A circle is inscribed in a quadrilateral $A B C D$ $∠ B = 90^{o} .$ If $A D = 24 c m, A B = 30 c m$ and $D S = 8 c m,$ find the radius of the circle.

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Solution

REF.image
given $\to ∠ B = 90^{\circ}$
Hence PB = r
$\Rightarrow A P = A B - P B$
$A P = 30 - r$
AD = 24
DS = 8
AS = AD - DS
AS = 16 cm
we know that tangents drawn from same
external pt are equal in lengths.
AP = AS
$\Rightarrow 30 - r = 16$
$\Rightarrow r = 14 c m$

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A circle is inscribed in a quadrilateral ABCD ∠B=90o. If AD=24 cm,AB=30 cm and DS=8 cm, find the radius of the circle.

REF.imagegiven →∠B=90∘Hence PB = r⇒AP=AB−PBAP=30−rAD = 24DS = 8AS = AD - DS AS = 16 cmwe know that tangents drawn from sameexternal pt are equal in lengths.AP = AS⇒30−r=16⇒r=14cm

A circle is inscribed in a quadrilateral $A B C D$ $∠ B = 90^{o} .$ If $A D = 24 c m, A B = 30 c m$ and $D S = 8 c m,$ find the radius of the circle.

REF.image
given $\to ∠ B = 90^{\circ}$
Hence PB = r
$\Rightarrow A P = A B - P B$
$A P = 30 - r$
AD = 24
DS = 8
AS = AD - DS
AS = 16 cm
we know that tangents drawn from same
external pt are equal in lengths.
AP = AS
$\Rightarrow 30 - r = 16$
$\Rightarrow r = 14 c m$