How do i do like how does it go cause i tryed doing it and i got the wrong answer each time
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Option C
The ratio for the volumes of two similar cylinders is 8 : 27
<h3><u>Solution:</u></h3>Let there are two cylinder of heights "h" and "H"
Also radius to be "r" and "R"
Where π = 3.14 , r is the radius and h is the height
Now the ratio of their heights and radii is 2:3 .i.e
<em><u>Ratio for the volumes of two cylinders</u></em>
Cancelling the common terms, we get
Substituting we get,
Hence, the ratio of volume of two cylinders is 8 : 27
Check the picture below, so, that'd be the square inscribed in the circle.
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
![\bf \textit{distance between 2 points}\\ \quad \\ \begin{array}{lllll} &x_1&y_1&x_2&y_2\\ % (a,b) &({{ -2}}\quad ,&{{ 5}})\quad % (c,d) &({{ -8}}\quad ,&{{ -3}}) \end{array}\qquad % distance value d = \sqrt{({{ x_2}}-{{ x_1}})^2 + ({{ y_2}}-{{ y_1}})^2} \\\\\\ \stackrel{diameter}{d}=\sqrt{[-8-(-2)]^2+[-3-5]^2} \\\\\\ d=\sqrt{(-8+2)^2+(-3-5)^2}\implies d=\sqrt{(-6)^2+(-8)^2} \\\\\\ d=\sqrt{36+64}\implies d=\sqrt{100}\implies d=10](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bdistance%20between%202%20points%7D%5C%5C%20%5Cquad%20%5C%5C%0A%5Cbegin%7Barray%7D%7Blllll%7D%0A%26x_1%26y_1%26x_2%26y_2%5C%5C%0A%25%20%20%28a%2Cb%29%0A%26%28%7B%7B%20-2%7D%7D%5Cquad%20%2C%26%7B%7B%205%7D%7D%29%5Cquad%20%0A%25%20%20%28c%2Cd%29%0A%26%28%7B%7B%20-8%7D%7D%5Cquad%20%2C%26%7B%7B%20-3%7D%7D%29%0A%5Cend%7Barray%7D%5Cqquad%20%0A%25%20%20distance%20value%0Ad%20%3D%20%5Csqrt%7B%28%7B%7B%20x_2%7D%7D-%7B%7B%20x_1%7D%7D%29%5E2%20%2B%20%28%7B%7B%20y_2%7D%7D-%7B%7B%20y_1%7D%7D%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Cstackrel%7Bdiameter%7D%7Bd%7D%3D%5Csqrt%7B%5B-8-%28-2%29%5D%5E2%2B%5B-3-5%5D%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B%28-8%2B2%29%5E2%2B%28-3-5%29%5E2%7D%5Cimplies%20d%3D%5Csqrt%7B%28-6%29%5E2%2B%28-8%29%5E2%7D%0A%5C%5C%5C%5C%5C%5C%0Ad%3D%5Csqrt%7B36%2B64%7D%5Cimplies%20d%3D%5Csqrt%7B100%7D%5Cimplies%20d%3D10)
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,
so, now we know the center coordinates and the radius, let's plug them in,
![\bf \textit{equation of a circle}\\\\ (x-{{ h}})^2+(y-{{ k}})^2={{ r}}^2 \qquad \begin{array}{lllll} center\ (&{{ h}},&{{ k}})\qquad radius=&{{ r}}\\ &-5&1&5 \end{array} \\\\\\\ [x-(-5)]^2-[y-1]^2=5^2\implies (x+5)^2-(y-1)^2=25](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bequation%20of%20a%20circle%7D%5C%5C%5C%5C%20%0A%28x-%7B%7B%20h%7D%7D%29%5E2%2B%28y-%7B%7B%20k%7D%7D%29%5E2%3D%7B%7B%20r%7D%7D%5E2%0A%5Cqquad%20%0A%5Cbegin%7Barray%7D%7Blllll%7D%0Acenter%5C%20%28%26%7B%7B%20h%7D%7D%2C%26%7B%7B%20k%7D%7D%29%5Cqquad%20%0Aradius%3D%26%7B%7B%20r%7D%7D%5C%5C%0A%26-5%261%265%0A%5Cend%7Barray%7D%0A%5C%5C%5C%5C%5C%5C%5C%0A%5Bx-%28-5%29%5D%5E2-%5By-1%5D%5E2%3D5%5E2%5Cimplies%20%28x%2B5%29%5E2-%28y-1%29%5E2%3D25)
so... hmm the diagonals for the square are the diameter of the circle, and keep in mind that the radius of a circle is half the diameter, so let's find the diameter.
that means the radius r = 5.
now, what's the center? well, the Midpoint of the diagonals, is really the center of the circle, let's check,
so, now we know the center coordinates and the radius, let's plug them in,