NCERT Solutions For Class 10 Math Chapter – 11 Exercise – 11.2

NCERT Solutions For Class 10 Math Chapter – 11 Exercise  – 11.2

 

Q1. Draw a circle of radius 6 cm . From a point 10 cm away from its center  , construct the pair of tangents to the circle and measure their lengths.

Solution:-

  

Ncert solutions class 10 Chapter 11-9

 

 

 

 

 

 

 

 

Steps of construction:

 

  • Take a point O as center 6 cm radius Draw a circle.
  • Take a point P such that OP = 10 cm
  • With O and P as centers and radius more than  half of OP draw arcs above and below OP to intersect at X and Y.
  • Join XY to intersect OP at M.
  • With M as center and OM as radius draw a circle to intersect the given circle at Q and R.
  • Join PQ and PR.

 

In right  ∆ PQO,

PQ2 = OP2 – OQ2

= (10)2 – (6)2

= 100 – 36

=  64

PQ = √64

=  8 cm.

 

Q2. 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:-

 

Ncert solutions class 10 Chapter 11-10

 

 

 

 

 

 

 

 

Steps of construction:

  • Take ‘O’ as center and radius 4 cm and 6 cm draw two circles.
  • Take a point ‘P’ on the bigger circle and join OP.
  • With ‘O’ and ‘P’ as center and radius more than half of OP draw arcs above and below OP to intersect at X and Y.
  • Join XY to  intersect OP at M.
  • With M as center and OM as radius draw a circle to cut the smaller circle at Q and R.
  • Join PQ and PR.

 

In the right ∆PQO,

PQ2 = OP2 – OQ2

=  62 –  42

=  36  – 16

PQ   = 20

 

PQ = √20

= 4.5 (Approx)

 

 

Q3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameters each at a distance of 7 cm from its center. Draw tangents to the circle from these two points P and Q.

Solution:-

Ncert solutions class 10 Chapter 11-11

 

 

 

 

 

 

 

 

 

Steps of construction:

 

  • Draw a circle with O as center  and radius is 3 cm.
  • Draw a diameter of it extend both the sides and take point P. Q  on the diameter such that OP = OQ = 7 cm.
  • Draw the perpendicular bisectors  of OP and OQ to intersect OP and OQ at M and N respectively.
  • With M as center and OM as radius draw a circle to cut the given circle at A and C. With N as center and ON as radius draw a circle to cut the given circle at B and D.
  • Join PA , PC QB , QD.

 

In right ∆PAO ∆QBO

PA2 = OP2– OA2

= (7)2 – (3)2

= 49  – 9

= 40

PA  = 6.3

 

 

QB2 = (OQ)2 – (OB)2

= (7)2 – (3)2

= 40

QB = 6.3

 

 

Q4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 600

Solution:-

 

Ncert solutions class 10 Chapter 11-12

 

 

 

 

 

 

 

 

 

Steps of construction:

 

  • With O as center and 5 cm as radius draw a circle.
  • Take a point A on circumference of the circle and join OA.
  • Draw AX perpendicular to OA.
  • Construct ∟AOB = 1200where  B lies on the circumference.
  • Draw BY perpendicular to OB.
  • Both AX and BY intersect at P.
  • PA and PB are the required tangents inclined at 600.

 

In quadrilateral OAPB,

 

∟APB = 3600 – [∟OAP + ∟OBP + ∟AOB]

= 3600  – [900 + 900 + 1200]

= 3600– 3000

=  600

Here PA and PB are the required tangents inclined at 600.

 

Q5. Draw a line segment AB of length 8 cm. Taking A as center , draw a circle of radius 4 cm and taking B as center, draw another circle of radius 3 cm Construct tangent to each from the center of the other circle.

Solution:-

 

Ncert solutions class 10 Chapter 11-13

 

 

 

 

 

 

 

 

 

Steps of construction:

 

  • Draw AB = 8 cm With A and B as centers 4 cm and 3 cm as radius respectively draw two circles.
  • Draw the perpendicular bisector of AB , intersecting AB at O.
  • With O as center and OA as radius draw a circle which intersects the two circles at P , Q , R and S.
  • Join BP , BQ AR and AS.
  • BP and BQ are the tangents from B to the circle with center A. AR and AS are the tangents from A to the circle with center B.

 

∟APB = ∟AQB = 900

AP Bisect PB and AQ Bisect QB.

Therefore , BP and BQ are the tangents to the circle with center A.

Similarly , AR and AS are the tangents to the circle with center B.

 

Q6. Let ABC be a right triangle in which AB = 6 cm , BC = 8 cm and ∟B = 900 BD is the perpendicular to AC.  The circle through B , C and D is drawn. Construct the tangents from A to this circle.

Solution:-

 

Ncert solutions class 10 Chapter 11-14

 

 

 

 

 

 

 

 

 

 

 

 

Steps of construction:

 

  • Draw BC = 8 cm Draw the perpendicular at B and cut BA = 6 cm on it Join AC right ∆ ABC is obtained.
  • Draw BD perpendicular to AC.
  • Since ∟BDC = 900and the circle has top pass through B , C and D BC must be a diameter of this circle So , take O as the midpoint of BC and with O as center and OB as radius draw a circle which will pass through B , C  and D.
  • To draw tangents from A to the circle with center O.
  • Join OA , and draw its perpendicular bisector to intersects OA at E.
  • With E as center and EA as radius draw a circle which intersects the previous circle at B and F.
  • Join AF.

 

∟ABO = ∟AFO = 900  (Angle in a semi – circle)

AB Bisect OB and AF Bisect OF.

Hence AB and AF are the tangents from A to the circle with center O.

 

 

Q7. 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:-

 

Ncert solutions class 10 Chapter 11-15

 

 

 

 

 

 

 

 

Steps of construction:

  • Draw any circle using a bangle.

To find its center.

 

  • Draw any two chords of the circle say AB and CD.
  • Draw the perpendicular bisectors of AB and CD to intersect at O.

 

  • Take any point P outside the circle and draw the perpendicular bisector of OP which meets at OP at O’.
  • With O’ as centers and OO’ as radius draw a circle which cuts the given circle at Q and R.
  • Join PQ and PR.

 

∟OQP = ∟ORP = 900

OQ Bisect QP and OR Bisect RP.

 

Hence , PQ and PR are the tangents to the given circle.