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

Question 1 of 20
Q1. 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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Q2. While constructing ∆PQR similar to ∆ABC such that AB:PQ = 3:5, we need to make an acute angle at A and divide it 
  • 3 times
  • 5 times
  • 8 times
  • 10 times
Construction of Similar Triangles

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Solution :

While constructing a triangle similar to the given triangle, such that their sides are in the ratio m:n, we need to make an acute angle and divide it max (m,n) times. 
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Q3. In the figure below, if AO = 29 cm, XO = 21 cm, then AY =?   
  • 25 cm
  • 24 cm
  • 22 cm
  • 20 cm

Solution :

Since the tangent is perpendicular to the radius, By Pythagoras' theorem, 292 = 212 + AX2 AX2 = 400 AY = AX = 20 cm 
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Q4. The centre of a circle can be found from a point of intersection of 
  • two diameters
  • two radii
  • the perpendicular of two chords
  • the perpendicular bisectors of two chords

Solution :

The centre of a circle can be found from a point of intersection of the perpendicular bisectors of two chords. 
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Q5. In the figure below, if ∠XAO = 30°, then what is the measure of ∠XAY?   
  • 30°
  • 45°
  • 60°
  • 90°

Solution :

The line joining the centre of a circle to the external point bisects the angle between two tangents and two radii.  2∠XAO = ∠XAY
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Q6. 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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Q7. To draw a perpendicular bisector, we need to measure
  • More than the length of the line segment
  • Less than half the length of the line segment
  • More than half the length of the line segment
  • Equal to the length of the line segment

Solution :

To draw a perpendicular bisector, we need to measure more than half the length of the line segment.
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Q8. The scale factor of ∆PQR, which is similar to ∆TUV, such that PR:TV = 6:2 is
  • 8
  • 6
  • 3
  • 2
Construction of Similar Triangles

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Solution :

The scale factor of ∆PQR, which is similar to ∆TUV, such that PR:TV = 6:2 is 3 .
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Q9. 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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Q10. 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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Q11. In the figure below, if ∠XAO = 30°, then what is the measure of ∠XOY?   
  • 90°
  • 100°
  • 120°
  • 145°

Solution :

∠XAO = 30°  ∠OXA = 90°  So, ∠XOA = 60° … (sum of the angles of a triangle) Since the line joining the centre of a circle and the external points bisects the angle between the two tangents and the two radii, we have ∠XOY = 2∠XOA = 120° 
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Q12. 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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Q13. 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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Q14. 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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Q15. 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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Q16. In the figure below, ∠XAO = 30°. What is the measure of ∠XOA?    
  • 30°
  • 45°
  • 60°
  • 90°

Solution :

∠XAO = 30° ∠OXA = 90° So, ∠XOA = 60° … (sum of the angles of a triangle) 
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Q17. To construct a line parallel to the given line, we use 
  • The alternate angles property
  • The corresponding angles property
  • Both A and B
  • None of these

Solution :

To construct a line parallel to the given line, we use the corresponding angles and alternate angles property. 
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Q18. If from a point we can draw no tangent to the given circle, then the point lies
  • inside the circle
  • outside the circle
  • on the circle
  • none of these

Solution :

We can draw no tangent to a circle from a point inside it.
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Q19. 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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Q20. 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)).
Still have doubt? Watch Video