Question

in a figure quadrilateral ABCD is circumscribed with center O and AD is perpendicular to AB .Radius of circle is 10 cm find the value of x​

Answer 1

Given

• O centre of circle .
• ABCD = Quadrilateral.
• radius of circle = 10cm.
• ∠BAD = 90°.
• AB, BC, CD and AD touch the inscribed circle at P, Q, R and S respectively.
• CR = 27cm.
• BC = 38cm.

To Find :-

• AB = x = ?

Solution :-

given that,

→ ∠BAD = 90°.

So,

→ OSAP is a square with side equal to radius = 10cm.

Than,

→ AP = 10cm.

Now,

→ CR = CQ. (Tangent segments to a circle from the same external point are congruent.)

So,

→ CQ = 27cm.

Than,

→ BQ = CB - CQ

→ BQ = 38 - 27

→ BQ = 11 cm.

again,

→ PB = BQ (Tangent segments to a circle from the same external point are congruent.)

Therefore,

→ PB = 11cm.

Hence,

→ AB = AP+BP =

→ x = 10 + 11

→ x = 21 cm.(Ans.)

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