Geometry MCQ Quiz - Objective Question with Answer for Geometry - Download Free PDF

Last updated on May 13, 2025

Geometry MCQs are one of the most common questions featured in competitive exams such as SSC CGL, Bank PO, MTS etc. Testbook is known for the quality of resources we provide and candidates must practice the set of Geometry Objective Questions and enhance their speed and accuracy. Geometry is the branch of mathematics concerned with the properties and relations of points, lines, surfaces, solids etc. With the Geometry Quizzes at Testbook, you will be able to understand and apply Geometry concepts. Get the option to ‘save’ Geometry MCQs to access them later with ease and convenience. For students to be truly prepared for the competitive exams that have a Geometry section in their syllabus, they must practice Geometry Questions Answers. We have also given some tips, tricks and shortcuts to solve Geometry easily and quickly. So solve Geometry with Testbook and ace the competitive examinations you’ve applied for!

Latest Geometry MCQ Objective Questions

Geometry Question 1:

In ΔABC, ∠C = 90° and CD ⊥ AB, also ∠A = 65°, then ∠CBA is equal to

22000

  1. 25°
  2. 35°
  3. 65°
  4. 40°
  5. None of the above

Answer (Detailed Solution Below)

Option 1 : 25°

Geometry Question 1 Detailed Solution

Given:

∠C = 90° 

CD ⊥ AB

∠A = 65°

Concept used:

The sum of the angles of a triangle is 180°.

Calculation:

In ΔABC,

∠BAC + ∠CBA  + ∠ACB  = 180°

⇒ 65° + 90° + ∠CBA  = 180°

⇒ ∠CBA  = 25°

Important Points

We can arrive at the solution based on the perpendicular as well. But it is more time-consuming. Since ∠A and ∠C are given, it is wise to directly apply them in the concept and get the solution.

Geometry Question 2:

The area of the figure ABCD formed by joining A(-1, 1) B, (5, 1), C(5, 6) and D(-1, 6) in order is a

  1. 40 sq. units
  2. 30 sq. units 
  3. 20 sq. units 
  4. 16 sq. units 

Answer (Detailed Solution Below)

Option 2 : 30 sq. units 

Geometry Question 2 Detailed Solution

 Given:

 A(-1, 1) B, (5, 1), C(5, 6) and D(-1, 6)

Formula used:

Area of the quadrilateral ABCD = Area of ∆ABC + Area of ∆ACD

Area of the triangle having vertices (x1, y1), (x2, y2) and (x3, y3)

=  \(\dfrac{1}{2}\) |x1(y2-y3) +x2(y3-y1) +x3(y1-y2) |

Calculations:

Area of ∆ABC =  \(\dfrac{1}{2}\)  | − 1[1 − 6)] + (5) (6 - 1) + 5 [1 - 1)] |

 \(\dfrac{1}{2}\)  | 5 + 25 |

= 15 sq. units

Area of ∆ACD =  \(\dfrac{1}{2}\) |− 1(6 – 6) + 5(6 – 1) + (-1)( 1 − 6) |

 \(\dfrac{1}{2}\) | 25 + 5 |

= 15 sq. units

Area of the quadrilateral ABCD = 15 + 15

= 30 sq. units

∴ the area of the quadrilateral is 30 sq. units.

∴ The answer is 30 sq.units.

Geometry Question 3:

If the points P(-2, 1), Q(α, β) and R(4, -1) are collinear and α - β = -3, then the value of (α + β) is: 

  1. 2
  2. -1
  3. 1
  4. 0

Answer (Detailed Solution Below)

Option 2 : -1

Geometry Question 3 Detailed Solution

Given:

Points: P(-2, 1), Q(α, β), R(4, -1)

α - β = -3

Formula used:

For collinear points, slope of PQ = slope of QR

Slope formula: (y2 - y1) / (x2 - x1)

Calculation:

Slope of PQ = (β - 1) / (α + 2)

Slope of QR = (-1 - β) / (4 - α)

Since P, Q, R are collinear:

⇒ (β - 1) / (α + 2) = (-1 - β) / (4 - α)

Cross-multiplying:

⇒ (β - 1)(4 - α) = (-1 - β)(α + 2)

⇒ 4β - β α - 4 + α = -α -2 - β α - 2β

⇒ 4β + α + α + 2β = -2 + 4

⇒ 3β + α = 1

Given that α - β = -3:

From α - β = -3 ⇒ α = β - 3

Substitute in 3β + α = 1:

⇒ 3β + (β - 3) = 1

⇒ 4β - 3 = 1

⇒ β = 1

So, α = 1 - 3 = -2

α + β = -2 + 1 = -1

∴ The correct answer is -1.

Geometry Question 4:

If a regular polygon has 10 sides, then the measure of its interior angle is greater than the measure of its exterior angle by how many degrees?

  1. 120
  2. 132
  3. 108
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 108

Geometry Question 4 Detailed Solution

Concept used:

If the number of sides of a regular polygon be n, then

The interior angle = \( \frac{(n - 2) \times 180^\circ }{n}\)

Calculation:

Let the required difference be x.

No. of sides of a regular polygon = 10

According to the question

\(\Rightarrow \frac{{\left( {n - 2} \right) \times 180^\circ }}{n} - \frac{{360^\circ }}{n} = x\)

\(\Rightarrow \frac{{\left( {10 - 2} \right) \times 180^\circ }}{{10}} - \frac{{360^\circ }}{{10}} = x\)

\(\Rightarrow \frac{{8 \times 180^\circ }}{{10}} - 36^\circ = x\)

⇒ 144 - 36° = x

∴ x = 108°

Geometry Question 5:

Find the ratio of the measure of the angle of a regular pentagon to the measure of the angle of a regular octagon.

  1. 7 : 8
  2. 5 : 6
  3. 4 : 5
  4. More than one of the above
  5. None of the above

Answer (Detailed Solution Below)

Option 3 : 4 : 5

Geometry Question 5 Detailed Solution

Given:

A regular pentagon has 5 sides, and a regular octagon has 8 sides.

The measure of the interior angle of a regular polygon is given by:

Interior angle = [(n - 2) × 180] / n, where n is the number of sides.

Formula used:

Ratio = Measure of the angle of the pentagon / Measure of the angle of the octagon

Calculations:

Step 1: Calculate the measure of the angle of the pentagon:

Angle = [(5 - 2) × 180] / 5

Angle = (3 × 180) / 5

Angle = 540 / 5

Angle = 108°

Step 2: Calculate the measure of the angle of the octagon:

Angle = [(8 - 2) × 180] / 8

Angle = (6 × 180) / 8

Angle = 1080 / 8

Angle = 135°

Step 3: Find the ratio:

Ratio = 108 / 135

Ratio = 4 / 5

The ratio of the measure of the angle of a regular pentagon to the measure of the angle of a regular octagon is 4:5.

Top Geometry MCQ Objective Questions

The area of the triangle whose vertices are given by the coordinates (1, 2), (-4, -3) and (4, 1) is:

  1. 7 sq. units
  2. 20 sq. units
  3. 10 sq. units
  4. 14 sq. units

Answer (Detailed Solution Below)

Option 3 : 10 sq. units

Geometry Question 6 Detailed Solution

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Given:-

Vertices of triangle = (1,2), (-4,-3), (4,1)

Formula Used:

Area of triangle = ½ [x(y- y3) + x(y- y1) + x(y- y2)]

whose vertices are (x1, y1), (x2, y2) and (x3, y3)

Calculation:

⇒ Area of triangle = (1/2) × [1(-3 – 1) + (-4) (1 – 2) + 4{2 – (-3)}]

= (1/2) × {(-4) + 4 + 20}

= 20/2

= 10 sq. units

In the triangle ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°. What is the value of the length of the side BC? 

F2 Savita SSC 1-2-23 D5

  1. 10 cm
  2. 7.13 cm
  3. 13.20 cm
  4. 11.13 cm

Answer (Detailed Solution Below)

Option 4 : 11.13 cm

Geometry Question 7 Detailed Solution

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Given:

In the triangle, ABC, AB = 12 cm and AC = 10 cm, and ∠BAC = 60°.

Concept used:

According to the law of cosine, if a, b, and c are three sides of a triangle ΔABC and ∠A is the angle between AC and AB then, a2 = b2 + c2 - 2bc × cos∠A

 Trigo

Calculation:

​According to the concept,

BC2 = AB2 + AC2 - 2 × AB × AC × cos60°

⇒ BC2 = 122 + 102 - 2 × 12 × 10 × 1/2

⇒ BC2 = 124

⇒ BC ≈ 11.13

∴ The measure of BC is 11.13 cm.

A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm. then what is the length of RS ?

  1. 7 cm
  2. 15 cm
  3. 9 cm
  4. 7.3 cm

Answer (Detailed Solution Below)

Option 3 : 9 cm

Geometry Question 8 Detailed Solution

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Given :

A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm

Calculations :

F1 Ashish S 25-10-21 Savita D1

If a circle touches all four sides of quadrilateral PQRS then, 

PQ + RS = SP + RQ

So,

⇒ 11 + RS = 8 + 12

⇒ RS = 20 - 11

⇒ RS = 9

∴ The correct choice is option 3.

AB and CD are two parallel chords of a circle of radius 13 cm such that AB = 10 cm and CD = 24cm. Find the distance between them(Both the chord are on the same side)

  1. 9 cm
  2. 11 cm
  3. 7 cm
  4. None of these

Answer (Detailed Solution Below)

Option 3 : 7 cm

Geometry Question 9 Detailed Solution

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Given

AB ∥ CD, and

AB = 10cm, CD = 24 cm

Radii OA and OC = 13 cm

Formula  Used

Perpendicular from the centre to the chord, bisects the chord.

Pythagoras theorem.

Calculation

F1 Vikash K 08-11-21 Savita D4

Draw OP perpendicular on AB and CD, and 

AB ∥ CD, So, the points O, Q, P are collinear.

We know that the perpendicular from the centre of a circle to a chord bisects the chord.

AP = 1/2 AB = 1/2 × 10 = 5cm

CQ = 1/2 CD = 1/2 × 24 = 12 cm

Join OA and OC

Then, OA = OC = 13 cm

From the right ΔOPA, we have

OP2 = OA2 -  AP2      [Pythagoras theorem]

⇒ OP2 = 132- 52

⇒ OP2 = 169 - 25 = 144

⇒ OP = 12cm

From the right ΔOQC, we have

OQ2 = OC2- CQ2      [Pythagoras theorem]

⇒ OQ2 = 13- 122

⇒ OQ2 = 169 - 144 = 25

⇒ OQ = 5 

So, PQ = OP - OQ = 12 -5 = 7 cm

∴ The distance between the chord is of 7 cm.

The ratio of the measures of each interior angle of a regular octagon to that of the regular dodecagon is:

  1. 8 : 12
  2. 9 : 10
  3. 12 : 8
  4. 4 : 5

Answer (Detailed Solution Below)

Option 2 : 9 : 10

Geometry Question 10 Detailed Solution

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Concept:

Octagon has eight sides.

Dodecagon has twelve sides.

Formula:

Interior angle of polygon = [(n – 2) × 180°] /n

Calculation:

Interior angle of octagon = [(8 – 2)/8] × 180° = 1080°/8 = 135°

Interior angle of dodecagon = [(12 – 2)/12] × 180° = 1800°/12 = 150°

∴ The ratio of the measures of the interior angles for octagon and dodecagon is 9 : 10

Two circles touch each other externally at P. AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40° . The measure of ∠ABP is:

  1. 45°
  2. 55°
  3. 50°
  4. 40°

Answer (Detailed Solution Below)

Option 3 : 50°

Geometry Question 11 Detailed Solution

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Given:

Two circles touch each other externally at P.

AB is a direct common tangent to the two circles, A and B are points of contact, and ∠PAB = 40°.

Concept used:

If two circles touch each other externally at some point and a direct common tangent is drawn to both circles, the angle subtended by the direct common tangent at the point where two circles touch each other is 90°.

Calculation:

F2 Madhuri SSC 13.02.2023 D1

According to the concept, ∠APB = 90°

Considering ΔAPB,

∠ABP

⇒ 90° - ∠PAB

⇒ 90° - 40° = 50°

∴ The measure of ∠ABP is 50°.

Two common tangents AC and BD touch two equal circles each of radius 7 cm, at points A, C, B and D, respectively, as shown in the figure. If the length of BD is 48 cm, what is the length of AC?

F1 SSC Arbaz  19-10-23 D1 v2

  1. 50 cm
  2. 40 cm
  3. 48 cm
  4. 30 cm

Answer (Detailed Solution Below)

Option 1 : 50 cm

Geometry Question 12 Detailed Solution

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Given:

Radius of each circle = 7 cm

BD = transverse common tangent between two circles = 48 cm

Concept used:

Length of direct transverse tangents = √(Square of the distance between the circle - Square of sum radius of the circles)

Length of the direct common tangents =(Square of the distance between the circle - Square of the difference between the radius of circles)

Calculation:

AC = Length of the direct common tangents

BD = Length of direct transverse tangents

Let, the distance between two circles = x cm

So, BD = √[x2 - (7 + 7)2]

⇒ 48 = √(x2 - 142)

⇒ 482x2 - 196  [Squaring on both sides]

⇒ 2304 = x2 - 196

⇒ x2 = 2304 + 196 = 2500

⇒ x = √2500 = 50 cm

Also, AC = √[502 - (7 - 7)2]

⇒ AC = √(2500 - 0) = √2500 = 50 cm

∴ The length of BD is 48 cm, length of AC is 50 cm

The complementary angle of supplementary angle of 130° 

  1. 50° 
  2. 30° 
  3. 40° 
  4. 70° 

Answer (Detailed Solution Below)

Option 3 : 40° 

Geometry Question 13 Detailed Solution

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Given:

One of the supplement angles is 130°.

Concept used:

For supplementary angle: The sum of two angles is 180°.

For complementary angle: The sum of two angles is 90°.

Calculation:

The supplement angle of 130° = 180° - 130° = 50°

The complement angle of 50° = 90° - 50° = 40°

∴ The complement angle of the supplement angle of 130° is 40°

Mistake PointsPlease note that first, we have to find the supplementary angle of 130° & after that, we will find the complementary angle of the resultant value.

To draw a pair of tangents to a circle which are inclined to each other at an angle of 75°, it is required to draw tangents at the end points of those two radii of the circle, the angle between whom is

  1. 65°
  2. 75°
  3. 95°
  4. 105°

Answer (Detailed Solution Below)

Option 4 : 105°

Geometry Question 14 Detailed Solution

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Concept:

Radius is perpendicular to the tangent at the point of contact

Sum of all the angles of a Quadrilateral = 360° 

Calculation:

F1 AbhishekP Madhuri 23.02.2022 D1

PA and PB are tangents drawn from an external point P to the circle.

∠OAP = ∠OBP = 90°  (Radius is perpendicular to the tangent at the point of contact)

Now, In quadrilateral OAPB,

∠APB + ∠OAP + ∠AOB + ∠OBP = 360° 

75° + 90° + ∠AOB + 90° = 360°

∠AOB = 105°

Thus, the angle between the two radii, OA and OB is 105°

Two circles touch each other externally at point X. PQ is a simple common tangent to both the circles touching the circles at point P and point Q. If the radii of the circles are R and r, then find PQ2.

  1. 3πRr/2
  2. 4Rr
  3. 2πRr
  4. 2Rr

Answer (Detailed Solution Below)

Option 2 : 4Rr

Geometry Question 15 Detailed Solution

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F1 Ashish.S 05-04-21 Savita D1

We know,

Length of direct common tangent = √[d2 - (R - r)2]

where d is the distance between the centers and R and r are the radii of the circles.

PQ = √[(R + r)2 - (R - r)2]

⇒ PQ = √[R2 + r2 + 2Rr - (R2 + r2 - 2Rr)]

⇒ PQ = √4Rr

⇒ PQ2 = 4Rr

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