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03 Jul 2022, 07:15
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Circle P is the base of a right circular cone whose apex is D. Points A and B lie on circle P such that line segment AB is a diameter of P. If ∠ADB is a right angle, and AD=3√2, then what is the volume of the cone?
A. 3π
B. 6π
C. 9π
D. 18π
E. 27π
Attachment:
1.png [ 8.67 KiB | Viewed 1541 times ]
03 Jul 2022, 07:44
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Volume of Cone: (Pie*r Square* h)/3
We get the value of r by solving for AB (Pythagoras theorem) which is 6, so r=3.
We get the value of h by solving for OD (Using Area of triangle. Compare area using two different set of sides), so h=3.
Volume= 9 pie
We get the value of r by solving for AB (Pythagoras theorem) which is 6, so r=3.
We get the value of h by solving for OD (Using Area of triangle. Compare area using two different set of sides), so h=3.
Volume= 9 pie
03 Jul 2022, 08:51
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To find the radius:
ADB is a 45-45-90 triangle, so corresponding lengths are 3\(\sqrt{2}\) - 3\(\sqrt{2}\) - 6, so radius = 3
To find the height:
Bisect ADB and you'll get another 45-45-90 triangle, using the same ratios (x - x - x\(\sqrt{2}\)), we get height = 3
Volume = \(\frac{1}{3}\)(\(\pi\)\(r^{2}\)h) = 9\(\pi\)
ADB is a 45-45-90 triangle, so corresponding lengths are 3\(\sqrt{2}\) - 3\(\sqrt{2}\) - 6, so radius = 3
To find the height:
Bisect ADB and you'll get another 45-45-90 triangle, using the same ratios (x - x - x\(\sqrt{2}\)), we get height = 3
Volume = \(\frac{1}{3}\)(\(\pi\)\(r^{2}\)h) = 9\(\pi\)
