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  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle. Given AB = 6 cm, BC = 8 cm and ∠ B = 90° and BD is perpendicular from B on AC. We draw circle through B, C, D Now, ∠ BDC = 90° Since ∠ BDC is angle formed  by chord BC in circle & its 90° ∴ BC is the diameter of circle. Thus, center of circle will be the bisector of line BC Steps of construction Bisect line BC. Let E be mid-point of BC. Thus, E is center of circle. We need to construct tangents from point A to the circle Join line AE and bisect it. Let M be mid point of AE Taking M as centre and AM as radius, draw a circle. Let it intersect the given circle at points B and P. Join AB and AP. Thus, AB and AP are the required tangents Justification: We need to prove that AG and AB are the tangents to the circle. Join EP. APE is an angle in the semi-circle of the blue circle And we know that angle in a semi-circle is a right angle. ∴ ∠EPA = 90° ⇒ OQ ⊥ PQ Since EP is the radius of the circle, AP has to be a tangent of the circle. Also, given ∠ B = 90° Since EB is the radius of the circle, AB is a tangent of the circle.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.