# Geometry Questions and Answers updated daily – Aptitude

Geometry Questions: Solved 321 Geometry Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Geometry Questions

21. Which of the following option is CORRECT for SAS similarity criterion for the triangle ABC and DEF?

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Correct Ans:All the options are correct

Explanation:

The SAS similarity theorem states that if two sides of a triangle are proportional to two sides of another triangle and the included angle in both are equal, then two triangles are similar.

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22. A trapezoid A, B, C, D, and E has bases of length 10 and 14 cm. If the height of the trapezoid above is 5 units, what is the length of the diagonal BC?

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Correct Ans:13

Explanation:

Image for the question is:

To find the diagonal, the top base must be subtracted from the bottom base:

14−10=4

This leaves us with 4, which is the sum of the distance to the left and right of the top base.

Taking half of that(42=2)gives us the length of the distance to only the left side.

The image reference:

This means that the base of a triangle that includes that diagonal is equal to 2+10=12

Height = 5,

**To calculate BC:**

By applying Pythagorean Theorum:

BC=√ 12^{2} + 5^{2}

= √144+25

=√169

= 13

Therefore, the hypotenuse BC is 13.

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23. A and B are the centres of two circles with radii 11 cm and 6 cm respectively. A common tangent touches these circles at O & D respectively. If AB = 13 cm, then the length of OD is ________.

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Correct Ans:12 cm

Explanation:

**From the image,**

OD= √d

^{2}- (r

_{1}-r

_{2})

^{2}

= √169-25

= √144

OD= 12 cm

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24. Let G be the centroid of the equilateral triangle ABC of perimeter 24 cm. Then the length of AG is ________.

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Correct Ans:8 √3 cm

Explanation:

**Given:**

The perimeter of the equilateral triangle ABC= 24 cm

**Note:**

Centroid divide each median in 2:1 ratio

Centroid divide each median in 2:1 ratio

Therefore, AG:GD= 2:1

Initially, AD should be found so that AG can be calculated.

An equilateral triangle is a triangle in which all three sides are equal.

From the above figure,

BC=24/3= 8 cm

AB=BC=CA=8cm

BD is half of BC

BD=8/2=4 cm

AD= √AB

^{2}- BD

^{2}

= √64-16

= 4√3

wkt,

AG:GD= 2:1

So, AG is the 2 parts of AD

AG= (4√3/3) * 2

AG= 8/√3

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25. The angle made by the line x + √3y - 6 = 0, with positive direction of x-axis is

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Correct Ans:150°

Explanation:

Given: x + √3y - 6 = 0

√3y = -x + 6

y = (-x/√3) + (6/√3)

WKT,

Slope of the line, m = -(1/√3)

m = tanθ = -(1/√3)

= -tan30°

= tan(180° - tan30°)

=

√3y = -x + 6

y = (-x/√3) + (6/√3)

WKT,

**Slope of line, m = tanθ**Slope of the line, m = -(1/√3)

m = tanθ = -(1/√3)

= -tan30°

= tan(180° - tan30°)

=

**tan150°.**
Workspace

26. If ABCD be a cyclic quadrilateral in which ∠A = 4x° , ∠B = 7x° , ∠C = 5y° , ∠D = y° , then x : y is

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Correct Ans:4 : 3

Explanation:

The sum of opposite angles of a concyclic quadrilateral is 180°.

**∠A + ∠C = 180°**

4x + 5y = 180° .... (1)

Similarly,

**∠B + ∠C = 180°**

7x + y = 180° .....(2)

By solving (1) and (2), we get

x = 720/31

y = 540/31

Therefore, x : y = 720/31 : 540/31

**x : y = 4 : 3.**

Workspace

27. âˆ†ABC is a right angle triangle where BD is perpendicular to AC. If AD = 12 cm and DC = 8, then BD = ?

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Correct Ans:4√6

Explanation:

**âˆ†ADB ∼ âˆ†BDC**

AD/BD = BD/DC

BD² = AD x DC

BD² = 12 x 8

BD² = 96

**BD = 4√6.**

Workspace

28. In the diagram given below, CD = BF = 10 units and ∠CED = ∠BAF = 30° .What would be the area of âˆ†AED?

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Correct Ans:50(√3 + 4)

Explanation:

Given: CD = BF = 10 units;

∠CED = ∠BAF = 30°

In âˆ†ECD,

tan 60° = ED/CD

√3 = ED/10

ED = 10√3

In âˆ†ABF,

tan60° = AB/BF

√3 = AB/10

AB = 10√3

In âˆ†BFC,

tan60° = BF/BC

√3 = 10/BC

BC = 10/√3

Area of âˆ†AED = 1/2(AD x ED)

= (1/2) x (AB + BC + CD) x 10√3

= (1/2) x 10√3 x (10√3 + 10/√3 + 10 )

= 50√3( √3 + 1/ √3 + 1)

= 50√3(3 + 1 + √3)/√3

= 50(4 + √3)

∠CED = ∠BAF = 30°

In âˆ†ECD,

tan 60° = ED/CD

√3 = ED/10

ED = 10√3

In âˆ†ABF,

tan60° = AB/BF

√3 = AB/10

AB = 10√3

In âˆ†BFC,

tan60° = BF/BC

√3 = 10/BC

BC = 10/√3

Area of âˆ†AED = 1/2(AD x ED)

= (1/2) x (AB + BC + CD) x 10√3

= (1/2) x 10√3 x (10√3 + 10/√3 + 10 )

= 50√3( √3 + 1/ √3 + 1)

= 50√3(3 + 1 + √3)/√3

= 50(4 + √3)

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29. Equation of line passing through (1, 4) and perpendicular to y = 2x + 3 is?

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Correct Ans:2y + x = 9

Explanation:

Given: y = 2x + 3

When two lines are perpendicular, m1m2 = -1

Where m1 and m2 are slopes of the perpendicular lines.

Slope of the given line = 2

Slope of line perpendicular to it = -1/2

The line passes through (1, 4).

Equation of line => (y - y1) = m(x - x1)

y - 4 = (-1/2)(x - 1)

y - 4 = (1 - x)/2

2y - 8 = 1 - x

2y + x = 9

Therefore, equation of line is

When two lines are perpendicular, m1m2 = -1

Where m1 and m2 are slopes of the perpendicular lines.

Slope of the given line = 2

Slope of line perpendicular to it = -1/2

The line passes through (1, 4).

Equation of line => (y - y1) = m(x - x1)

y - 4 = (-1/2)(x - 1)

y - 4 = (1 - x)/2

2y - 8 = 1 - x

2y + x = 9

Therefore, equation of line is

**2y + x = 9.**
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30. A line DE parallel to the side BC intersects the other two sides of triangle at points D and E such that AD = (1/6)AB and AE = (1/6)AC. If the value of BC is 18 cm, calculate the value of DE (in cm).

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Correct Ans:3

Explanation:

DE || BC,

Therefore, ∠ADE = ∠ABC

∠AED = ∠ACB

By AA - similarity,

∆AED = ∆ADE

AB/AD = BC/DE

Given, AD = 1/6 AB; BC = 18cm

6 = 18/DE

DE = 18/6

DE = 3 cm.

Workspace

31. The two lines 3x - 8y =16 and 2x + 4y = 6 intersect at (a, b). Find the value of (aÂ² - 4bÂ²).

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Correct Ans:15

Explanation:

3x - 8y =16 .....(i)

2x + 4y = 6 .....(ii)

By solving equation (i) and (ii), we get

x = 4 = a

y = -1/2 = b

Therefore,

aÂ² - 4bÂ² = (4)Â² - 4(-1/2)Â²

= 16 - 4(1/4)

= 16 - 1

= 15

The value of (aÂ² - 4bÂ²) = 15.

2x + 4y = 6 .....(ii)

By solving equation (i) and (ii), we get

x = 4 = a

y = -1/2 = b

Therefore,

aÂ² - 4bÂ² = (4)Â² - 4(-1/2)Â²

= 16 - 4(1/4)

= 16 - 1

= 15

The value of (aÂ² - 4bÂ²) = 15.

Workspace

32. In a triangle ABC, AB = AC, BA is produced to D in such a manner that AC = AD. The circular measure of ∠(BCD) is

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Correct Ans:π/2

Explanation:

Given: AB = AC

Therefore, ∠(ABC) = ∠(ACB) ....(i)

Also given: AC = AD

Therefore, ∠(ACD) = ∠(ADC) ....(ii)

In a triangle,

∠(ABC) + ∠(ADC) + ∠(DCB) = 180°

∠(ABC) + ∠(ADC) + ∠(ACB) + ∠(ACD) = 180°

2∠(ACB) + 2∠(ACD) = 180°

From equation (i) & (ii)

Therefore, ∠(BCD) = 90° or π/2

Therefore, ∠(ABC) = ∠(ACB) ....(i)

Also given: AC = AD

Therefore, ∠(ACD) = ∠(ADC) ....(ii)

In a triangle,

∠(ABC) + ∠(ADC) + ∠(DCB) = 180°

∠(ABC) + ∠(ADC) + ∠(ACB) + ∠(ACD) = 180°

2∠(ACB) + 2∠(ACD) = 180°

From equation (i) & (ii)

Therefore, ∠(BCD) = 90° or π/2

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33. If the line DE is drawn parallel to the base of a triangle ABC by intersecting the other two sides, then which of the following is the CORRECT equation for this case.

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Correct Ans:AD/DB = AE/EC

Explanation:

If line DE is parallel to base BC of triangle ABC to internally divide the remaining two sides, then

AD/DB = AE/EC

As ∆ADE ~ ∆ABC

AD/DB = AE/EC

As ∆ADE ~ ∆ABC

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34. Consider the circle shown below having angle AOB as 135º and the shaded portion is the x part of the circular region. Calculate the value of x.

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Correct Ans:1/4

Explanation:

∠AOC = 180º - 135º = 45

∠AOC = ∠BOD = 45º

Shaded part is xth part of circular region.

∴ x = 90/360

= 1/4

Therefore, value of x is 1/4.

∠AOC = ∠BOD = 45º

Shaded part is xth part of circular region.

∴ x = 90/360

= 1/4

Therefore, value of x is 1/4.

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35. Consider the circle shown in the figure and choose the correct option for this case.

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Correct Ans:QC || PB

Explanation:

∠AQC = ∠APB = 90º (angle in semi-circle)

∴ AC = AB = Diameter

∴ ∠QCA = ∠PBA

∴ QC âˆ¥ PB

∴ AC = AB = Diameter

∴ ∠QCA = ∠PBA

∴ QC âˆ¥ PB

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36. AB and CD are two parallel chords on the opposite sides of the center of the circle. If AB = 10 cm, CD = 24 cm and the radius of the circle is 13 cm, the distance between the chords is

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Correct Ans:17 cm

Explanation:

From O, draw OL perpendicular to AB and OM perpendicular to CD.

Then, join OA and OC.

AL = (1/2)AB = (1/2)10

AL = 5 cm

OA = radius of the circle = 13 cm

From âŠ¿OAL, (OL)

= (13)

(OL)

OL = 12 cm

Then, CM = (1/2)*CD = (1/2)*24 = 12 cm

And OC = radius of the circle = 13 cm

From âŠ¿OMC, (OM)

= (13)

(OM)

OM = 5 cm

Therefore, distance between the chords, ML = OM + OL

ML = 5+ 12

ML = 17 cm.

Then, join OA and OC.

AL = (1/2)AB = (1/2)10

AL = 5 cm

OA = radius of the circle = 13 cm

From âŠ¿OAL, (OL)

^{2}= (OA)^{2}- (AL)^{2}= (13)

^{2}- (5)^{2}= 169 - 25(OL)

^{2}= 144OL = 12 cm

Then, CM = (1/2)*CD = (1/2)*24 = 12 cm

And OC = radius of the circle = 13 cm

From âŠ¿OMC, (OM)

^{2}= (OC)^{2 }- (CM)^{2}= (13)

^{2}- (12)^{2}= 169 - 144(OM)

^{2}= 25OM = 5 cm

Therefore, distance between the chords, ML = OM + OL

ML = 5+ 12

ML = 17 cm.

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37. In a âˆ†ABC, ∠A + ∠B = 65° and ∠B + ∠C = 140°. Then, ∠B is equal to

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Correct Ans:25°

Explanation:

Given, ∠A + ∠B = 65°

∠B + ∠C = 140°

(∠A + ∠B) + (∠B + ∠C) = (65° + 140°) = 205°

(∠A + ∠B + ∠C) + ∠B = 205°

180° + ∠B = 205°

∠B = 205° - 180° = 25°

∠B + ∠C = 140°

(∠A + ∠B) + (∠B + ∠C) = (65° + 140°) = 205°

(∠A + ∠B + ∠C) + ∠B = 205°

180° + ∠B = 205°

∠B = 205° - 180° = 25°

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38.

Find the center of the circle whose equation is x^2 + y^2 -10x + 12y -10 = 0

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Correct Ans:(5, -6)

Explanation:

For general format ofcircle:

The center-radius form of the circle equation is in the format

Given eqn is:

Group the

=>(x^2 - 10x) + (y^2 + 12y) = 10

Take the

=> (x^2 - 10x + 25) + (y^2 + 12y + 36) = 10 + 25 + 36

=> (x – 5)^2 + (y + 6)^2 = 71

=> (x – 5)^2 + [y – (-6)]^2 = 71

On comparing it with center-radius form of the circle equation: (

=>

*ax*^{2}+*by*^{2}+*cx*+*dy*+*e*= 0The center-radius form of the circle equation is in the format

**(**, with the*x*–*h*)^{2}+ (*y*–*k*)^{2}=*r*^{2}**center being at the point(**and the radius being "*h, k*)*r*".Given eqn is:

**x^2 + y^2 -10x + 12y -10 = 0****=> x^2 + y^2 -10x + 12y = 10**Group the

*x*-stuff together. And then*y*-stuff together.=>(x^2 - 10x) + (y^2 + 12y) = 10

Take the

*x*-term coefficient, multiply it by one-half, square it, and then add this to both sides of the equation, as shown. Do the same with the*y*-term coefficient.=> (x^2 - 10x + 25) + (y^2 + 12y + 36) = 10 + 25 + 36

=> (x – 5)^2 + (y + 6)^2 = 71

=> (x – 5)^2 + [y – (-6)]^2 = 71

On comparing it with center-radius form of the circle equation: (

*x*–*h*)^{2}+ (*y*–*k*)^{2}=*r*^{2}=>

**The center is at (h, k) = (5, -6)**
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39. Find the distance between the points (2,2) and (-1,6)

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Correct Ans:5 units

Explanation:

Given, two points, (x1,y1) = (2,2)

And (x2,y2) = (-1,6)

= √{(-1 – 2)^2 + (6 – 2)^2}

= √{9 + 16}

= √(25)

Thus,

And (x2,y2) = (-1,6)

**Distance between the points = √{(x2 – x1)^2 + (y2 – y1)^2}**= √{(-1 – 2)^2 + (6 – 2)^2}

= √{9 + 16}

= √(25)

**= 5**Thus,

**Distance between the points = 5**
Workspace

40. Find the distance between the points (2,3) and (3,4)

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Correct Ans:sqrt(2)

Explanation:

Given two points, (x1,y1) = (2,3)

And (x2,y2) = (3,4)

= sqrt {(3 – 2)^2 + (4 – 3)^2}

= sqrt {1 + 1}

Thus,

And (x2,y2) = (3,4)

**Distance between the points = sqrt {(x2 – x1)^2 + (y2 – y1)^2}**= sqrt {(3 – 2)^2 + (4 – 3)^2}

= sqrt {1 + 1}

**= sqrt(2)**Thus,

**Distance between the points = sqrt(2)**
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