Bunuel wrote:
Joy111 wrote:
A cylinder has a base with a circumference of 20pi meters and an equilateral triangle inscribed on the interior side of the base. A marker is dropped into the tank with an equal probability of landing on any point on the base. If the probability of the marker landing inside the triangle is (Sqrt 3)/4 , what is the length of a side of the triangle?
A. 3(sqrt 2pi)
B. 3(sqrt 3pi)
C. 10 pi
D. 10(sqrt 3pi)
E. 20 pi
Since the probability of the marker landing on the portion of the base
inside the triangle is \(\frac{\sqrt{3}}{4}\) then the portion of the base (circle) inside the triangle must be \(\frac{\sqrt{3}}{4}\) of the area of the base.
Next: \(circumference=20\pi=2\pi{r}\) --> \(r=10\) --> \(area_{base}=\pi{r^2}=100\pi\);
The area of the equilateral triangle is \(\frac{\sqrt{3}}{4}\) of the base: \(area_{equilateral}=\frac{\sqrt{3}}{4}*100\pi\) --> also the ares of the equilateral triangle is \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}\), where \(a\) is the length of a side --> \(area_{equilateral}=a^2*\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{4}*100\pi\) --> \(a=10{\sqrt\pi}\).
Answer: C.
Similar question to practice:
a-cylindrical-tank-has-a-base-with-a-circumference-of-105453.htmlHope it helps.
Hello All,
See my solution below and please tell me where I am going wrong:
Radius of base = 10 (as derived by you)
Now, if we draw the equilateral triangle inscribed in a circle (as shown in my attachment which is not to scale), then:
OA = radius = 10
and O is the centroid
Centroid divides a median in the ratio 2:1. Hence, OD = OA / 2 = 5
Now, AD is the height of the triangle and in equilateral triangle,
[math]height = a * sqrt (3) /2
where a = side of triangle
Hence, 15 = a * sqrt (3) /2
Hence a = 10 * sqrt (3)
Where am I going wrong?
Attachments
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