Geometry 61 MCQ part b

MULTIPLE CHOICE QUESTIONS (Part b)

1. The content of 3 cylinders each of radius 2 cm and height 1 cm is to be poured into a cone of height 9 cm. What is the radius of the cone? a. 2 cm b. 4 cm c. 6 cm d. 1 cm Sol: Correct option is (a)



Volume of 3 cylinders = 3    (2) 2  1  12 cm 2 Volume of cone =

1    r 2  9  3 r 2 3



 3 r 2  12  r 2  4  r  2 cm



2. A 90o sector is cut out of a circular metal sheet of radius 4 cm and bent to form the lateral surface of a cone. What is the volume of the cone? (   3.14 )? a. 3.85 cm3 b. 9.34 cm3 c. 6.74 cm3 d. 4.05 cm3 Sol: Correct option is (d) 90 1 Area of sector =    (4) 2     16  4 360 4 Lateral area of cone =  rl



  rl  4  rl  4

Also l  4  r  1 Also h  l 2  r 2  16  1  15 Hence volume =

1    (1) 2  15 3



1   3.14  3.87  4.05cm3 3



3. In  PQR , PQ=3, QR=4 and PR=5. The triangle is rotated about QR in space. What is the volume of solid formed? a. 8 b. 9 c. 16 d. 25 Sol: Correct option is (c)



The shape of a solid formed will be cone. Here r =4 and h =3

1  Volume     (4) 2  3  16 3



4. What is the volume of a sphere whose radius is 3 cm?



a. 12 cm3 b. 36 cm3 c. 24 cm 3 d. 8 cm3 Sol: Correct option s (b)

4 3 r 3



Volume =



4     (3)3  36 cm3 3



5. What is the area of the sphere whose radius is 4 cm? a. 32 cm 2 b. 16 cm 2 c. 4 cm 2 d. 64 cm 2 Sol: Correct option is (d)



Area = 4 r 2  4    (4) 2  64 cm2 6. What is the volume of the sphere whose area is 144cm 2 ? a. 144 cm3 b. 72 cm 3 c. 288 cm 3 d. 36 cm3



Sol: Correct option is (c) Area = 4 r 2

 4 r 2  144  r 2  36  r  6



Volume =



4 3 4  r     (6) 3  288 cm3 3 3



7. What is the area of the sphere whose volume is 36 cm3 ? a. 36 cm 2 b. 72 cm 2 c. 144 cm 2 d. 18 cm 2 Sol: Correct option is (a)

4 3 r 3



Volume =



4   r 3  36 3  r3  36  3  27 4



r 3



Area = 4 r 2  4 r 2  4    (3)2  36 cm 2 8. A circle with radius 5 cm is rotated about its diameter. What is the volume of the solid formed?



a. 134.62 b. 181.9 c. 166.66 d. 125 Sol: Correct option is (c)



The shape of solid formed is that of a sphere Volume =

4 3 4  r     (5)3  166.66 3 3



9. A cylinder of height 6 cm is inscribed in a sphere of radius 5 cm. What is the volume of the cylinder? a. 96 cm3 b. 16 cm3 c. 24 cm 3 d.

8 3 cm 



Sol:



Correct option is (a)



Radius of cylinder =



52  32  4cm



Hence volume =   (4) 2  6  96 cm3 10. What is the area of the shaded portion?



a. 10 b. 15 c. 8 d. 20



Sol:



Correct option is (d)



Area of shaded portion =  (6) 2   (4) 2   (36  16)  20 11. A circle of radius 4 cm is inscribed inside a square PQRS. What is the area of shaded portion?



a. 12.82cm 2



b. 13.76cm 2 c. 18cm 2 d. 19.8cm 2



Sol:



Correct option is (b)



Area of square = (8)2  64cm2 Area of circle =  (4) 2  16 cm2  50.24cm 2 Area of shaded portion = 64  50.24  13.76cm 2



12.



In above figure PQ=5 and PS=3. What is the value of RQ? a. 4 b. 6 c. 8 d. 10 Sol: Correct option is (c)



QS= PQ 2  PS 2  52  32  16  4 Also QS=QR  SR  4  QR  8



13.



 In above figure if m PSR  130o . What is the value of x? a. 50o b. 65o c. 75o d. 90o Sol:

x



Correct option is (b)



1 1  1 PQR  mPSR   130  65o 2 2 2



14.



In above figure PQR  90o and PR=8 cm. What is the value of QP? a. 4 2cm b. 2 2cm c. 4 3cm d. 3cm Sol: Correct option is (a)



Here PS=SR=4 cm Also QPR  QRP  45o

 tan 45o  QS  QS  PS  4cm PS



Hence QP  42  42  4 2cm 15. In below figure SO= 10 2 cm. What is the value of SR?



a. 10



b. 20 c. 25 d. 43



Sol:



Correct option is (b)



Here OSR  45o ST 1 OS 10 2   cos 45o   ST   OS 2 2 2  ST  10  SR  10  10  20 16. What are the values of x and y in below figure?



a. 60, 75



b. 20, 60 c. 60,100 d. 100, 20 Sol:

x  60 o



Correct option is (c)



Also SPR  20 o



x  y  SPR  180o

Also



 y  180  60  20  100o

17. What are the values of y and z in above below figure respectively?



a. 40, 60 b. 60, 40 c. 80, 60 d. 80, 40 Sol: Correct option is (d)



y = 2  40  800



Also z =



1 y  40o 2



18. What is the value of a and b in below figure respectively?



a. 65o , 40o b. 55o , 40 o c. 50o , 65o d. 65o , 75o Sol: a  Correct option is (c) 1 (2  50o )  50o 2



a  50o

1 PRQ   130  65o 2



b  180  50  65 b  65o 19. What are the values of a and b in below figure respectively? PQ=QR



a. 75o , 65o b. 65o , 65o c. 45o , 75o d. 45o , 45o Sol: Correct option is (b)



1 PQR   100  50o 2



As PQ=QR  QPR  QRP

 QPR  a  65o b  65o 1 (180  50)  65 2



20. What is the value of x and y in the below figure respectively?



a. 140o , 75o b. 50o , 75o c. 75o ,140o d. 50o ,140 o Sol: Correct option is (d)



1 x   100  50o 2 PRQ  180  60  50  70o y  2PRQ  2  70  140o



21. What is the value of x in the below figure?



a. 60 o b. 35o c. 30o d. 45 o Sol:

x



Correct option is (c)



1 1 (100  40)  (60)  30o 2 2



22. What is the value of y in the below figure?



a. 60 o b. 35o c. 30o d. 45 o Sol: y Correct option is (a) 1 (130  (360  130  120)) 2 1 1 (130  110)  (20)  10o 2 2



y



23. What is the value of x in below figure?



a. 4 b. 8 c. 6 d. 2 Sol: Correct option is (c)



x  6  18  2 x 18  2 6 6



24. What is the value of x in the below figure?



a. 2



b. 4 c. 9 d. 1 Sol: Correct option is (d)



Here (6+2) 2  (15  x ) x

16  15x  x 2



 x 2  15 x  16  0  ( x  1)( x  16)  0  x 1  x  16



25. What is the value of x in the below figure?



a. 4 b. 6 c. 3 d. 2 Sol: Correct option is (b)



Here (5  4)  4  x 2



 x2  9  4  x 2  36  x  6 26. What is the value of x in the below figure?



8 3 4 b. 3



a.



c.



8 5 3 5



d.



Sol:



Correct option is (a)



Here (8 x  x) x  82

 9 x  x  82 82 8 x 9 3



 x2 



27. What is the value of x in the below figure if PQ  QR ?



a. 4 b. 2 c. 6 d. 8



Sol:



Correct option is (c)



102  x 2  ( x  2)2 x 2  x 2  4  4 x  100 2 x 2  4 x  96  0 x 2  2 x  48  0 ( x  6)( x  8)  0 x6  x  8 28. What is the value of x in the below figure?



a. 1 b. 3 c. 5 d. 2 Sol: Correct option is (d)



x  8  42  x  8  16 x 16 2 8



29. What is the value of x in the below figure?



a. 8 b. 6 c. 5 d. 3 Sol: Correct option is (b)



Here PR 2  RS  RQ  x 2  4  (5  4)  x 2  36 x6 30. What area the values of a, b and c respectively in the below figure? a. 4 2, 2, 4 b. 4, 4 2, 4 2 c. 4, 2, 4 2 d. 2,3, 4 2



Sol: Here



Correct option is (b)



b2  4  4  b  4



Also a 2  8  4  32  a  32  4 2 c 2  8  4  32  c  32  4 2



31. What is the value of x in the below figure?



a. 4 2 b. 8 c. 8 2 d. 8 Sol: Correct option is (c)

2  leg  8  2  8 2



Hypotenuse =



32. What is the value of x in the below figure?



a. 8 b. 2 c. 6 d. 4 Sol: Correct option is (a)



IN figure 30 o  60 o  90 o is triangle Hypotenuse = 2  shorterleg  2  4  8 33. What is the value of x in the below figure?



a. 60 o b. 90o c. 30o d. 45 o Sol: sin x  Correct option is (c) PQ 5 1   PR 10 2



 x  30o 34. PQRS is a parallelogram. What are the values of x and y respectively in the below figure?



a. 30o ,8 b. 45o ,5 c. 45o ,8 d. 60o ,3 Sol: Correct option is (c)



Here SPR  90o

 x  180  90  45  45o



Also y= 5+3 =8 35. What is the value of x and y in the below figure if PQRS is a parallelogram?



a. 4, 2 b. 3, 4 c. 2, 3 d. 1, 4 Sol: Correct option is (b)



Here 2x+y=10 and 2+2y =11

 2 x  4 y  22  3 y  12  y  4  x  3



36. In the figure given below what is the value of x if PQRS is a trapezoid and PT=TQ and SU=UR



a. 6 b. 3 c. 2 d. 9 Sol:

x



Correct option is (d)



1 1 (8  10)  (18)  9 2 2



37. PQRS is a square of side 8 cm. What is the area of the shaded portion?



a. 32cm 2 b. 64cm 2 c. 16cm 2 d. 2cm 2 Sol: Correct option is (a)



Area of square = 8  8  64cm 2 Area of triangle =

1  8  8  32cm 2 2



Area of shaded portion = 32cm 2 101. What is the distance between the points (2, 5) and (5, 9)? a. 2 b. 4 c. 5 d. 11 Sol: Correct option is (c)



d  (5  2) 2  (9  5) 2  32  42  d  25  5

38. What is the center of the circle whose equation is given as ( x  2) 2  ( y  5)2  16 ? a. 1, 5 b. 4, 2 c. 2, 5 d. 5, 4 Sol: Correct option is (c)



In equation ( x  a ) 2  ( y  b)2  r 2 (a, b) is the center Hence (2, 5) is the center. 39. What is the radius of the circle whose equation is ( x  1) 2  ( y  3) 2  25 ? a. 3 b. 2 c. 1 d. 6 Sol: Correct option is (d)



In equation ( x  a ) 2  ( y  b)2  r 2 , r is the radius of the circle Here

r 2  25  r  5



40. What is the equation of the circle having origin as center and 4 as radius?



a. x 2  ( y  1) 2  16 b. x 2  y 2  16 c. x 2  ( y  2) 2  4 d. x 2  y 2  4 Sol: Correct option is (b)



Equation can be written as



( x  0)2  ( y  0) 2  42  x 2  y 2  16

41. What is the distance between the centers of the circles whose equations are ( x  1) 2  ( y  3) 2  25 and ( x  9) 2  ( y  9) 2  144 ? a. 10 b. 25 c. 15 d. 7 Sol: Correct option is (a)



Center of 1st circle is (1, 3) Center of 2nd circle is (9,9) Distance between them =

(9  1) 2  (9  3) 2  64  36  100  10



42. Find the area of the square whose 2 corresponding vertices are (3, 7) and (7, 10)?



a. 16 b. 25 c. 64 d. 100 Sol: Correct option is (b)

(7  3) 2  (10  7) 2  42  32  5



Length of side = Area = (5) 2  25



43. What is the slope of the line which passes through (1, 3) and (5, 5)? a. 4 b.

1 3



c. 1 d.

1 2



Sol:



Correct option is (d)

y2  y1 5  3 2 1    x2  x1 5  1 4 2



Slope =



44. What is the value of tan x in the figure given below?



a.



4 3 3 2



b.



c. 3 d.

1 3



Sol:



Correct option is (a)



tan x o = Slope of line



Slope =



y2  y1 3  ( 1) 4   3 0 3 x2  x1



45. What is the slope of the line which is parallel to the line having slope 2? a. 1 b. 3 c. 4 d. 2 Sol: Correct option is (d)



As lines are parallel hence they have equal slope, hence slope of both lines is 2.



1 46. What is the slope of line l which is perpendicular to the line n having slope ? 2



a. -1 b. -2 c. 2 d.  Sol:

1 3



Correct option is (b)



Let slope of line l  m1 Let slope of line n  m2 Here m1m2  1 Also m2 

1 2



m1 



1  1 2



 m1  2

47. What is the slope of altitude AT of  ABC , where A (2, 4), B (0, 0) and C (1, 8)? a. 

1 4 1 8



b. 



c. 4 d. 2 Sol: Correct option is (b)



Here Slope of BC = Let slope of AT = m

 m  8  1 m 1 8



80 8 1 0



48. What is the mid point of the segment joining points (2, 3) and (8, 11)? a. (5, 7) b. (7, 8) c. (4, 7) d. (3, 11) Sol: Correct option is (a)



 28 x'   5  2   11  3  y'   7  2 



Hence mid point is (5, 7) 49. If the mid point of the segment joining A (4, 7) and B is (5, 8) then what are the coordinates of b?



a. (2, 3)



b. (4, 8) c. (6, 9) d. (4, 9) Sol: Correct option is (c)



Let the coordinates of B be (x, y) Then

5 x4  x  10  4  6 2



Also

8 y7  x  16  7  9 2



Coordinates of B are (6, 9) 50. What are the coordinates of the point at which two lines having equations x-y =3 and 2x+y=24 intersect each other a. (6, 4) b. (9, 6) c. (8, 4) d. (9, 12) Sol: Correct option is (b)



Coordinates of the point can be derived by solving both equations x-y =3 and 2x = y =24 Solving these equations x = 9 and y = 6 Hence coordinates are (9, 6) 51. What are the x and y intercepts of the line 3x + 4y = 12?



a. 3, 2 b. 4, 2 c. 4, 3 d. 2, 6 Sol: Correct option is (c)



For x- intercept y =0



 3x  12  x  4

For y intercept x = 0

 4 y  12  y  3



52. What is the slope of the line 2x + 3y = 8?



a.



2 3 1 2 3 2 2 5



b.



c.



d.



Sol:



Correct option is 9a)



Here equation of line is 2x + 3y = 8



 3 y  8  2x y 2 8 x 3 3 2 3



m



53. What is the equation of the line passing through (2, 4) and having slope 3? a. y = 4x +1 b. y = 3x-2 c. y = 3x-5 d. y = x-9 Sol: Correct option is (b)



Equation of line is



y  y1  m( x  x1 ) y  4  3( x  2)  y  3x  6  4  y  3x  2

54. What is the equation of line which passes through (2, 5) and is parallel to y = 5x-8? a. y = 2(x-3) b. y = x +4 c. y = 2x -1 d. y = 5 (x-1) Sol: Correct option is (d)



Equation of line is y  y1  m( x  x1 ) m=5

y  5  5( x  2)  y  5 x  10  5  y  5x  5  y  5( x  1)



55. What is the equation of line passing through (2, 5) and perpendicular to 1 line y  x  7 ? 2 a. x + 2y + 3 b. 2x – y = 10 c. 2x + y = 9 d. 2x + 3y = 14 Sol: Correct option is (c)



Equation of line is y  y1  m( x  x1 ) y1  5, x1  2 Also m = -2

 y  5  2( x  2)  y  2 x  4  5  2 x  y  9



56. What is the equation of line passing through points (2, 7) and (6, 3)?



a. x + y =9 b. x + 2y = 3 c. x + y = 11 d. 2x – y = 13 Sol: Correct option is (a)

73 4   1 2  6 4



Here m =



y  7  1( x  2)  y  x  2  7  x y 9



57. What is the equation of horizontal line through (5, 4)? a. x = 8 b. y =4 c. y = 3x d. y +x =9 Sol: Correct option is (b)



As the line is horizontal, hence m = 0

 y  4  0( x  5)  y4



58. What is the equation of a vertical line passing through (7, 11)?



a. y =11 b. x =11 c. y = 2x d. x = 7 Sol: Correct option is (d)



As the line is vertical, hence the value of x remains constant  x  7 59. What is the equation of the line passing through (9, 12) and which makes a 45 o angle measured counter clockwise from the positive x-axis? a. y = x + 3 b. 2x + y = 11 c. y = 2x – 3 d. x = y -4 Sol: Correct option is (a)



Here m = tan 45o  1 Hence equation of line is y -12 = 1 (x-9)

 y  x3



60. What is the equation of the perpendicular bisector of the segment joining (2, 4) and (4, 8)?



a. 2x + 3 Y = 16 b. x + 2y = 15 c. x – y = 4 d. x = 3y Sol: Correct option is (b)

84 2 42 1 2



Here slope of segment =



Slope of perpendicular bisector =



Perpendicular bisector passes through (3, 6) Hence equation of perpendicular bisector is

y6  1 ( x  3) 2



 2 y  12   x  3  x  2 y  15



61. What is the equation of the perpendicular bisector of the segment joining (3, 7) and (5, 13)? a. x + 3y = 34 b. 2x + y = 19 c. x = 3y d. y = 24 Sol: Correct option is (a)

13  7 6  3 53 2



Slope of segment =



Slope of perpendicular bisector =



1 3



Also perpendicular bisector passes through (4, 10) Hence equation of perpendicular bisector is

y  10  1 ( x  4) 3



 3 y  30   x  4  x  3 y  34