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Ex 11.2, 6

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

  1. 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
  2. Join line AE and bisect it. Let M be mid point of AE
  3. Taking M as centre and AM as radius, draw a circle.
  4. Let it intersect the given circle at points B and P.
  5. 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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  1. Chapter 11 Class 10 Constructions
  2. Concept wise
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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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