NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.1
Question 5

Question.  5

5. At one end \(A\) of a diameter \(AB\) of a circle of radius 5 cm, a tangent \(XAY\) is drawn. The length of the chord \(CD\) parallel to \(XY\) and at a distance 8 cm from \(A\) is

(A)

4 cm

(B)

5 cm

(C)

6 cm

(D)

8 cm

Detailed Answer with Explanation:

Step 1: Draw a circle with centre \(O\) and radius \(5\,\text{cm}\). Mark diameter \(AB\). At point \(A\), draw the tangent \(XAY\).

Step 2: We are told that a chord \(CD\) is drawn parallel to the tangent \(XY\). The perpendicular distance of this chord from point \(A\) is \(8\,\text{cm}\).

Step 3: To calculate chord length, we need the perpendicular distance of the chord from the centre \(O\). The total distance from \(O\) to point \(A\) is the radius = \(5\,\text{cm}\). The chord is \(8\,\text{cm}\) away from \(A\). Therefore, distance from \(O\) to the chord = \(|OA - 8| = |5 - 8| = 3\,\text{cm}|.

Step 4: Formula for chord length: \[ \text{Chord length} = 2 \sqrt{r^2 - d^2} \] where \(r\) is radius and \(d\) is distance from centre to the chord.

Step 5: Substitute values: \(r = 5\,\text{cm}, \; d = 3\,\text{cm}\) \[ \text{Chord length} = 2 \sqrt{5^2 - 3^2} = 2 \sqrt{25 - 9} = 2 \sqrt{16} = 2 \times 4 = 8\,\text{cm}. \]

Final Answer: The chord length = 8 cm. (Option D)

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 9: Circles – Exercise 9.1 | Detailed Answers