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Prove that the parallelogram circumscribing a circle is a rhombus

Solution:

ABCD is a parallelogram. Therefore, opposite sides are equal.

AB = CD

BC = AD

According to Theorem 10.2: The lengths of tangents drawn from an external point to a circle are equal.

Therefore,

BP = BQ (Tangents from point B)…… (1)

CR = CQ (Tangents from point C)…… (2)

DR = DS (Tangents from point D)…… (3)

AP = AS (Tangents from point A)……. (4)

Adding (1) + (2) + (3) + (4)

BP + CR + DR + AP = BQ + CQ + DS + AS

On re-grouping,

BP + AP + CR + DR = BQ + CQ + DS + AS

AB + CD = BC + AD

Substitute CD = AB and AD = BC since ABCD is a parallelogram, then

AB + AB = BC + BC

2AB = 2BC

AB = BC

∴ AB = BC = CD = DA

This implies that all the four sides are equal.

Therefore, the parallelogram circumscribing a circle is a rhombus.

☛ Check: NCERT Solutions for Class 10 Maths Chapter 10

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Video Solution:

Prove that the parallelogram circumscribing a circle is a rhombus.

NCERT Solutions Class 10 Maths Chapter 10 Exercise 10.2 Question 11

Summary:

It has been proved that the parallelogram ABCD circumscribing a circle with center O is a rhombus.

☛ Related Questions:

  • A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 8 cm and 6 cm respectively (see Fig. 10.14). Find the sides AB and AC.
  • Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the center of the circle.
  • From a point Q, the length of the tangent to a circle is 24 cm and the distance of Q from the centre is 25 cm. The radius of the circle is(A) 7 cm(B) 12 cm(C) 15 cm(D) 24.5 cm
  • In Fig. 10.11, if TP and TQ are the two tangents to a circle with centre O so that ∠ POQ = 110°, then ∠ PTQ is equal to(A) 60°(B) 70°(C) 80°(D) 90°

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