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Explore popular questions from Geometry for IBPS. This collection covers Geometry previous year IBPS questions hand picked by experienced teachers.

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Q 1. If (5, 1), (x, 7) and (3, -1) are 3 consecutive verticles of a square then x is equal to :

-3

B

-4

C

5

D

6

E

None of these

Explanation

For the verticles to form a square, we know that the length of each side of the square should be equal. Therefore,
(x - 5){tex}^2{/tex} + (7 - 1){tex}^2{/tex} = (x - 3){tex}^2{/tex} + (7 + 1){tex}^2{/tex}
[x{tex}^2{/tex} + 5{tex}^2{/tex} - 2 (x) (5) ] + [36] = [x{tex}^2{/tex} + 3{tex}^2{/tex} - 2 (x) (3) ] + [64]
[5{tex}^2{/tex} + 36] - [9 + 64] = (10 - 6) x
x = -{tex}\frac{12}{4}{/tex} = - 3
This gives the side of the square x = - 3.

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Q 2. What is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between two sides is 150°?

24 sq units

B

48 sq units

C

24{tex}\sqrt 3{/tex}

D

48{tex}\sqrt 3{/tex}

E

Such a triangle does not exist

Explanation

If two sides of a triangle and the included angle 'y' is known, then the area of the triangle can be found using the formula
{tex}\frac{1}{2}{/tex}* (product of sides) * sin y
Substituting the values in the formula, we get {tex}\frac{1}{2}{/tex}* 8 * 12 * sin 150 = 24 sq units

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Q 3. What is the measure of the radius of the circle that circumscribes a triangles whose sides measure 9, 40 and 41?

A

6

B

4

C

24.5

20.5

E

12.5

Explanation

From the measure of the length of the sides of the triangle 9, 40 and 41 we can infer that the triangle is a right angled triangle. 9, 40, 41 is a Pythagorean triplet.
In a right angled triangle, the radius of the circle that circumscribes the triangle is half the hypotenuse.
In the given triangle, the hypotenuse = 41
Therefore, the radius of the circle that circumscribes the triangle ={tex}\frac{41}{2}{/tex} = 20.5 units

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Q 4. Verticles of a quadrilateral ABCD are A (0, 0), B (4, 5), C (9, 9) and D (5, 4). What is the shape of the quadrilateral?

A

Square

B

Rectangle but not a square

Rhombus

D

Parallelogram but not a rhombus

E

None of these

Explanation

The lengths of the four sides AB, BC, CD and DA are all equal to {tex}\sqrt{41}{/tex}.
Hence, the given quadrilateral is either a Rhombus or a Square.
Now let us compute the lengths of the two diagonals AC and BD.
The length of AC is {tex}\sqrt{162}{/tex} and the length of BD is {tex}\sqrt{2}{/tex}
As the diagonals are not equal and the sides are equal, the given quadrilateral is a Rhombus.

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Q 5. If the sum of the interior angles of a regular polygon measures upto 1440 degrees, how many sides does the polygon have?

10 sides

B

8 sides

C

12 sides

D

9 sides

E

None of these

Explanation

We know that the sum of an exterior angle and an interior angle of a polygon = 180°
We also know that sum of all the exterior angles of a polygon = 360°
The question states that the sum of all interior angles of the given polygon = 1440°
Therefore, sum of all the interior and exterior angles of the polygon = 1440 + 360 = 1800
If there are ‘n' sides to this polygon, then the sum of all the exterior and interior angles = 180 x n = 10

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Q 6. What is the radius of the in circle of the triangle whose sides measure 5, 12 and 13 units?

2 units

B

12 units

C

6.5 units

D

6 units

E

7.5 units

Explanation

The triangle given is a right angled triangle as its sides are 5, 12 and 13 which is one of the Pythagorean triplets.
Note: In a right angled triangle, the radius of the incircle is given by the following relation
{tex}\frac{sum\ of\ perpendicular\ sides - hypotenuse}{2}{/tex}
As the given triangle is a right angled triangle, radius of its incircle = {tex}\frac{5+12-13}{2}{/tex}= 2 units

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Q 7. How many diagonals does a 63 sided convex polygon have?

A

3780

1890

C

3843

D

3906

E

1953

Explanation

The number of diagonals of an n-sided convex is {tex}\frac{n(n-3)}{2}{/tex}
This polygon has 63 sides. Hence, {tex}n{/tex} = 63
Therefore, number of diagonals = {tex}\frac{63\times60}{2}{/tex} = 1890

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Q 8. If 10, 12 and ‘x' are sides of an acute angled triangle, how many integer values of ‘x' are possible?

A

7

B

12

9

D

13

E

11

Explanation

For any triangle sum of any two sides must be greater than the third side.
The sides are 10, 12 and ‘x'.
From Rule 2, x can take the following values : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21 - A total of 19 values.
When x = 3 or x = 4 or x = 5 or x = 6, the triangle is an OBTUSE angled triangle.
The smallest value of x that satisfies both conditions is 7. (10{tex}^{2}{/tex} + 7{tex}^{2}{/tex} > 12{tex}^{2}{/tex})
The highest value of x that satisfies both conditions is 15. (10{tex}^{2}{/tex} + 12{tex}^{2}{/tex} + 15{tex}^{2}{/tex})
When x = 16 or x = 17 or x = 18 or x = 19 or x = 20 or x = 21, the triangle is an OBTUSE angled triangle.
Hence, the values of x that satisfy both the rules are x = 7, 8, 9, 10, 11, 12, 13, 14, 15. A total of 9 values.

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Q 9. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 6 inches and 8 inches.

10 inches

B

11 inches

C

18 inches

D

20 inches

E

None of these

Explanation

Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.
6 : 8 : ?= 3 (2) : 4 (2) : ?
Yes, it is a 3-4-5 triangle for n =
Calculate the third side 5n = 5 x 2 = 10
The length of the hypotenuse is 10 inches.

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Q 10. Find the length of one side of a right triangle if the length of the hypotenuse is 15 inches and the length of the other side is 12 inches.

A

8 inches

B

7 inches

9 inches

D

13 inches

E

None of these

Explanation

Test the ratio of the lengths to see if it fits the 3n : 4n : 5n ratio.
? : 12 : 15 = ? : 4 (3) : 5 (3)
Yes, it is a 3-4-5 triangle for n = 3
Calculate the third side 3n = 3 x 3 = 9
The length of the side is 9 inches.

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Q 11. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are both 3 inches.

A

{tex}5{/tex} inches

B

{tex}3\sqrt{4}{/tex} inches

C

{tex}6{/tex} inches

{tex}3\sqrt{2}{/tex} inches

E

None of these

Explanation

This is a right triangle with two equal sides so it must be a 45° - 45° - 90° triangle. You are given that the both the sides are 3. If the first and second value of the ratio n : n : n{tex}\sqrt{2}{/tex} is 3 then the length of the third side is 3{tex}\sqrt{2}{/tex}
The length of the hypotenuse is 3{tex}\sqrt{2}{/tex} inches.

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Q 12. Find the lengths of the other two sides of a right triangle if the length of the hypotenuse is 4{tex}\sqrt{2}{/tex} inches and one of the angles is 45°.

4 inches

B

9 inches

C

8 inches

D

7 inches

E

None of these

Explanation

This is a right triangle with a 45° so it must be a 45° - 45° - 90° triangle.
You are given that the hypotenuse is 4{tex}\sqrt{2}{/tex} . If the third value of the ratio n : n : n{tex}\sqrt{2}{/tex} is 4{tex}\sqrt{2}{/tex} then the lengths of the other two sides must 4.
The lengths of the two sides are both 4 inches.

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Q 13. Find the length of the hypotenuse of a right triangle if the lengths of the other two sides are 4 inches and 4{tex}\sqrt{3}{/tex} inches.

8 inches

B

9 inches

C

10 inches

D

11 inches

E

None of these

Explanation

Test the ratio of the lengths to see if it fits the n : n{tex}\sqrt{3}{/tex} : 2n ratio.
4 : 4{tex}\sqrt{3}{/tex} : ? n : n{tex}\sqrt{3}{/tex} : 2n
Yes, it is a 30° - 60° - 90° triangle for n = 4
Calculate the third side
2n = 2 x 4 = 8
The length of the hypotenuse is 8 inches.

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Q 14. What is the area of the following square, if the length of BD is 2{tex}\sqrt{2}{/tex} ?

A

1

B

2

C

3

4

E

5

Explanation


We need to find the length of the side of the square in order to get the area.
The diagonal BD makes two 45° - 45° - 90° triangles with the sides of the square.
Using the 45° - 45° - 90° special triangle ratio n: n : n{tex}\sqrt{2}{/tex}. If the hypotenuse is 2{tex}\sqrt{2}{/tex} then the legs must be 2. So, the length of the side of the square is 2.
Area of square = 5{tex}^{2}{/tex} = 2{tex}^{2}{/tex} = 4

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Q 15. In the figure below, what is the value of y?

A

40

B

50

60

D

100

E

120

Explanation

Vertical angles being equal allows us to fill in two angles in the triangle that y° belongs to.
Sum of angles in a triangle = 180°
So, y° + 40° + 80° = 180°
y° + 120° = 180°
y° = 60°

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Q 16. Two circles both of radii 6 have exactly one point in common. If A is a point on one circle and B is a point on the other circle, what is the maximum possible length for the line segment AB?

A

12

B

15

C

18

D

20

24

Explanation


Sketch the two circles touching at one point.
The furthest that A and B can be would be at the two ends as shown in the above diagram.
If the radius is 6 then the diameter is 2 x 6 = 12 and the distance from A to B would be 2 x 12 = 24

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Q 17. Note: Figures not drawn to scale.
In the figures above, x = 60, How much more is the perimeter of triangle ABC compared with the triangle DEF.

0

B

2

C

4

D

6

E

8

Explanation

Note: Figures not drawn to scale Since x = 60°, triangle ABC is an equilateral triangle with sides all equal. The sides are all equal to 8. Perimeter of triangle ABC = 8 + 8 + 8 = 24 Triangle DEF has two angles equal, so it must be an isosceles triangle. The two equal sides will be opposite the equal angles
So, the length of DF = length of DE = 10
Perimeter of triangle DEF = 10 + 10 + 4 = 24 Subtract the two perimeters.
24 - 24 = 0

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Q 18. An equilateral triangle has one side that measures 5 in. What is the size of the angle opposite that side?

A

55°

B

70°

C

110°

60°

E

None of these

Explanation

Since it is an equilateral triangle all its angles would be 60°. The size of the angle does not depend on the length of the side.
The size of the angle is 60°.

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Q 19. An isosceles triangle has one angle of 96°. What are the sizes of the other two angles?

A

24°

B

34°

42°

D

96°

E

None of these

Explanation

Since it is an isosceles triangle it will have two equal angles. The given 96° angle cannot be one of the equal pair because a triangle cannot have two obtuse angles.
Let x be one of the two equal angles. The sum of all the angles in any triangle is 180°.
x + x + 96° = 180°
2x = 84°
x = 42°
The sizes of the other two angles are 42° each.

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Q 20. Find the circumference of the circle with a diameter of 8 inches?

A

25 inches

25.163 inches

C

29.45 inches

D

35.62 inches

E

None of these

Explanation

Formula C = {tex}\pi {/tex}d
C = 8{tex}\pi {/tex}
The circumference of the circle is 8{tex}\pi {/tex} = 25.163 inches

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Q 21. Find the area of the circle with a diameter of 10 inches?

A

55.78 sq. inches

B

99.75 sq. inches

C

92 inches

78.55 sq. inches

E

None of these

Explanation

Formula A = {tex}\pi {/tex}r{tex}^{2}{/tex}
Change diameter to radius r = {tex}\frac{1}{2}{/tex}d = {tex}\frac{1}{2}{/tex}x 10 = 5
Plug in the value: A = {tex}\pi {/tex}5{tex}^{2}{/tex} = 25 {tex}\pi {/tex}
The area of the circle is 25{tex}\pi {/tex}
78.55 sq. inches

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Q 22. Find the area of the circle with a radius of 10 inches?

314.2 sq. inches

B

115 inches

C

320.29 sq. inches

D

56.12 sq. inches

E

None of these

Explanation

Formula A = {tex}\pi {/tex}r2
Plug in the value A = {tex}\pi {/tex}102 = 100 {tex}\pi {/tex}
The area of the circle is 100 {tex}\pi {/tex}
314.2 sq. inches