NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 9: Circles - Exercise 9.4
Question 1

Question. 1

If a hexagon ABCDEF circumscribes a circle, prove that

\(AB + CD + EF = BC + DE + FA\).

Answer:

Relation is proved: \(AB + CD + EF = BC + DE + FA\).

Detailed Answer with Explanation:

Step 1: Recall the property of tangents.

If two tangents are drawn to a circle from the same external point, their lengths are equal.

For example, from point A the tangents are to sides AB and AF. So, we can say:

Let \(AP = AF'\) and \(AQ = AB'\). Actually, just note: AP = AF and AQ = AB.

Step 2: Apply this property at each vertex of the hexagon.

  • From A: tangent lengths are equal → mark them as x.
  • From B: tangent lengths are equal → mark them as y.
  • From C: tangent lengths are equal → mark them as z.
  • From D: tangent lengths are equal → mark them as p.
  • From E: tangent lengths are equal → mark them as q.
  • From F: tangent lengths are equal → mark them as r.

Step 3: Write the sides of the hexagon in terms of these tangent lengths.

  • \(AB = x + y\)
  • \(BC = y + z\)
  • \(CD = z + p\)
  • \(DE = p + q\)
  • \(EF = q + r\)
  • \(FA = r + x\)

Step 4: Now, add the sides according to the relation we need to prove.

Left side: \(AB + CD + EF\)

= (x + y) + (z + p) + (q + r)

= x + y + z + p + q + r

Right side: \(BC + DE + FA\)

= (y + z) + (p + q) + (r + x)

= x + y + z + p + q + r

Step 5: Compare both sides.

They are equal: \(AB + CD + EF = BC + DE + FA\).

Final Result: The relation is proved.

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 9: Circles – Exercise 9.4 | Detailed Answers