NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 10: Construction - Exercise 10.4
Question 3

Question. 3

Draw two concentric circles of radii \(3\,\text{cm}\) and \(5\,\text{cm}\). From a point \(T\) on the outer circle, construct the pair of tangents to the inner circle. Measure the length of a tangent and verify it by calculation.

Answer:

Final answer: Each tangent has length \(4\,\text{cm}.\)

Detailed Answer with Explanation:

Step-by-step Construction

  1. Take a point \(O\) on your paper. This will be the common centre.
  2. Using a compass, draw a circle with radius \(3\,\text{cm}\). This is the inner circle.
  3. From the same centre \(O\), draw another circle with radius \(5\,\text{cm}\). This is the outer circle. Now you have two concentric circles.
  4. Mark a point \(T\) anywhere on the outer circle (radius \(5\,\text{cm}\)).
  5. Join the line \(OT\).
  6. Now, to construct the tangents from \(T\) to the inner circle:
    • Draw the circle with diameter \(OT\).
    • This new circle will intersect the inner circle (radius \(3\,\text{cm}\)) at two points. Let those points be \(A\) and \(B\).
    • Join \(TA\) and \(TB\). These are the required tangents.

Step-by-step Verification

We know that a radius drawn to the point of contact is perpendicular to the tangent.

So, in triangle \(OTA\):

  • \(OA = 3\,\text{cm}\) (radius of inner circle)
  • \(OT = 5\,\text{cm}\) (radius of outer circle)
  • \(\angle OAT = 90^\circ\)

By Pythagoras theorem:

\[ TA^2 = OT^2 - OA^2 = (5)^2 - (3)^2 = 25 - 9 = 16 \] \[ TA = \sqrt{16} = 4\,\text{cm} \]

Hence, the length of each tangent is \(4\,\text{cm}\).

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