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=√ 122 + 52
= √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= √d2 - (r1-r2)2
= √169-25
= √144
OD= 12 cm
OD= √d2 - (r1-r2)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
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= √AB2 - BD2
= √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
The perimeter of the equilateral triangle ABC= 24 cm
Note:
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= √AB2 - BD2
= √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
Workspace
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 line, m = tanθ
Slope of the line, m = -(1/√3)
m = tanθ = -(1/√3)
= -tan30°
= tan(180° - tan30°)
= tan150°.
√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°.
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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.
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27. ∆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:
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∆ADB ∼ ∆BDC
AD/BD = BD/DC
BD² = AD x DC
BD² = 12 x 8
BD² = 96
BD = 4√6.
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28. In the diagram given below, CD = BF = 10 units and ∠CED = ∠BAF = 30° .What would be the area of ∆AED?
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Correct Ans:50(√3 + 4)
Explanation:
Given: CD = BF = 10 units;
∠CED = ∠BAF = 30°
In ∆ECD,
tan 60° = ED/CD
√3 = ED/10
ED = 10√3
In ∆ABF,
tan60° = AB/BF
√3 = AB/10
AB = 10√3
In ∆BFC,
tan60° = BF/BC
√3 = 10/BC
BC = 10/√3
Area of ∆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 ∆ECD,
tan 60° = ED/CD
√3 = ED/10
ED = 10√3
In ∆ABF,
tan60° = AB/BF
√3 = AB/10
AB = 10√3
In ∆BFC,
tan60° = BF/BC
√3 = 10/BC
BC = 10/√3
Area of ∆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 2y + x = 9.
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.
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31. The two lines 3x - 8y =16 and 2x + 4y = 6 intersect at (a, b). Find the value of (a² - 4b²).
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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,
a² - 4b² = (4)² - 4(-1/2)²
= 16 - 4(1/4)
= 16 - 1
= 15
The value of (a² - 4b²) = 15.
2x + 4y = 6 .....(ii)
By solving equation (i) and (ii), we get
x = 4 = a
y = -1/2 = b
Therefore,
a² - 4b² = (4)² - 4(-1/2)²
= 16 - 4(1/4)
= 16 - 1
= 15
The value of (a² - 4b²) = 15.
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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 ∥ PB
∴ AC = AB = Diameter
∴ ∠QCA = ∠PBA
∴ QC ∥ 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 ⊿OAL, (OL)2 = (OA)2 - (AL)2
= (13)2 - (5)2 = 169 - 25
(OL)2 = 144
OL = 12 cm
Then, CM = (1/2)*CD = (1/2)*24 = 12 cm
And OC = radius of the circle = 13 cm
From ⊿OMC, (OM)2 = (OC)2 - (CM)2
= (13)2 - (12)2 = 169 - 144
(OM)2 = 25
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 ⊿OAL, (OL)2 = (OA)2 - (AL)2
= (13)2 - (5)2 = 169 - 25
(OL)2 = 144
OL = 12 cm
Then, CM = (1/2)*CD = (1/2)*24 = 12 cm
And OC = radius of the circle = 13 cm
From ⊿OMC, (OM)2 = (OC)2 - (CM)2
= (13)2 - (12)2 = 169 - 144
(OM)2 = 25
OM = 5 cm
Therefore, distance between the chords, ML = OM + OL
ML = 5+ 12
ML = 17 cm.
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37. In a ∆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: ax2+by2+cx+dy+e= 0
The center-radius form of the circle equation is in the format(x–h)2+ (y–k)2=r2, with the center being at the point(h, k) and the radius being "r".
Given eqn is: x^2 + y^2 -10x + 12y -10 = 0
=> x^2 + y^2 -10x + 12y = 10
Group thex-stuff together. And theny-stuff together.
=>(x^2 - 10x) + (y^2 + 12y) = 10
Take thex-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 they-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=r2
=> The center is at (h, k) = (5, -6)
The center-radius form of the circle equation is in the format(x–h)2+ (y–k)2=r2, with the center being at the point(h, k) and the radius being "r".
Given eqn is: x^2 + y^2 -10x + 12y -10 = 0
=> x^2 + y^2 -10x + 12y = 10
Group thex-stuff together. And theny-stuff together.
=>(x^2 - 10x) + (y^2 + 12y) = 10
Take thex-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 they-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=r2
=> 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)
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
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
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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)
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)
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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