Hushar Mulga
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In the given figure, ▢ABCD is a parallelogram. It circumscribes the circle with centre T

Chapter 3 – Circle – Text Book Solution

Problem Set 3 | Q 5 | Page 85

In the given figure, ▢ABCD is a parallelogram. It circumscribes the circle with centre T. Point E, F, G, H are touching points. If AE = 4.5, EB = 5.5, find AD.

In figure 3.85, c ABCD is a parallelogram
solution

ABCD is a parallelogram.

∴ AB = CD       .....(1)       (Opposite sides of parallelogram are equal)

AD = BC          .....(2)       (Opposite sides of parallelogram are equal)

Tangent segments drawn from an external point to a circle are congruent.

AE = AH           .....(3)

DG = DH          .....(4)

BE = BF            .....(5)

CG = CF           .....(6)

Adding (3), (4), (5) and (6), we get

AE + BE + CG + DG = AH + DH + BF + CF

⇒ AB + CD = AD + BC      .....(7)

From (1), (2) and (7), we have

2AB = 2AD

⇒ AB = AD

∴ AD = AB = AE + EB = 4.5 + 5.5 = 10 units

Explanation:-

The given problem involves a parallelogram ABCD, where tangent segments are drawn from an external point to a circle. The aim is to find the value of AD, which is one of the sides of the parallelogram.

Given information:

  • AB = CD (opposite sides of parallelogram are equal) …(1)
  • AD = BC (opposite sides of parallelogram are equal) …(2)
  • AE = AH, DG = DH, BE = BF, CG = CF (tangent segments drawn from an external point to a circle are congruent)
  • Adding (3), (4), (5), and (6), we get: AE + BE + CG + DG = AH + DH + BF + CF AB + CD = AD + BC …(7)

Using equations (1), (2), and (7), we can write: 2AB = 2AD AB = AD

Therefore, we can conclude that AD is equal to AB, which is also equal to AE + EB. We are given that AE = 4.5 and EB = 5.5. Hence, AD = AB = AE + EB = 4.5 + 5.5 = 10 units.

Thus, the value of AD is 10 units.

Chapter 3 – Circle – Text Book Solution

Problem Set 3 | Q 5 | Page 84

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