Geometry Questions and Answers updated daily – Aptitude

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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
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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
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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 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:


∆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)
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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.
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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.
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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
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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

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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.
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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
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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.
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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°
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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(xh)2+ (yk)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: (xh)2+ (yk)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
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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)
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