Full fledge assessment (Please watch the above videos before attempting this assessment)

Question 1 of 20
Q1.    If the angle between the tangents is 50o (as shown in the figure), then the angles between the two radii will be 
  • 120°
  • 125°
  • 130°
  • 135°

Solution :

In ∆AXO,  ∠XAO = 25° then ∠XOA = 180° − 90° − 25° = 65° ∠XOY = 130° 
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Q2. A tangent and radius of the circle make an angle of measure 
  • 45°
  • 90°
  • 180°
  • 270°

Solution :

A tangent and radius of the circle make an angle of measure 90°.
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Q3. If the length of the tangent from the external point P is 12 cm, and the distance of P from the centre is 13 cm, then the radius of the circle is 
  • 5 cm
  • 6 cm
  • 7 cm
  • 8 cm

Solution :

By Pythagoras' theorem,  132 = 122 + radius2 radius2 = 25 radius = 5 cm
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Q4. While constructing ∆XYZ similar to ∆STU such that SU:XZ = 6:7, the smaller triangle is 
  • ∆XYZ
  • ∆STU
  • Both are equal
  • None of these
Construction of Similar Triangles

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Solution :

While constructing ∆XYZ similar to ∆STU such that SU:XZ = 6:7, the smaller triangle is DSTU.   
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Q5. If PQ is divided into 3 equal parts at A and B, then PB:BQ is 
  • 1:3
  • 2:3
  • 2:1
  • 1:2

Solution :

P-A-B-Q If PA = x, PB = 2x, BQ = x.
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Q6.    In the figure above, if the angle between the tangents is 70o, then the angles between the two radii will be
  • 110o
  • 125o
  • 130o
  • 135o

Solution :

In ∆AXO, ∠XAO = 35°  then ∠XOA = 180°− 90° − 35° = 55°  ∠XOY = 110° … The line joining the centre of a circle and an external point bisects the angle between the tangents and the radius.
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Q7. While constructing ∆ABC similar to ∆TUV such that AB:TU = 1:5, the scale factor is 
  • 1/5
  • 6
  • 5
  • 1/6
Construction of Similar Triangles

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Solution :

While constructing ∆ABC similar to ∆TUV such that AB:TU = 1:5, the scale factor is 1/5.
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Q8. If PQ is divided into 5 equal parts at A, B, C and D, then PB:BQ is 
  • 2:3
  • 3:2
  • 2:5
  • 5:2

Solution :

P-A-B-C-D-Q, PB = 2x, BQ = 3x
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Q9. The number of tangents which can be drawn from a point inside a circle to it is
  • 0
  • 1
  • 2
  • 3

Solution :

No tangents can be drawn from a point inside a circle.
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Q10. While constructing a similar triangle such that its corresponding sides are in the ratio 5:2, we need to construct an acute angle at any one of the bases of the given triangle and make 
  • 7 equal divisions
  • 6 equal divisions
  • 5 equal divisions
  • 2 equal divisions

Solution :

While constructing a similar triangle, we need to construct an acute angle at any one of bases of the given triangle and make 5 equal divisions (max (5 and 2)).
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Q11. The angle drawn in a semicircle from the end points of the diameter is 
  • 30°
  • 45°
  • 60°
  • 90°

Solution :

The angle drawn in a semicircle from the end points of the diameter is 90°. 
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Q12. If AB is divided into 3 equal parts at P and R, then PB =?
  • 1/4 AR
  • 1/2 BA
  • 1/4 AB
  • 2/3 AB

Solution :

A-P-R-B If PB = 2x, then BA = 3x.
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Q13. PQ is divided into 2 equal parts at A, then PA:AQ is 
  • 1:3
  • 1:1
  • 2:1
  • 1:2

Solution :

P-A-C, PA = x, AQ = x 
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Q14. A tangent cuts the circle at 
  • one point
  • two points
  • no point
  • three points

Solution :

A tangent cuts the circle at one point. 
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Q15. If two lines meet a transversal and they make equal corresponding angles, then the two lines are 
  • Perpendicular
  • Intersecting
  • Parallel
  • None of these

Solution :

If two lines meet a transversal and they make equal corresponding angles, then the two lines are parallel. 
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Q16. To divide a line segment in 5:3, we have to make
  • 5 equal divisions
  • 3 equal divisions
  • 2 equal divisions
  • 8 equal divisions

Solution :

To divide a line segment in m:n, we have to make m+n equal divisions.
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Q17. The number of tangents which can be drawn through a point on the circle is
  • one
  • two
  • three
  • many

Solution :

Only one tangent can be drawn through a point on the circle. 
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Q18. If AB is divided into 4 equal parts at P, Q and R, then PQ =? 
  • ¼ AR
  • ¼ BR
  • ¼ AB
  • ¼ BA

Solution :

A-P-Q-R-B If PQ = x, BA = 4x. 
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Q19. To divide a line segment in 4 equal parts, we need to draw 
  • perpendicular bisector two times
  • perpendicular bisector three times
  • perpendicular bisector four times
  • none of these

Solution :

To divide a line segment in 4 equal parts, we need to draw a perpendicular bisector three times.
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Q20. The centre of a circle lies on the
  • Perpendicular
  • perpendicular bisector
  • perpendicular of its chord
  • perpendicular bisector of its chord

Solution :

The centre of a circle lies on the perpendicular bisector of its chord.
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