In the given figure, crescent is formed by two circles which touch
at the point A, O is the centre of the bigger circle. If AB=9 cm
and ED =5 cm, find the area of the shaded region. [Take pi = 3.14]
Answers
Let r be the radius of smaller circle
and R be radius of bigger circle.
Given: BD= 9 cm , EF= 5cm
therefore, CD= (R-9)cm, CA = R cm , EC= R-5 cm
By theorem of intersecting chords,
BC2=CD*CA
(R-5)^2=(R-9)R
R^2+25-10R=R^2 - 9R
25=10R-9R
R=25 cm
now
AB=BD+AD
2R=9+2r
2*25=9+2r
50-9=2r
2r=41
r=41/2
area of shaded region = π(R^2-r^2)
=π[(25)^2-(41/2)^2]
=π*819/4
=3.14*819/4
=642.91cm^2
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Let r be the radius of smaller circle
and R be radius of bigger circle.
Given: BD= 9 cm , EF= 5cm
therefore, CD= (R-9)cm, CA = R cm , EC= R-5 cm
By theorem of intersecting chords,
BC2=CD*CA
(R-5)^2=(R-9)R
R^2+25-10R=R^2 - 9R
25=10R-9R
R=25 cm
now
AB=BD+AD
2R=9+2r
2*25=9+2r
50-9=2r
2r=41
r=41/2
area of shaded region = π(R^2-r^2)
=π[(25)^2-(41/2)^2]
=π*819/4
=3.14*819/4
=642.91cm²
please mark it brainliest