Question

In the figure, quadrilateral Abcd is circumscribing a circle with O and Ad perpendicular AB. if radius of incircle is 10 cm then value of x is ​

Answer 1

Answer:

• Value of x is 21 cm.

Step-by-step explanation:

Construction :-

• Join RO.
• Join OP.
• Join OQ.

Solution :-

• Here, x = AB = AP + PB. We need value of AP and PB. Beacuse we don't have direct method for finding value of x.

We know,

Tangent at any point of circle is perpendicular to the radius through point of contact.

So,

• ∠OSA = ∠OPA = 90°

And, ∠A = 90° [Given]

$\leadsto$ ∠OSA = ∠OPA = ∠A = 90°

If three angles of quadrilateral ASOP are 90°. Then, ∠SOP will also 90°.

Thus, Quadrilateral ASOP is a square.

All sides of Square are equal.

So,

$\leadsto$ AP = 10 cm

ROQC is a quadrilateral and C is an exterior point of circle.

We know,

Lengths of tangents drawn from an external point to a circle are equal.

So,

$\leadsto$ RC = QC = 27 cm.

Similarly,

In quadrilateral POQB, B is an external point.

So,

• PB = QB

$\leadsto$ QB = BC - CQ

$\leadsto$ QB = 38 - 27

$\leadsto$ QB = 11

QB = PB

So,

PB = 11 cm

Now,

$\leadsto$ x = AP + PB

$\leadsto$ x = 10 + 11

$\leadsto$ x = 21

Therefore,

Value of x is 21 cm.

Answer 2

Given :-

Quadrilateral - ABCD

Centre of circle = O

Perpendiculars = AD and AB

To Find :-

x

Solution :-

According to the question

Value of x = AP + PB [since there is no method for finding x

Since, tangents of the given circle are :-

∠OSA

∠OPA

They both will have the value of 90.

Now

We may observe that the 3 angles are 90 degree. So

360 = 90 + 90 + 90 + x

360 = 270 + x

360 -270 = x

90 = x

Angles are 90,90,90,90

Now

All angles are same so its a square

AP = 10 cm

So,

The length of tangents at external point of circle are equal in their size.

RC  = 27 cm

QC = 27 cm.

PB = QB

QB = 38 - 27 = 11

Now

At earlier we discussed PB = QB

x = 10 + 11 = 21 cm

$\sf$