Question

In the figure, quadrilateral ABCD is circumscribing a circle with centre O

and AD⊥AB. If radius of incircle is 10cm, then the value of x is​

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.

Then,

→ AP = 10cm.

Now,

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

So,

→ CQ = 27cm.

Then,

→ 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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Answer 2

Answer:

X=21

Step-by-step explanation:

Given

Quadrilateral ABCD

CR=27cm

CB=38cm

Circumscribing a circle of center O

AD Perpendicular to AB

Radius of the circle= 10cm

To Find

x( AB)

Construction

Draw OP, at the point of contact of the tangent AB

Solution

CR=CQ ( Tangents from an External Point)

CB= CQ+BQ

38=27+BQ

38-27=BQ

11cm=BQ

BQ=BP (Tangents from an External Point)

BP=11cm

OS=10cm ( Radius of the circle)

AS=AP (Tangents from an External Point)

OP= 10cm (Radius of the circle)

Angle OSA=90

Angle OPA =90

Since two sides are equal and two opposite angles are 90 degree.

We can conclude that OSAP is Square.

Since, OSAP is a Square, then AP= 10cm

X=AB=AP+BP=11+10=21cm

X=21cm

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