MCQ: Geometry - 1 - SSC CGL MCQ

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

We need to find 
⇒AB² + BC² + CD² + DE²  
By putting the value from the equation (1), we get,
⇒ AC² + CD² + DE²
⇒ By putting the value from the equation (2), we get,
⇒ AD2 + DE²
By putting the value from the equation (3), we get,
⇒ AE²
Hence, the correct answer is AE².

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

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

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.

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

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°.

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.

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

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

3P + P180°
⇒ P = 45°

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°.

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

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°.

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°

Let us assume 2∠A = 3∠B = 6∠C = K

As we know the formula,

Since ∠1 = ∠2

But BD = CD (given)

AB = AC
∴ the given ∆ is isosceles