Full fledge assessment videos for constructions

Understand to construct how to divide a line segment in a given ratio. Con...

This video explains the construction of tangents to a given circle from a p...

Understand to construct how to divide a line segment in a given ratio. Con...

Construction of tangents to a given circle from a point on the circle and f...

Understand to construct how to divide a line segment in a given ratio. Con...

Construction of tangents to a given circle from a point on the circle and ...

This video covers an application on similar triangles

This video covers an application on tangents from external point

This video covers an application based on construction related to section f...

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
Submit

Solution :

The centre of a circle lies on the perpendicular bisector of its chord.
Still have doubt? Watch Video
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

" >
Submit

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. 
Still have doubt? Watch Video
Q3. In the figure below, if AO = 29 cm, XO = 21 cm, then AY =?  
  • 25 cm
  • 24 cm
  • 22 cm
  • 20 cm
Submit

Solution :

Since the tangent is perpendicular to the radius, By Pythagoras' theorem, 292 = 212 + AX2 AX2 = 400 AY = AX = 20 cm 
Still have doubt? Watch Video
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
Submit

Solution :

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

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
Still have doubt? Watch Video
Q6. The number of tangents which can be drawn from a point inside a circle to it is
  • 0
  • 1
  • 2
  • 3
Submit

Solution :

No tangents can be drawn from a point inside a circle.
Still have doubt? Watch Video
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
Submit

Solution :

To draw a perpendicular bisector, we need to measure more than half the length of the line segment.
Still have doubt? Watch Video
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

" >
Submit

Solution :

The scale factor of ∆PQR, which is similar to ∆TUV, such that PR:TV = 6:2 is 3 .
Still have doubt? Watch Video
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
Submit

Solution :

To divide a line segment in m:n, we have to make m+n equal divisions.
Still have doubt? Watch Video
Q10. If AB is divided into 4 equal parts at P, Q and R, then PQ =? 
  • ¼ AR
  • ¼ BR
  • ¼ AB
  • ¼ BA
Submit

Solution :

A-P-Q-R-B If PQ = x, BA = 4x. 
Still have doubt? Watch Video
Q11. In the figure below, if ∠XAO = 30°, then what is the measure of ∠XOY?  
  • 90°
  • 100°
  • 120°
  • 145°
Submit

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° 
Still have doubt? Watch Video
Q12. PQ is divided into 2 equal parts at A, then PA:AQ is 
  • 1:3
  • 1:1
  • 2:1
  • 1:2
Submit

Solution :

P-A-C, PA = x, AQ = x 
Still have doubt? Watch Video
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
Submit

Solution :

P-A-B-C-D-Q, PB = 2x, BQ = 3x
Still have doubt? Watch Video
Q14. The angle drawn in a semicircle from the end points of the diameter is 
  • 30°
  • 45°
  • 60°
  • 90°
Submit

Solution :

The angle drawn in a semicircle from the end points of the diameter is 90°. 
Still have doubt? Watch Video
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
Submit

Solution :

By Pythagoras' theorem,  132 = 122 + radius2 radius2 = 25 radius = 5 cm
Still have doubt? Watch Video
Q16. In the figure below, ∠XAO = 30°. What is the measure of ∠XOA?   
  • 30°
  • 45°
  • 60°
  • 90°
Submit

Solution :

∠XAO = 30° ∠OXA = 90° So, ∠XOA = 60° … (sum of the angles of a triangle) 
Still have doubt? Watch Video
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
Submit

Solution :

To construct a line parallel to the given line, we use the corresponding angles and alternate angles property. 
Still have doubt? Watch Video
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
Submit

Solution :

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

" >
Submit

Solution :

While constructing ∆ABC similar to ∆TUV such that AB:TU = 1:5, the scale factor is 1/5.
Still have doubt? Watch Video
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
Submit

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
×

Okay