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  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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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. The tangent can be constructed on the given circle as follows. Taking any point O on the given plane as centre, draw a circle of 3 cm radius. Extend diameter of circle to a distance of 7 cm from centre on both sides. Let these points be P and Q. Where OP = OQ = 7 cm Bisect OR and OS. Let M and N be the mid-points of OP and OQ respectively. .Taking M as centre and MO as  radius, draw a circle. Do the same for point N. Let it intersect at points A, B and C, D respectively Join PA, PB, PC and PD Thus, PA, PB, QC, QD are the required tangents Justification: We need to prove that PA, PB, QC, QD are the tangents to the circle. Join OA and OB. ∠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, PB is a tangent of the circle. Similarly, we can prove QC and QD are tangents

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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