NCERT Solutions Maths Class 10 Exercise 11.2

NCERT Solutions Maths Class 10 Exercise 11.2

 

Maths Class 10 Exercise 11.2

 

In each of the following, give also the justification of the construction:

1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.

Solution: To construct: A circle of radius 6 cm and two tangents to the circle from a point 10 cm away from its centre.


Steps of construction:

           1.      Draw a circle with centre O and radius 6 cm.

           2.      Draw a line segment OP = 10 cm.

3.      Construct the perpendicular bisector of OP which intersects OP at point O’.

4.      Take O’P as radius and draw another circle to intersect the previous circle at Q and R.

5.      From point P, draw line segments PQ and PR.

6.      PQ and PR are the required tangents.

7.      On measuring, PQ = PR = 8 cm.

Justification: Radius, tangent and distance between centre and external point (from which tangent is drawn) make a right-angled triangle.

Using Pythagoras theorem, we have,

PQ2 = OP2 – OQ2

       = 102 – 62

       = 100 – 36

       = 64

Or, PQ = 8 cm

 

2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Solution: To construct: A tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm.

Steps of construction:

           1.      Construct two concentric circles with centre O and radii 4 cm and 6 cm.

           2.      Draw the radius OP of bigger circle.

3.      Construct a perpendicular bisector of OP which intersects OP at point O’.

4.      Taking O’P as radius, draw another circle to intersect the smaller circle at points Q and R.

5.      From point P, draw tangents PQ and PR as shown in the figure.

6.      On measuring, PQ = PR = 4.8 cm.

Justification: Radius, tangent and distance between centre and external point (from which tangent is drawn) make a right-angled triangle. 

Using Pythagoras theorem, we have

PQ2 = OP2 – OQ2

       = 62 – 42

       = 36 – 16 = 20

Or, PQ = 2√5 cm

            = 4.8 cm

 

3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q.

Solution: To construct: A circle of radius 3 cm and tangents to the circle from two points P and Q, which are situated on one of its extended diameter each at a distance of 7 cm from its centre.

Steps of construction:

           1.      Construct a circle with centre O and radius 3 cm. Extend the diameter to the points A and B on either side.

  1. Construct the perpendicular bisectors of OA and OB.
  2. The perpendicular bisector of OA intersects it at point O’ and the perpendicular bisector of OB intersects it at point O”.
  3. Taking O’A as radius, draw the second circle.
  4. Taking O”B as radius, draw the third circle.
  5. From point A, draw tangents AP and AQ.
  6. From point B, draw tangents BR and BS.

 

4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60o.

Solution: To construct: A pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60o.

Steps of construction:

           1. Construct a circle with radius 5 cm.

  1. Draw a radius OA and a line AP perpendicular to radius OA.
  2. Construct OB making an angle of 120o with OA. (Because angle at centre is double the angle between two tangents).
  3. Join A and B to P, to get two tangents.
  4. Here, APB = 60o

 

5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.

Solution:

Steps of construction:

           1.      Construct a line segment AB = 8 cm.

           2.      Taking A as centre, draw a circle with radius 4 cm.

3.      Taking B as centre, draw another circle with radius 3 cm.

4.      Construct the perpendicular bisector of AB.

5.      Taking mid-point of AB as centre and AB as diameter, draw the third circle.

6.      From point A, draw tangents AR and AS.

7.      From point B, draw tangents BP and BQ.


6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and B = 90o. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangent from A to this circle.

Solution:

Steps of construction:

           1. Construct a line segment AB = 6 cm.

  1. Construct a right angle at point B and draw BC = 8 cm. Join AC.
  2. Draw a perpendicular BD to AC.
  3. Taking BC as diameter, draw a circle which passes through the points B, C and D.
  4. Join A to O and taking AO as diameter, draw second circle.
  5. From point A, draw tangents AB and AP.

 

7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.

Solution: To construct: A circle with the help of a bangle and two tangents to the circle from a point outside the circle.

Steps of construction:

           1.      Draw a circle with the help of a bangle.

           2.      To find the centre of the circle, draw any chord of the circle and draw its perpendicular bisector to find the diameter of the circle. Then mark its centre as O.

           3.      Take any point P outside the circle and draw a line segment OP.

4.      Construct the perpendicular bisector of OP which intersects OP at point O’.

5.      Take O’P as radius and draw another circle to intersect the previous circle at Q and R.

6.      From point P, draw line segments PQ and PR.

7.      PQ and PR are the required tangents.

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