The radii of two circles are 8 cm and 6 cm respectively. Find the radius of the circle having area equal to the sum of the area of the two circles
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Take the radius of new circle as R.
Given condition is π×R^2 = π×8^2 + π×6^2.
Thus, R = √(8^2+6^2) = 10 cm
Given condition is π×R^2 = π×8^2 + π×6^2.
Thus, R = √(8^2+6^2) = 10 cm
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area of 1st circle = 3.14 × r^2
= 3.14 × 8 ×8
=3.14×64
=200.96 cm^2
area of 2nd circle = 3.14× r^2
=3.14×6×6
=3.14×36
=113.04
sum = 200.96+113.04
=314 cm^2
now area = 3.14 × r^2
314=3.14×2×r
314/2=3.14×r
157/3.14=r
50 cm = r
= 3.14 × 8 ×8
=3.14×64
=200.96 cm^2
area of 2nd circle = 3.14× r^2
=3.14×6×6
=3.14×36
=113.04
sum = 200.96+113.04
=314 cm^2
now area = 3.14 × r^2
314=3.14×2×r
314/2=3.14×r
157/3.14=r
50 cm = r
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