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

Answer :

hey here is your answer

Given: AB=30cm AD=24 cm DS= 8 cm   (let here S = G,in the diagram) To find: r solution: 1) AH=AE 2) DG = DH                              ( tangents from an external point are equal) 3) CG  = CF4)BE=BF let O be the centre of the circlejoin O to E and O to F thus we can say that OEBF   is a square.(angle B=900,BE=BF, OE=OF{radii} ) now, 5)DH=DG=8 cmAD=24 AH=24-8 AH= 16 cm from equation 1, AE= 16 cm now, AB= 30 cmBE = 30-16 BE=14 cm since we know that OEBF is a square, all sides will be equaltherefore, OE=OF=BE=BF= 14 CMtherefore radius of circle is 14cm

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Answer:

Given:

AB=30cm

AD=24 cm

DS= 8 cm   (let here S = G,in the diagram)

To find: r

solution:

1) AH=AE

2) DG = DH                              

( tangents from associate degree external purpose square measure equal)

3) CG  = CF

4)BE=BF

let O be the centre of the circle

join O to E and O to F

thus we are able to say that OEBF could be a sq..(angle B=90°,BE=BF, OE=OF )

now,

5)DH = DG=8 cm

AD=24

AH=24-8

AH= 16 cm

from equation one,

AE= 16 cm

now, AB= 30 cm

BE = 30-16

BE=14 cm

since we all know that OEBF could be a sq., all sides are equal

therefore, OE=OF=BE=BF= fourteen CM

therefore radius of circle is 14cm

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@GauravSaxena01

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