All Exams > SSC CGL > SSC CGL Mathematics Previous Year Paper (Topic-wise) > MCQ Questions
Geometry MCQs for SSC CGL Exam
It covers all Important Questions with answers on Geometry for the SSC CGL exam. The questions are based on important topics. Details about the questions:- Topic: Geometry
- Type of Questions: MCQs with solutions
- Number of Questions: 50
- You can attempt them on EduRev to score high in SSC CGL exam.
Two parallel chords of a circle of diameter 20 cm are 12 cm and 16 cm long. If the chords are in the same side of the centre, then the distance between them is (SSC CGL 1st Sit. 2014)- a)28 cm
- b)2 cm
- c)4 cm
- d)8 cm
Correct answer is option 'B'. Can you explain this answer?
Two parallel chords of a circle of diameter 20 cm are 12 cm and 16 cm long. If the chords are in the same side of the centre, then the distance between them is (SSC CGL 1st Sit. 2014)
a)
28 cm
b)
2 cm
c)
4 cm
d)
8 cm
 | Ssc Cgl answered |
In ΔADO,
distance between chords = OC – OD
= 2CM
In a parallelogram PQRS, angle P is four times of angle Q, then the measure of ∠R is (SSC CGL 1st Sit. 2015)- a)36°
- b)72°
- c)130°
- d)144°
Correct answer is option 'D'. Can you explain this answer?
In a parallelogram PQRS, angle P is four times of angle Q, then the measure of ∠R is (SSC CGL 1st Sit. 2015)
a)
36°
b)
72°
c)
130°
d)
144°
 | Mira Sharma answered |
P = 4Q
P + Q = 180°
4Q + Q = 180°
∵ 180/5 = 36°
So, R = 180° – 36° = 144°
P + Q = 180°
4Q + Q = 180°
∵ 180/5 = 36°
So, R = 180° – 36° = 144°
The sides of a triangle having area 7776 sq. cm are in the ratio 3 : 4 : 5. The perimeter of the triangle is (SSC CGL 1st Sit. 2015)- a)400 cm
- b)412 cm
- c)424 cm
- d)432 cm
Correct answer is option 'D'. Can you explain this answer?
The sides of a triangle having area 7776 sq. cm are in the ratio 3 : 4 : 5. The perimeter of the triangle is (SSC CGL 1st Sit. 2015)
a)
400 cm
b)
412 cm
c)
424 cm
d)
432 cm
| | Mira Sharma answered |
Let sides of Δ be 3x, 4x, 5x
7776 = 6x2
∴ x = 36
Sides of Δ will be 108, 144 and 180
Perimeter of Δ is 108 +144 + 180 = 432 cm
7776 = 6x2
∴ x = 36
Sides of Δ will be 108, 144 and 180
Perimeter of Δ is 108 +144 + 180 = 432 cm
In ΔABC, AB = BC = K, AC = √2 K, then ΔABC is a: (SSC CHSL 2015)- a)Isosceles triangle
- b)Right angled triangle
- c)Equilateral triangle
- d)Right isosceles triangle
Correct answer is option 'D'. Can you explain this answer?
In ΔABC, AB = BC = K, AC = √2 K, then ΔABC is a: (SSC CHSL 2015)
a)
Isosceles triangle
b)
Right angled triangle
c)
Equilateral triangle
d)
Right isosceles triangle
 | Vikram Mehta answered |
In ΔABC
AC = 2K
AC2 = 2K2
AC2 = AB2 + BC2
So ΔABC is right angled triangle
So, in ΔABC
So, In triangle ABC, ∠B = 90°; ∠C =45°; ∠A = 45°
Hence, triangle ABC is right isoscles triangle.
AC = 2K
AC2 = 2K2
AC2 = AB2 + BC2
So ΔABC is right angled triangle
So, in ΔABC
So, In triangle ABC, ∠B = 90°; ∠C =45°; ∠A = 45°
Hence, triangle ABC is right isoscles triangle.
The measure of an angle whose supplement is three times as large as its complement, is (SSC CGL 1st Sit. 2015)- a)30°
- b)45°
- c)60°
- d)75°
Correct answer is option 'B'. Can you explain this answer?
The measure of an angle whose supplement is three times as large as its complement, is (SSC CGL 1st Sit. 2015)
a)
30°
b)
45°
c)
60°
d)
75°
| | Mira Sharma answered |
Let ‘x’ be the measure of an angle.
Then its complement angle = 90° – x
and its supplement angle = 180° – x
According to question
(180° – x) = 3(90° – x)
180° – x = 270° – 3x
2x = 90°
x = 45°
Then its complement angle = 90° – x
and its supplement angle = 180° – x
According to question
(180° – x) = 3(90° – x)
180° – x = 270° – 3x
2x = 90°
x = 45°
G is the centroid of ΔABC. The medians AD and BE intersect at right angles. If the lengths of AD and BE are 9 cm and 12 cm respectively; then the length of AB (in cm) is? (SSC CGL 1st Sit. 2015)- a)10
- b)10.5
- c)9.5
- d)11
Correct answer is option 'A'. Can you explain this answer?
G is the centroid of ΔABC. The medians AD and BE intersect at right angles. If the lengths of AD and BE are 9 cm and 12 cm respectively; then the length of AB (in cm) is? (SSC CGL 1st Sit. 2015)
a)
10
b)
10.5
c)
9.5
d)
11
 | Malavika Rane answered |
A triangle ABC if and only if AG = BG = CG, where G is the point of intersection of the medians of triangle ABC.
AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is: (SSC CGL 1st Sit. 2013)- a)10 cm
- b)11 cm
- c)12 cm
- d)13 cm
Correct answer is option 'D'. Can you explain this answer?
AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and distance between them is 17 cm, then the radius of the circle is: (SSC CGL 1st Sit. 2013)
a)
10 cm
b)
11 cm
c)
12 cm
d)
13 cm
 | Dhrutindra Pradhan answered |
D.13
Two sides of a triangle are of length 4 cm and 10 cm. If the length of the third side is 'a' cm. then (SSC CGL 1st Sit. 2012)- a)a > 5
- b)6 ≤ a ≤ 12
- c)a < 5
- d)6 < a < 14
Correct answer is option 'D'. Can you explain this answer?
Two sides of a triangle are of length 4 cm and 10 cm. If the length of the third side is 'a' cm. then (SSC CGL 1st Sit. 2012)
a)
a > 5
b)
6 ≤ a ≤ 12
c)
a < 5
d)
6 < a < 14
| | Vikram Mehta answered |
The sum of any two sides of a triangle is greater than third side and their difference is less than third side.
10 – 4 < a < 10 + 4
6 < a < 14
10 – 4 < a < 10 + 4
6 < a < 14
O is the orthocentre of ΔABC , and if ∠BOC = 110° then ∠BAC will be (SSC CGL 1st Sit. 2016)- a)110°
- b)70°
- c)100°
- d)90°
Correct answer is option 'B'. Can you explain this answer?
O is the orthocentre of ΔABC , and if ∠BOC = 110° then ∠BAC will be (SSC CGL 1st Sit. 2016)
a)
110°
b)
70°
c)
100°
d)
90°
 | Bank Exams India answered |
For Orthocentre ∠BAC = 180 – ∠BOC
= 180 – 110 = 70°
= 180 – 110 = 70°
If D, E and F are the mid points of BC, CA and AB respectively of the ΔABC then the ratio of area of the parallelogram DEFB and area of the trapezium CAFD is: (SSC CGL 1st Sit. 2015)- a)1 : 3
- b)1 : 2
- c)3 : 4
- d)2 : 3
Correct answer is option 'D'. Can you explain this answer?
If D, E and F are the mid points of BC, CA and AB respectively of the ΔABC then the ratio of area of the parallelogram DEFB and area of the trapezium CAFD is: (SSC CGL 1st Sit. 2015)
a)
1 : 3
b)
1 : 2
c)
3 : 4
d)
2 : 3
| | Vikram Mehta answered |
Here AC = 2 AE = 2FD
If the altitude of an equilateral triangle is 12√3 cm, then its area would be: (SSC CGL 1st Sit. 2015)- a)12 cm2
- b)72 cm2
- c)36√3 cm2
- d)144√3 cm2
Correct answer is option 'D'. Can you explain this answer?
If the altitude of an equilateral triangle is 12√3 cm, then its area would be: (SSC CGL 1st Sit. 2015)
a)
12 cm2
b)
72 cm2
c)
36√3 cm2
d)
144√3 cm2
| | Mira Sharma answered |
Let ΔABC is a equilateral triangle with AD as an altitude from A on side BC.
Let AB = BC = AC = x
From question AD = 12√3 cm.
then from ΔABD,
(AD)2 + (BD)2 = (AB)2
If the measure of three angles of a triangle are in the ratio 2 : 3 : 5, then the triangle is: (SSC CGL 1st Sit. 2015)- a)equilateral
- b)isocsceles
- c)Obtuse angled
- d)right angled
Correct answer is option 'D'. Can you explain this answer?
If the measure of three angles of a triangle are in the ratio 2 : 3 : 5, then the triangle is: (SSC CGL 1st Sit. 2015)
a)
equilateral
b)
isocsceles
c)
Obtuse angled
d)
right angled
 | Pranab Goyal answered |
Explanation:
Given:
- The measure of three angles of a triangle are in the ratio 2 : 3 : 5.
Solution:
Step 1: Find the total sum of angles in a triangle
- Let the three angles be 2x, 3x, and 5x (since they are in the ratio 2 : 3 : 5).
- According to the property of triangles, the sum of all angles in a triangle is 180 degrees.
- Therefore, 2x + 3x + 5x = 180.
- Solving the equation, we get x = 12.
Step 3: Determine the type of triangle based on the angles
- Now we substitute the value of x back into the angles: 2x = 24, 3x = 36, and 5x = 60.
- The angles of the triangle are 24°, 36°, and 60°.
- Since the angles are 24°, 36°, and 60°, they form a right-angled triangle.
- Therefore, the triangle is a right-angled triangle.
Therefore, the correct answer is option 'D' - right-angled triangle.
Given:
- The measure of three angles of a triangle are in the ratio 2 : 3 : 5.
Solution:
Step 1: Find the total sum of angles in a triangle
- Let the three angles be 2x, 3x, and 5x (since they are in the ratio 2 : 3 : 5).
- According to the property of triangles, the sum of all angles in a triangle is 180 degrees.
- Therefore, 2x + 3x + 5x = 180.
- Solving the equation, we get x = 12.
Step 3: Determine the type of triangle based on the angles
- Now we substitute the value of x back into the angles: 2x = 24, 3x = 36, and 5x = 60.
- The angles of the triangle are 24°, 36°, and 60°.
- Since the angles are 24°, 36°, and 60°, they form a right-angled triangle.
- Therefore, the triangle is a right-angled triangle.
Therefore, the correct answer is option 'D' - right-angled triangle.
ABCD is a cyclic quadrilateral AB and DC are produced to meet at P. If ∠ADC = 70° and ∠DAB = 60°, then the ∠PBC + ∠PCB is (SSC CGL 2nd Sit. 2013)- a)130°
- b)150°
- c)155°
- d)180°
Correct answer is option 'A'. Can you explain this answer?
ABCD is a cyclic quadrilateral AB and DC are produced to meet at P. If ∠ADC = 70° and ∠DAB = 60°, then the ∠PBC + ∠PCB is (SSC CGL 2nd Sit. 2013)
a)
130°
b)
150°
c)
155°
d)
180°
| | Mira Sharma answered |
As ABCD is a cyclic quadrilateral.
In which
∠ADC = 70°
∠ABC = 180° – 70° = 110°
⇒ ∠PBC = 180° – 110° = 70°
And ∠DAB = 60°
∠BCD = 180° – 60° = 120°
⇒ ∠PCB = 180° – 120° = 60°
∴ ∠PBC + ∠PCB = 70° + 60° = 130°
A tangent is drawn to a circle of radius 6cm from a point situated at a distance of 10 cm from the centre of the circle.
The length of the tangent will be (SSC CGL 1st Sit. 2015)- a)4 cm
- b)5 cm
- c)8 cm
- d)7 cm
Correct answer is option 'C'. Can you explain this answer?
A tangent is drawn to a circle of radius 6cm from a point situated at a distance of 10 cm from the centre of the circle.
The length of the tangent will be (SSC CGL 1st Sit. 2015)
The length of the tangent will be (SSC CGL 1st Sit. 2015)
a)
4 cm
b)
5 cm
c)
8 cm
d)
7 cm
| | Mira Sharma answered |
AB2 + OA2 = OB2 ⇒ AB2 = OB2 = OA2
AB2 = (10)2 – (6)2 = 100 – 36 = 64
AB = 8cm
Let C1 and C2 be the inscribed and circumscribed circles of a triangle with sides 3 cm, 4 cm and 5 cm then area of C1 to area of C2 is (SSC CGL 1st Sit. 2015)- a)9/16
- b)9/25
- c)4/25
- d)16/25
Correct answer is option 'C'. Can you explain this answer?
Let C1 and C2 be the inscribed and circumscribed circles of a triangle with sides 3 cm, 4 cm and 5 cm then area of C1 to area of C2 is (SSC CGL 1st Sit. 2015)
a)
9/16
b)
9/25
c)
4/25
d)
16/25
| | Mira Sharma answered |
Let ΔABC has three sides BC, AB and AC equal to 3 cm, 4 cm and 5 cm respectively.
Now, as, (5)2 = (3)2 + (4)2 i.e. (AC)2 = (AB)2 + (BC)2
∴ ΔABC is a right angle triangle
Then, for circumcircle C2, radius = AC/2 = 5/2 = 2.5
AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intersecting, AB produced at C such that BC = OD. If ∠BCD = 20°, then ∠AOD =? (SSC CGL 2nd Sit. 2013)- a)20°
- b)30°
- c)40°
- d)60°
Correct answer is option 'D'. Can you explain this answer?
AB is the chord of a circle with centre O and DOC is a line segment originating from a point D on the circle and intersecting, AB produced at C such that BC = OD. If ∠BCD = 20°, then ∠AOD =? (SSC CGL 2nd Sit. 2013)
a)
20°
b)
30°
c)
40°
d)
60°
| | Ssc Cgl answered |
Here BC = OD = radius {given}
In ΔBOC,
BC = OB (radius).
∴ ∠BOC = ∠BCO = 20°
∠OBA = ∠BOC + ∠BCO = 20° + 20° = 40°
Again, In DAOB,
AO = BO = radius
∠OAB = ∠OBA = 40°
∴ ∠AOB = 180° – 40° – 40° = 100°
Again,
∠AOD + ∠AOB + ∠BOC = 180°
∠AOD + 100° + 20° = 180°
∠AOD = 180° – 120° = 60°
Two circles of radii 4 cm and 9 cm respectively touch each other externally at a point and a common tangent touches them at the points P and Q respectively. They the area of a square with one side PQ, is (SSC CGL 1st Sit. 2012)- a)97 sq. cm
- b)194 sq. cm
- c)72 sq. cm
- d)144 sq. cm
Correct answer is option 'D'. Can you explain this answer?
Two circles of radii 4 cm and 9 cm respectively touch each other externally at a point and a common tangent touches them at the points P and Q respectively. They the area of a square with one side PQ, is (SSC CGL 1st Sit. 2012)
a)
97 sq. cm
b)
194 sq. cm
c)
72 sq. cm
d)
144 sq. cm
 | Notes Wala answered |
r1 + r2 = 13 cm
r2 – r1 = 9 – 4 = 5 cm
∴ Area of square of side PQ = 12 × 12 = 144 sq. cm.
In a triangle ABC, ∠A = 90°, ∠C = 55°, AD ⊥ BC . What is the value of ∠BAD? (SSC CGL 1st Sit. 2013)- a)45°
- b)55°
- c)35°
- d)60°
Correct answer is option 'B'. Can you explain this answer?
In a triangle ABC, ∠A = 90°, ∠C = 55°, AD ⊥ BC . What is the value of ∠BAD? (SSC CGL 1st Sit. 2013)
a)
45°
b)
55°
c)
35°
d)
60°
 | Ishaan Roy answered |
Understanding the Triangle
In triangle ABC, we know the following:
- ∠A = 90° (Right angle)
- ∠C = 55°
Using the triangle angle sum property, we can find ∠B:
Calculating ∠B
- Sum of angles in a triangle = 180°
- ∠B = 180° - ∠A - ∠C
- ∠B = 180° - 90° - 55°
- ∠B = 35°
Now, we have the measures of all angles in triangle ABC:
- ∠A = 90°
- ∠B = 35°
- ∠C = 55°
Finding ∠BAD
Point D is where AD is perpendicular to BC, creating two right triangles: ABD and ACD.
- In triangle ABD:
- ∠BAD + ∠ABD = 90° (since AD is perpendicular to BC)
- Let's denote ∠BAD as x. Hence, ∠ABD = 90° - x.
Using the angle sum in triangle ABD:
- ∠BAD + ∠ABD + ∠B = 180°
- x + (90° - x) + 35° = 180°
- 90° + 35° = 180°
- This confirms our angles are consistent.
Now, since ∠C = 55° in triangle ACD:
- ∠CAD = ∠C = 55°.
Since the angles around point A must sum to 90°:
- ∠BAD + ∠CAD = 90°
- x + 55° = 90°
- x = 90° - 55°
- x = 35°
Therefore, the value of ∠BAD is:
Final Answer
- ∠BAD = 35°
- The correct answer is option B: 55°.
In triangle ABC, we know the following:
- ∠A = 90° (Right angle)
- ∠C = 55°
Using the triangle angle sum property, we can find ∠B:
Calculating ∠B
- Sum of angles in a triangle = 180°
- ∠B = 180° - ∠A - ∠C
- ∠B = 180° - 90° - 55°
- ∠B = 35°
Now, we have the measures of all angles in triangle ABC:
- ∠A = 90°
- ∠B = 35°
- ∠C = 55°
Finding ∠BAD
Point D is where AD is perpendicular to BC, creating two right triangles: ABD and ACD.
- In triangle ABD:
- ∠BAD + ∠ABD = 90° (since AD is perpendicular to BC)
- Let's denote ∠BAD as x. Hence, ∠ABD = 90° - x.
Using the angle sum in triangle ABD:
- ∠BAD + ∠ABD + ∠B = 180°
- x + (90° - x) + 35° = 180°
- 90° + 35° = 180°
- This confirms our angles are consistent.
Now, since ∠C = 55° in triangle ACD:
- ∠CAD = ∠C = 55°.
Since the angles around point A must sum to 90°:
- ∠BAD + ∠CAD = 90°
- x + 55° = 90°
- x = 90° - 55°
- x = 35°
Therefore, the value of ∠BAD is:
Final Answer
- ∠BAD = 35°
- The correct answer is option B: 55°.
If D is the mid-point of the side BC of ΔABC and the area of ΔABD is 16 cm2, then the area of ΔABC is (SSC CGL 2nd Sit. 2012)- a)16 cm2
- b)24 cm2
- c)32 cm2
- d)48 cm2
Correct answer is option 'C'. Can you explain this answer?
If D is the mid-point of the side BC of ΔABC and the area of ΔABD is 16 cm2, then the area of ΔABC is (SSC CGL 2nd Sit. 2012)
a)
16 cm2
b)
24 cm2
c)
32 cm2
d)
48 cm2
 | Knowledge Hub answered |
Area of ΔABD = 16 cm2
Area of ΔABC = 2 × Area of ΔABD [∵ In triangle, the midpoint of the opposite side, divides it into two congruent triangles. So their areas are equal and each is half the area of the original triangle]
⇒ 32 cm2
Area of ΔABC = 2 × Area of ΔABD [∵ In triangle, the midpoint of the opposite side, divides it into two congruent triangles. So their areas are equal and each is half the area of the original triangle]
⇒ 32 cm2
In a triangle, if three altitudes are equal, then the triangle is (SSC CGL 1st Sit. 2013)- a)uilateral
- b)Isoceles
- c)Obtuse
- d)EqRight
Correct answer is option 'A'. Can you explain this answer?
In a triangle, if three altitudes are equal, then the triangle is (SSC CGL 1st Sit. 2013)
a)
uilateral
b)
Isoceles
c)
Obtuse
d)
EqRight
| | Bank Exams India answered |
Let ΔABC is a equilateral triangle of side AB = BC = AC = 2a unit.
AD, BE and CF are three altitudes in the DABC,
In equilateral triangle ∠A = ∠B = ∠C = 60°
Similarly, In ΔACF,
Here, we get AD = BE = CF
Hence, three altitude are equal
Thus, triangle must be a equilateral triangle.
Chapter doubts & questions for Geometry - SSC CGL Mathematics Previous Year Paper (Topic-wise) 2026 is part of SSC CGL exam preparation. The chapters have been prepared according to the SSC CGL exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for SSC CGL 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.
Chapter doubts & questions of Geometry - SSC CGL Mathematics Previous Year Paper (Topic-wise) in English & Hindi are available as part of SSC CGL exam. Download more important topics, notes, lectures and mock test series for SSC CGL Exam by signing up for free.
