A 20 point QUESTION......!!!! a circle drawn in a sector of a larger circle of radius 24 cm as shown in the figure. The smaller circle is tangent to the bonding radii and arc of the sector. The radius of the smaller circle is ...please give answer step by step.... IN THE TRIANGLE SHOWN IN THE FIGURE ..ONE ANGLE IS 60 ANOTHER IS 30 AND ANOTHER IS 90..
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see figure, ∆AOT is right angled isosceles triangle.
from Pythagoras theorem,
OA = √{R² + R²} = R√2
now, AP = radius of big circle = 24cm
AP = OA + OP
here, OP = radius of small circle,
24cm = R√2 + R
R(1 + √2) = 24cm
R = 24/(√2 + 1)
= 24(√2 - 1)/2
= 12(√2 - 1) cm
hence, radius of small circle is 12(√2 - 1) cm
from Pythagoras theorem,
OA = √{R² + R²} = R√2
now, AP = radius of big circle = 24cm
AP = OA + OP
here, OP = radius of small circle,
24cm = R√2 + R
R(1 + √2) = 24cm
R = 24/(√2 + 1)
= 24(√2 - 1)/2
= 12(√2 - 1) cm
hence, radius of small circle is 12(√2 - 1) cm
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