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Ex 11.2, 2

Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also verify the measurement by actual calculation.

Tangents on the given circle can be drawn as follows.

  1. Draw a circle of 4 cm radius with  centre as O on the given plane.
  2. Draw a circle of 6 cm radius taking O as its centre.
  3. Locate a point P on this circle and join OP.
  4. Bisect OP. Let M be the mid-point of OP
  5. Take M as its centre and MO as its radius, draw a circle. Let it intersect the given circle at the points Q and R.
  6. Join PQ and PR.

PQ and PR are the required tangents.

By measuring,

Lengths of PQ and PR is 4.47 m

Finding lengths of PQ and PR

Join OQ and OR

Since tangent is perpendicular to radius

∠ PQO = 90° and ∠ PRO = 90°

Thus, Δ PQO is a right angled triangle,

Also,

PO = radius of bigger circle = 6 cm

and OQ = radius of smaller circle = 4 cm

By Pythagoras theorem

PO2 = PQ2 + OQ2

62 = PQ2 + 42

36 = PQ2 + 16

PQ2 = 36 – 16

PQ2 = 20

  PQ = √20 = √(5 ×4) = √4 × √5 = 2√5

  PQ = 2 × 2.236

  PQ = 4.47 cm

Similarly, PR  = 4.47 cm

Justification:

We need to prove that PQ and PR are the tangents to the circle.

Join OQ and OR.

∠PQO is an angle in the semi-circle
of the blue circle
And we know that angle in a
semi-circle is a right angle.

∴ ∠PQO = 90°

⇒ OQ ⊥ PQ

Since OQ is the radius of the circle,

PQ has to be a tangent of the circle.

Similarly, PR 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 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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