## Full fledge assessment videos for constructions

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.
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

" >

### 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

### 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
• 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.
Still have doubt? Watch Video
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
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

### 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

### 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

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### 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

### 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

### 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°

### 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

### 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

### 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°

### 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

### Solution :

Still have doubt? Watch Video
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)
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

### 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

### 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

" >

### 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

### 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
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