In Fig 2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB=29 cm, AD=23 cm, $∠ 90^{\circ}$ and DS= 5 cm, then the radius of the circle (in cm.) is:

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Solution

Given (i) $∠ B = 90^{\circ}$
(ii) AD=23 cm
(iii) AB = 29 cm
(iv) DS= 5 cm
Here
$O Q ⊥ A B$ (Radius is perpendicular to the tangent)
$O P ⊥ B C$
OQ=OP (radius of a circle)
$\Rightarrow$ BQ=BP (tangent from point B)
Therefore OPBQ is a square.
$\Rightarrow O P = r$
Now RD=DS (tangent from point B)
Therefore OPBQ is a square.
$\Rightarrow O P = r$
Now RD=DS (tangent from point D)
$\Rightarrow R D = 5 c m = D S$ (given)
Therefore AR=AD-RD
=23-5 = 18 cm ........(1)
Also AR=AQ (tangent from point A)
$\Rightarrow A Q = 18 c m$ [given equation (1)]
Now, AB=AQ+BQ
$\Rightarrow 29 = 18 + r$
$\Rightarrow r = 11 c m$
So, the radius of the circle is 11 cm.

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In Fig 2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB=29 cm, AD=23 cm, ∠ 90∘ and DS= 5 cm, then the radius of the circle (in cm.) is:

In Fig 2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB=29 cm, AD=23 cm, $∠ 90^{\circ}$ and DS= 5 cm, then the radius of the circle (in cm.) is: