometry - 1 - Question 12

In fig., AB || CD, ∠a is equal to:

Detailed Solution for MCQ: Geometry - 1 - Question 12

CD || AB (Given)
Produce RQ to meet AB in S
∠CRS = ∠PSR (at. int. ∠s)
But ∠CRS = 55°
∴ ∠PSR = 55°
Now in QSP
∠QSP + ∠QPS + ∠PQS = 180°
55° + 38° + ∠SQP = 180°
∴ ∠SQP = 180° – 93° = 87°
But angle a and ∠PQS are linear
∠a = 180° – 87°
∠a = 93°

MCQ: Geometry - 1 - Question 13

In a ΔABC, if 2∠A = 3∠B = 6∠C, Then ∠A is equal to:

Detailed Solution for MCQ: Geometry - 1 - Question 13

Let us assume 2∠A = 3∠B = 6∠C = K

MCQ: Geometry - 1 - Question 14

The sides AB and AC of ΔABC have been produced to D and E respectively. The bisectors of ∠CBD and ∠BCE meet at O. If ∠A = 40°, then ∠BOC is equal to:

Detailed Solution for MCQ: Geometry - 1 - Question 14

As we know the formula,

MCQ: Geometry - 1 - Question 15

If the bisector of an angle of Δ bisects the opposite side, then the Δ is :

Detailed Solution for MCQ: Geometry - 1 - Question 15

Since ∠1 = ∠2

But BD = CD (given)

AB = AC
∴ the given ∆ is isosceles

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## Quantitative Aptitude for SSC CGL

314 videos|170 docs|185 tests