Solution:
Steps of construction:
- Draw a circle with O as centre and radius as 3 cm.
- Draw a diameter of it and extend both the sides and mark the points as P, Q such that OP = OQ = 7 cm.
- Draw the perpendicular bisectors of OP and OQ to intersect PQ at M and N respectively.
- With M as centre and OM as radius draw a circle to cut the given circle at A and C. With N as the centre and ON as radius draw a circle to cut the given circle at B and D.
- Join PA, PC, QB, QD.
PA, PC and QB, QD are the required tangents from P and Q respectively.
Proof:
∠PAO = ∠QBO = 90° (Angle in a semi-circle)
∴ PA ⊥ AO, QB ⊥ BO
In right angle triangle PAO,
OP = OQ = 7cm (By construction)
OA = OB = 3cm (radius of the given circle)
PA² = (OP)² - (OA)²
= (7)² - (3)²
= 49 - 9
= 40
PA = √40
= 6.3 (approx.)
Also, in right triangle QBO,
QB² = (OQ)² - (OB)²
= (7)² - (3)²
= 49 - 9
= 40
QB = √40
= 6.3 (approx)
☛ Check: Class 10 Maths NCERT Solutions Chapter 11
Video Solution:
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
NCERT Solutions Class 10 Maths Chapter 11 Exercise 11.2 Question 3
Summary:
A circle of radius 3 cm is drawn. Two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre has been taken and the tangents PA and QB each of length 6.3 cm has been constructed.
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Draw a circle of radius 3 cm. Take two points P and Q on one of its diameters extended on both sides, each at a distance of 7 cm on opposite sides of its centre. Draw tangents to the circle from these two points P and Q.
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Last updated at July 14, 2020 by
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Ex 11.2, 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. Steps of construction Draw a circle of radius 3 cm Draw diameter of circle, and extend it and mark points P and Q, 7 cm from the center Let’s first draw tangent from point P 3. Make perpendicular bisector of PO Let M be the midpoint of PO. 4 Taking M as centre and MO as radius, draw a circle. 5. Let it intersect the given circle at points A and B. 6. Join PA and PB. Now, we draw tangent from point Q Similarly we draw tangent from point Q ∴ QC and QD are the tangents from point Q Justification We need to prove that PA, PB, QC, QD are the tangents to the circle. Join OA, OB, OC and OD Now, ∠PAO is an angle in the semi-circle of the blue circle And we know that, Angle in a semi-circle is a right angle. ∴ ∠PAO = 90° ⇒ OA ⊥ PA Since OA is the radius of the circle, PA has to be a tangent of the circle. Similarly, we can prove PB, QC, QD are tangents of the circle.