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

1. ABCD is a cyclic quadrilateral and AB is the diameter of the circle. If ∠CAB = 48 °, then what is the value (in degrees) of ∠ADC? 




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Correct Ans:138°
Explanation:


given: AB is a diameter
Therefore, ∠ACB = 90°
Also, given that, ∠CAB = 48°
∠ABC = 180° - (90° + 48°)
= 42°
ABCD is a cyclic quadrilateral
∠ADC = 180° - ∠ABC
= 180° - 42°
= 138°
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2. The area of quadrilateral ABCD whose vertices in order are A(1, 1) B(7, -3), C(12, 2) and D(7, 21) is




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Correct Ans:132 sq.units
Explanation:
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3. In the given diagram O is the centre of the circle and CD is a tangent. ∠CAB and ∠ACD are supplementary to each other ∠OAC = 30°. Find the value of ∠OCB.





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Correct Ans:30°
Explanation:
Given:
∠CAB and ∠ACD are supplementary to each other.
∠CAB + ∠ACD = 180°
In the given diagram, AB || CD
∠DCB = ∠ABC
Also given, ∠OAC = 30°
∠OAC = ∠OCA = 30°
Therefore, ∠AOC = 120°
∠ABC = 60°
Since, ∠DCB = ∠ABC
∠DCB = 60°

∠OCD = 90°
∠OCB = ∠OCD - ∠DCB = 90° - 60°
∠OCB = 30°.
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4. If two medians BE and CF of a triangle ABC, intersect each other at G and if BG = CG, ∠BGC = 60°, BC = 8 cm, then area of the triangle ABC is




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Correct Ans:48√3 cm²
Explanation:


Here, ∆BGC is an equilateral triangle.
Area of ∆BGC = √3/4 a2
= (√3/4)(8)2
= (√3/4)64
= 16√3

Area of ∆BGC = (1/3)Area of ∆ABC
16√3 = (1/3) x Area of ∆ABC
Area of ∆ABC = 48√3 cm².
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5. In the given figure D, E and F are mid points of AB, AC and BC respectively. P, Q and R are mid points of DE, DF and EF. Find ratio of area of triangle PQR to that of parallelogram ADFE.





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Correct Ans:(1:8)
Explanation:
Triangle ADE, DEF, BDF and CEF are congruent to each other. Hence share equal area of ∆ABC.

----> Area (∆ADE) = (1/4) * Area (∆ABC);
----> Area (∆DEF) = (1/4) * Area (∆ABC);
----> Area (ADFE) = (1/2) * Area (∆ABC);
Similarly, triangle PQR, DPQ, PER and QRF are congruent to each other and share equal are of ∆DEF

----> Area (∆PQR) = (1/4) * Area (∆DEF);
----> = (1/4) * (1/4) * Area (∆ABC); = (1/16) * Area (∆ABC);
----> (Area (∆PQR)/ Area (ADFE) ) = ( (1/16) * Area (∆ABC);/(1/2) * Area (∆ABC); )
----> = (1/8)

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6. A triangle ABC is inscribed inside a circle. Bisectors of the angle ∠A,∠B and ∠C meet the circle at P,Q and R respectively. Then ∠PQR = ?




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Correct Ans:90° - (1/2)∠ABC
Explanation:



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7. The graph of the equation 4x - 5y = 20 intersects the X-axis at the point.




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Correct Ans:(5, 0)
Explanation:
When a straight line cuts X-axis, the coordinates of point of intersection = (x, 0),
i.e. y = 0
Putting y = 0 in 4x – 5y = 20
4x = 20
x = 5
So, point of intersection at X-axis = (5, 0).
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8. ∆ABC is similar to ∆PQR. Length of AB is 36 cm and length of the corresponding side PQ is 16 cm. If area of ∆ABC is 1296 sq cm, what is the area of ∆PQR?     




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Correct Ans:256 sq cm
Explanation:
----> AB= 36 and PQ= 16cm
----> Area of triangle ABC= 1296
-----> We know that when two triangle are similar then ratio of area of tringles = ratio of square of corresponding side
----> Then , (1296/Area of triangle PQR) = ((362)/(162))
---->= (1296/256)

----> So area of triangle PQR = 256sqcm
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9. In the given figure, area of isosceles triangle ABE is 72 cm2 and BE = AB and AB = 2 AD, AE II DC, then what is the area (in cm2) of the trapezium ABCD?





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Correct Ans:144
Explanation:


Area of â–³ABE = 1/2 * x * x = 72
(AB = BE) = x = 12
AD = AB/2 = 6
BC = 12 + 6 = 18 cm
Area of trapezium ABCD = 1/2 * h * (AD + BC)
= 1/2 * 12 * 24
= 144
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10. Two circles of equal radius of 'r' intersect each other in such a way that both pass through center of each other. What is the length of common chord?




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Correct Ans:r√3
Explanation:


OA = PA = OP = r
OM = MP = r/2
AB = 2*AM
AB = 2*√(OA2 - OM2)
AB = 2*√(r2 - r2/4)
AB = 2*√3r2/4
AB = 2r/2 * √3 = r√3
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11. The line passing through (4,3) and (y,0) is parallel to the line passing through (-1,-2) and (3,0). Find y?




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Correct Ans:-2
Explanation:
Slope of line passing through (4,3) and (y,0)
WKT, m1 = (y2 - y1)/(x2 -x1)
m1 = (0 - 3)/(y - 4)
= -3/(y - 4)
Slope of line passing through (-1,-2) and (3,0)
m2 = (y2 - y1)/(x2 -x1)
= (0 - (-2))/(3 - (-1))
= 2/4
= 1/2

If two lines are parallel, then slopes are equal.
m1 = m2
-3/(y - 4) = 1/2
y - 4 = -3 x 2
y = -2.
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12. If ∆ABC and ∆DEF are similar triangles and BC = 4 cm, EF = 7 cm, area of ∆ABC is 144 cm2 then find the are of ∆DEF     




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Correct Ans:441 cm2
Explanation:




Hence the Answer is : 441 cm2
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13. ∆ABC is a right angle triangle, ∠B = 90°, BD is perpendicular to AC. If AC = 14 cm, BC= 12 cm, find the length of CD. 




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Correct Ans:10(2/7) cm
Explanation:


Let CD = x
In ∆BDC, cosC = x/12
In ∆ABC, cosC = 12/14
equate cosθ => x/12 = 12/14
x = 12*12 / 14 = 72/7
CD = 10(2/7) cm
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14. Two circles of radii 10cm and 8 cm intersect and the length of the common chord is 12 cm. find of the common chord is 12 cm. Find the distance between their centers.




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Correct Ans:13.29 cm
Explanation:


Let A and B be the centre of the circle of radii 10 cm and 8 cm.
PQ = 12 cm
Where, PL = (1/2)PQ = (1/2)*12 = 6 cm.

In right triangle ALP, we have
AP2 = AL2 + LP2
AL = √[AP2 - LP2]
AL = √[102 - 62]
AL = √(64) = 8 cm.

In right triangle BLP, we have
BP2 = BL2 + LP2
BL = √[BP2 - LP2]
BL = √[82 - 62]
BL = √(28) = 5.29 cm
Distance between the centres = AL + BL
= 8 + 5.29
= 13.29 cm.
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15.

What is the average of angles x and y?




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Correct Ans:95°
Explanation:


∠a = 40°(Vertically opposite angles)
∠b = 130°(Vertically opposite angles)
Since, the sum of angles in a trapezoid is 360°.
x + y + a + b = 360°
x + y + 40° + 130° = 360°
x + y = 190°
Average of angles x and y = 190°/2 = 95°.
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16. Triangle ABC is similar to triangle PQR and their areas are in ratio 1:4 respectively. If PQ = 6, QR = 8 and PR = 10, find the length of AB




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Correct Ans:3
Explanation:
Reference:
Triangle ABC is similar to triangle PQR and their areas are in ratio 1:4 respectively. If PQ = 6, QR = 8 and PR = 10,

Let find the length of AB:
(AB/PQ) = (√(1/4))
----------> = (1/2)
----------> Summit the PQ value,
---------> (AB/6) = (1/2)
---------> AB = 3

Hence the answer is :AB = 3
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17. In a triangle ABC, the lengths of the sides AB, AC and BC are 3, 5 and 6 cm respectively. If a point D on BC is drawn such that the line AD bisects the ∠A internally, then what is the length of BD?




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Correct Ans:2.25 cm
Explanation:


BD/AB = DC/AC
Let BD = x
So, DC = 6 - x
x/3 = (6 - x)/5
5x = 18 - 3x
8x = 18
x = 9/4 = 2.25 cm
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18. The length of the chord of a circle is 8 cm and perpendicular distance between centre and the chord is 3 cm. Then the radius of the circle is equal to:




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Correct Ans:5 cm
Explanation:


Chord, AB = 8 cm
Then, AC = CB = 4 cm
Perpendicular distance between centre 'O' and chord AB, ie., OC = 3 cm

By Pythagoras theorem in right angled triangle OAC,
Now, radius of the circle = OA = √[OC2 + AC2]
= √[32 + 42]
= √[9 + 16]
= √25
= 5 cm
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19. O is the incentre of ∆ABC and ∠A = 30° then ∠BOC is




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Correct Ans:105°
Explanation:


∠B + ∠C = 180° - 30° = 150°
∠OBC + ∠OCB = 150°/2 = 75°
∠BOC = 180° - 75° = 105°
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20. In a ∆ABC, incentre is O and ∠BOC = 110°, then the measure of ∠BAC is




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Correct Ans:40°
Explanation:
∠BOC = 90° + A/2



110° = 90° + A/2
A/2 = 110 - 90 = 20
A = 2 * 20 = 40°
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