Geometry 63 MCQ part a

Lines and angles, Circles, Right Triangles, Area of Plane figures, Volume of solids and Coordinate Geometry



MULTIPLE CHOICE QUESTIONS (Part a)

1. The measure of an angle is 35o . What is the measure of the complement of this angle? a. 40o b. 60o c. 55o d. 145o Sol: Correct option is (c)



Let the angle given be x

x  35o



Let the complement of x be y Then x  y  90o  y  90o  x  y  55o 2. The measure of an angle is 75o . What is the measure of the supplement of this angle? a. 15o b. 105o c. 110o d. 50o



Sol:



Correct option is (b)



Let the angle given be x

x  75o



Let the complement of x be y Then x  y  180o  y  180o  x  y  105o 3. The measure of an angle is five times of its complement. What is the measure of the angle? a. 50o b. 70o c. 30o d. 75o Sol: Correct option is (d)



Let the angle be x and its complement be y It is given that x  5 y Also x  y  90o

 5 y  y  90o  6 y  90o  y  15o  x  75o



4. An angle is half of its supplement. What is the measure of this angle? a. 20 o b. 60o c. 90o d. 110o Sol: Correct option is (b)



Let the angle be x and its supplement be y Then x  y  180o and x   y  2x  x  2 x  180o  3 x  180o  x  60o 5. If the two complementary angles are congruent, then what is the measure of each angle? a. 60o b. 30o c. 45o d. 90o Sol: Correct option is (c)

y 2



Let one angle be x and its complement be y As both are congruent, then x  y Also x  y  90o



 x  x  90o  2 x  90o



 x  45o  y  45o 6. If the two supplementary angles are congruent, then what is the measure of each angle? a. 60o b. 45o c. 30o d. 90o Sol: Correct option is (d)



Let one angle be x and its supplement be y As both are congruent x  y Also x  y  180o



 x  x  180o  2 x  180o  x  90o  y  90o



7. Find the value of x in the above figure a. 8 b. 3 c. 5 d. 9 Sol: Correct option is (a)



As vertically opposite angles are equal, therefore

x  3  2 x  11  x  2x  8  8  2x  x  x 8



8. What is the value of x in the above figure? a. 40 b. 50 c. 60 d. 10 Sol: Correct option is (b)



As both marked angles are supplementary angles, therefore

(3 x  15)  45  180



 3 x  30  180  3 x  150  x  50o



9. 9.What is the value of y in the above diagram? a. 60o b. 30o c. 90o d. 65o Sol: Correct option is (d)



As 2 x  50 and x  40 are both supplementary angels, therefore

(2 x  50)  ( x  40)  180  3 x  90  180  x  90o



Also



2 y  2 x  50  2 y  180  50  2 y  130  y  65o

10. The supplement of an angle is four times its complement. What is the measure of the angle?



a. 60o b. 70o c. 90o d. 105o Sol: Correct option is (a)



Let the angle be x and its supplement be y and its complement be z Then x  y  180 and x  z  90

 y  z  90 Also y  4 z  4 z  z  90  z  30  y  120



Also x  y  180  x  180  120  60o



11. In above figure aIIb and c is a transversal. What are the angles whose measure is same as that of 5 a. 1, 3, 6 b. 7, 3, 2 c. 6, 8 d. 1, 2



Sol:



Correct option is (b)



7  5 As both are vertically opposite angles



Also 7  3  5  3 Also 3  2  5  2



12. In above diagram lIIm and n is the transversal. What is the value of x in the above diagram? a. 40 b. 60 c. 50 d. 30 Sol: Correct option is (c)



Here y  2 x  1 As y and 2 x  1 are vertically opposite angles Also (2 x  10  (2 x  19)  180

 4 x  20  180  4 x  200  x  50



13. If in above diagram lIIm and nIIo , then which of the following statements are true? a. 1  4  3 b. 1  3  6 c. 1  4  10 d. 3  6  11 Sol: Correct option is (d)



3  6 As both are alternate interior angles.

Also 3  11 As both are corresponding angles. Hence 3  6  11



14. In above diagram if AB=AC, then which of the following is ture? a. no bisects lAB b. BAC  90o c. ABC  BAC d. none of the above Sol: Correct option is (a)



As AB=AC  ABC  ACB Also ABC  ACB  lAB  lAB  2ABC Also ABC  BAn  lAB  2BAn  BAn 

1 lAB 2



15. In above diagram SR II PQ and SM  PQ. What is the value of x? a. 40 o b. 30o c. 70o d. 60o



Sol:



Correct option is (c)



SMP  90o



We know that SPM  SMP  20  180



 SPM  180  20  90  SPM  70



Also x  SPM as both are corresponding angles Hence x  70o



16. In above diagram PQ II RS and RQ II TS. What is the value of x? a. 78o b. 85o c. 66o d. 36o Sol: Correct option is (a)



RSQ  36o As both are corresponding angles



Also RST  QRS  66o Also x  RST  RSQ  180o

 x  180  66  36  78o



17. In above figure PQ II RS II UT. What are the values of x and y respectively. a. 20, 30 b. 10, 40 c. 10, 30 d. 30, 40 Sol: Correct option is (c)



Here PQR QRS  180



 130  2 x  y  180  2 x  y  50

Also RST  STU  180

 120  3 x  y  180  30 x  y  60



Solving above equations of x and y

x  10 and y  30



18. In above figure PQ II SR and PS II QR. What are the values of x and y respectively a. 50, 100 b. 50, 70 c. 60, 100 d. 60, 70 Sol: Correct option is (a)



QRU  RQP  2 x  40  x  10  x  50



Also SPQ  RQP  180

 y  20  x  10  180  y  180  120  10  x  y  150  x  y  150



19. In above figure RT II SU. What is the value of y in the above figure? a. 130 b. 160



c. 75 d. 120 Sol: Correct option is (d)



Here 80  40  x  2 x  180



 3 x  60  x  20

We know that

PQT  40  x  180  PQT  120



Also PQT  y

 y  120



20. In above figure SQ  PQ . What are the values of x and y respectively. a. 20, 30 b. 60, 40 c. 30, 60 d. 40, 60 Sol: Correct option is (c)



Here (2 x  y )  60  180



 2 x  y  120



Also y  x  60  90  180



 y  x  30

Solving above equations

x  30 and y  60



21. What is the area of the above figure? a. 42 b. 35 c. 38 d. 36 Sol: Correct option is (b)



Area of above figure can be split into 2 parts Area = 9  (4  1)  8 1  27  8  35



22. What is the area of the above figure? a. 18 b. 72 c. 36 d. 24 Sol: Correct option is (c)



Area = 3  2  6  2  9  2

 6  12  18  36



23. What is the area of the above figure? Given PQRS is a rhombus and POQ  90o



a. 100 b. 180 c. 150 d. 120 Sol: Correct option is (d)

1  d1  d 2 2



Area of rhombus =



Here d1  12  12  24 Also PO  PQ 2  OQ 2  132  122  25  5



 d 2  5  5  10

1  Area   24 10  120 2



24. What is the area of equilateral triangle whose perimeter is 24. a. 27.68 b. 30.21 c. 4864 d. 29.60 Sol: Correct option is (a)



Here PQ=QR=RP=8 Also  PQS  PRS  QS  4

 PS  PQ 2  QS 2  82  42  48  6.92



Area =



1  8  6.92  27.68 2



25. What is the area of isosceles triangle with sides 13, 13 and 10 a. 80 b. 120 c. 60 d. 40 Sol: Correct option is (c)



Here  PQS  PRS

 QS  5



Also PS  132  52

 PS  144  12 1  Area  10  12  60 2



26. What is the area of a isosceles right triangle whose hypotenuse is 50. a. 694.61 b. 587.38 c. 798.6 d. 624.81 Sol: Correct option is (d)



Here PQ=QR=a Also a 2  a 2  50 2  2a 2  2500

 a  35.35



Area =



1  35.35  35.35  624.81 2



27. What is the area of an equilateral triangle whose height is 100? a. 4832.81 b. 5773. 67 c. 8192.74 d. 384.96



Sol:



Correct option is (b)



PS=100 Let PS=2a  QS=a

 PS  (20) 2  a 2  3a 2



 3a  100



a



100 3 1 200 100  2 3



Area =



Area = 5773.67 28. A square PQRS is inscribed in a circle of radius 4. What is the area of square? a. 64 b. 16 c. 4 d. 32 Sol: Correct option is (d)



Each diagonal of square = 8 Hence Area =

1  8  8  32 2



29. One diagonal of a rhombus is 4 times the other diagonal and the area of diagonal is 72. What is the measure of diagonals? a. 24, 12 b. 4, 16 c. 8, 32 d. 6, 24 Sol: Correct option is (d)



Let one diagonal = d Other diagonal = 4d Area =

1  d  4d  72 2



 d 2  36  d  6

Hence measure of first diagonal = 6 and measure of second diagonal = 24 30. The height of a triangle is 3 times its base. What is the length of base if area is 54?



a. 8 b. 4 c. 6 d. 3 Sol: Correct option is (c)



Let height = h and base = b h=3b

1 3  Area   b  3b  b 2 2 2 3 54  2  b 2  54  b 2   b 2  36  b  6 2 3



31. What is the area of a triangle whose sides are 4, 6, 8? a. 12.68 b. 18.92 c. 11.61 d. 7.99 Sol: Correct option is (c)



A  S ( S  a)( S  b)( S  c )

Here S 

a b c 4 68  9 2 2



A  9(9  4)(9  6)(9  8)  9(5)(3)(1)  11.61 32. What is the area of a trapezoid with height 11 and base 14 and 6



a. 55 b. 150 c. 19 d. 110 Sol: Correct option is (d)



1 Area   h  (b1  b2 ) 2 1 1  11 (14  6)  11 20  110 2 2



33. The median of a trapezoid divides into two regions. What is the ration of the areas of these two regions? a. 2 b. 3 c. 1 d. 4 Sol: Correct option is (c)



Here ST is median Also SM  PQ and let SM =h



1 Area of SPT   PQ  h 2



Let PQ=2a Also area of STQR 

1 (a  a )  h  ah 2



1 Also area of SPT   2a  h  ah 2



Hence ratio =



ah 1 ah



34. What is the area of above drawn trapezoid? a. 49 b. 38 c. 32 d. 78 Sol: Correct option is (b)



PQ=TR=8, SR=11  ST=3  PT  52  32  16  4 Here QR=4

1  Area   (8  11)  4  38 2



35. What is the area of a trapezoid with legs 10, 10 and bases 17 and 5? a. 88 b. 102 c. 72 d. 34 Sol: Correct option is (a)



Here TU=5, also RT=US and RT+US=12  RT  US  6 Here PT = 102  62  64  8

1  Area   8  (17  5)  88 2



36. What is the area of a trapezoid whose bases are 12 and 26. The base angles of the trapezoid are 45o ? a. 109 b. 84 c. 79 d. 133



Sol:



Correct option is (d)



Here TU=12, also RT=TS=7

PT  tan 45o  1  PT  RT  7 RT 1 Hence  Area   7  (12  26)  133 2



37. An isosceles trapezoid has bases 8 and 14 and its area is 44. What is the height of the trapezoid? a. 2 b. 4 c. 17 d. 9 Sol: Correct option is (b)



Here Area = 44



1  Area   h  (8  14)  11h 2

 11h  44  h  4



38. An isosceles trapezoid has bases 8 and 14. Its area is 44. What is the perimeter of trapezoid? a. 16 b. 82 c. 32 d. 28 Sol: Correct option is (c)



Here Area = 44

1  Area   h  (14  8)  11h 2

 11h  44  h  4



Now PT =4, also RT=US=3



 PR  42  32  25  5

Hence perimeter = 8+14+5+5=32 39. The complement of an angle is 50o more than the angle itself. What is, the measure of the angle?



a. 30o b. 20o c. 40o d. 15o Sol: Correct option is (b)



Let angle be x and its complement be y



x  y  90

Also y  x  50

 x  x  50  90  x  20o



40. The supplement of an angle is 40o more than the angle itself, what is the measure of the angle? a. 90o b. 130o c. 70o d. 60o Sol: Correct option is (c)



Let the measure of angle = x Let the measure of its supplement = y

x  y  180



Also y  x  40



 x  x  40  180  x  70o



41. What is the area of a circle whose diameter is 20? a. 215 b. 38 c. 97 d. 100 Sol: Correct option is (d)



Here diameter = 20

 radius  20  10 2



 Area   r 2   (10) 2  100



42. What is the circumference of the circle whose radius is 8? a. 16 b. 8 c. 9 d. 3 Sol: Correct option is (a)



Circumference = 2 r  2 (8)  16 43. What is the area of a circle whose circumference is 14 ? a. 37 b. 36 c. 16 d. 49



Sol:



Correct option is (d)



Area =  r 2 Circumference = 2 r  14



r 7

 Area   r 2   (7)2  49



44. PQRS is a square with side 8. QTRQ is a semicircle with diameter QR. What is the area of the whole region? a. 92.82 b. 67.74 c. 89.12 d. 64 Sol: Correct option is (c)



Area of square = a 2  8  8  64 Area of semicircle =

QR 8  4 2 2 1  (4) 2  8  25.12 2 1 2 r 2



Here r =



Area of semicircle =



Area of whole region = 64+25.12=89.12



45. In above figure PQ=6 and QR=8. Semicircles are drawn on each side of right angled triangle. What is the area of the whole region? (   3.14 ) a. 82.7 b. 39.6 c. 39.1 d. 78.5 Sol: Correct option is (d)

1  6  6  24 2 1    (3) 2  4.5  14.13 2 1    (4)2  8  25.12 2



Area of triangle =



Area of 1st semicircle =



Area of 2nd semicircle =



Also PR  82  62  100  10 Area of 3rd semicircle =

1    (5) 2  12.5  39.25 2



Area of whose region = 14.13+25.12+39.25=78.5 46. A wheel has radius 4. How far will the car travel in 5 minutes, if the wheel rotates once every 30 seconds? (   3.14 )



a. 251.2 b. 217.84 c. 79.38 d. 118.60 Sol: Correct option is (a)



Circumference = 2    r  8  25.12 No of rotations in 5 minutes = 2 rotations/minute  5 minutes = 10 rotations Hence distance traveled = 25.12 x 10 = 251.2 a



47. What is the area of shaded portion in above figure? a. 20.82 b. 25.64 c. 50.24 d. 96



Sol:



Correct option is (b)



Area of rectangle = 12  8  96 Area of semicircle =

1 2 1  r   3.14  16  25.12 2 2



Area of circle =  r 2    (16)  50.24 Area of shaded region = 96-25.12-50.24=20.64 48. What is the area of shaded region?



a. 96 b. 314 c. 200 d. 114 Sol: Correct option is (d)

90    (20) 2  314 360



Area of sector PQR=



1 Area of  PQR   20  20  200 2 Area of shaded region = 314=200=114



49. What is the length of arc  ? ADB a. 36.64 b. 42.81 c. 39.82 d. 7.54



Sol:



Correct option is (a)



Circumference = 2 r  2  3.14  10  62.8

150 Length of arc   AB  62.8  26.16 360



Hence length of arc   62.80  26.16  36.64 ADB 50. A circle has area 100 m 2 . If the sector of the circle has area 20 m 2 . What is the measure of the angle of the sector? a. 36o b. 72o c. 150o d. 60o Sol: Correct option is (b)



Area of sector = 20 m 2 Let measure of angle = x Then area of sector =

x 100 m 2 360







x 100  20 360



 x  72o



51. What is the lateral area of a cylinder having radius 6 cm and height 8 cm? a. 48 cm 2 b. 96 cm2 c. 112 cm2 d. 39 cm 2 Sol: Correct option is (b)



Lateral area = 2 rh  2    6  8  96 cm 2 52. What is the total area of the cylinder having radius 7 cm and height 3 cm? a. 45 cm 2 b. 80 cm 2 c. 160 cm2 d. 140 cm2 Sol: Correct option is (d)



Total area = Lateral area + 2 x Area of base Lateral area = 2 rh  2    7  3  42 cm2 Area of base =  r 2    (7)2  49 cm 2 Total area = (42  2  49 )  140cm2 53. What is the volume of cylinder whose radius is 5 cm and height is 6 cm? a. 50 cm3 b. 36 cm3 c. 150 cm3 d. 25 cm3



Sol:



Correct option is (c)



Volume =  r 2 h    6  5  5  150 cm3 54. What is the lateral area of a cone whose radius is 4 cm and height is 3 cm? a. 10 cm2  b. 5 cm 2 c. 17 cm2 d. 20 cm 2 Sol: Correct option is (d)



Slant height l  32  42  25  5 Lateral area =  rl    4  5  20 cm 2 55. What is the total area of a cone whose radius is 3 cm and height is 4 cm? a. 24 cm2 b. 15 cm2 c. 6 cm 2 d. 18 cm2 Sol: Correct option is (a)



Slant height l  32  42  25  5 Total area = Lateral area + 2 x Area of base Lateral area =  rl    3  5  15 Area of base =  r 2    (3)2  9 Total area = 15  9  24 cm 2 56. What is the volume of a cone whose radius is 3 cm and height is 7 cm?



a. 7 cm3 b. 63 cm3 c. 9 cm3 d. 21 cm3 Sol: Correct option is (d)

1 2 1  r h     (3) 2  7  21 cm3 3 3



Volume of cone =



57. The radius of cylinder is twice the height of cylinder. What is the height of cylinder if its volume is 32 cm3 ? a. 6 cm b. 3 cm c. 7 cm d. 2 cm Sol: Correct option is (d)



Volume =  r 2 h Also r = 2h  h 

r 2

3



 r  r  Volume   r 2    2 2





 r3  32  r 3  64  r  4 2

r 4  h   2cm 2 2



Also h 



58. The total area of a cylinder is 8 cm2 . The height of a cylinder is 3 times the radius of the cylinder. What is the height of the cylinder?



a. 4 cm b. 3 cm c. 8 cm d. 1 cm Sol: Correct option is (b)



Total area = 2 rh  2 r 2 Also h=3r

 Total area = 2 r (3r )  2 r 2  6 r 2  2 r 2  8 r 2



 8 r 2  8  r2  1 r  1 Also h=3r  h  3  1  3cm 59. What is the ratio of the volume of the cylinder to that of volume of a cone having same radius and height as that of a cylinder? a. 4:1 b. 1:2 c. 3:1 d. 1:1 Sol: Correct option is (c)



Volume of cylinder =  r 2 h Volume of cone =

1 2 r h 3



Ratio =



 r 2h 3   3 :1 1 2 r h 1 3



60. The radius of the cylinder is increased to 2 times but its height is decreased to half. What is the ratio of the new volume of the cylinder to the old one? a. 2:1 b. 3:1 c. 4:1 d. 1:4 Sol: Correct option is (a)



V1   r 2 h V2   (r ' ) 2 h '



r '  2r and h ' 

 V2   (2r ) 2 

 V2 2   2 :1 V1 1



h 2 h  2 r 2 h  2V1 2



61. Oil is being poured into a cylindrical vessel at the rate of 3 m3 / min ute . In how much minutes the cylinder of radius 2 m and height 2.2 m will get failed? (   3.14 ) a. 86 min b. 74 min c. 93 min d. 113 min Sol: Correct option is (c)



Volume =  r 2 h  3.14  2  2  22  276.32m3 No of minute = 276.32m3  92.1min utes  93min utes 3m3 / min ute



62. Two cylinders of same height and radius 3 cm and 4 cm are containing water inside them. This water is to be poured into a new cylinder of same height but different radius. What should be the radius of new cylinder? a. 5 cm b. 3 cm c. 2 cm d. 7 cm Sol: Correct option is (a)



Let height of all cylinders be (h) Volume of two cylinders combined =  (3)2 h   (4) 2 h   25h  25 h Volume of new cylinder =  r 2 h



  r 2 h  25 h  r 2  25  r  5 cm

63. Water contained in the cone of radius 4 cm and height 3 cm is to be poured into a cylinder of height 1 cm. What should be the radius of the cylinder? a. 2 cm b. 4 cm c. 8 cm d. 1 cm Sol: Correct option is (b)

1 2 r h 3



Volume of cone =

1 =  (4) 2 h  16 3



Volume of cylinder =  r 2 h



  r 2 h  16  r 2 h  16  r 2 (1)  16  r  4cm