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:-
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:-
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:-
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:-
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:-
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:-
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:-
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.