Solution: Steps of construction:
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 QNCERT 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. ☛ Related Questions: Math worksheets and > Suggest Corrections 2
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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. |