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Hey, I need some help with quite challenging maths exercises. The solutions of how to find those would be very needed, because my mind is close to blowing up :D
1. Cilinder height is 2 m and bottom radius is 7 m. Inside the cilinder is a square, which is inclined by the axis and its corners are located on both bottom circles. Find the square side length. (I literally have no clue on where to even start here)
2. Cone height is 10 cm and bottom area is 25 pi square-cm. Find the area of the section which is parallel with the bottom, if the section is 4 cm away from the bottom of the cone. (I found the bottom radius with 25 pi=pi*r² and the r=5cm. With the pythagoras I found the side of the triangle which came as square root of 125(cm). From that point I am stuck)
3. Triangles sides are 12 cm, 17 cm and 25 cm and its spinning around its closest side. Find the area of the new object. (I tried to use Heron formula, but I have no clue what to continue with there)
If you can, please post imgur link here, I would appreciate it sooooo much!
Thanks in advance and have a wonderful day!

Hope you get help with this subject haha... These exercises seem hard af mate.

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Everybody knows if you seriously needed help with Math, you wouldn't come to HLTV.

Hltv is a surprising place! Dont underestimate it

+1

but still, I would rather go to a Math forum to get instant help.

on the first one you know that the square's diagonal equals to the cylinder's height. all you have to do is apply the pythagoras's theorem, since you know the hypothenuse's value and the other two unkown values are equal

do you really think people on hltv can help you

Ask Pythagore to help ya

1. Draw a picture (I'm not quite sure what this problem is talking about without one)
2. Use a similiar triangles, one small triangle from above the frustum, and one large triangle which has the entire height of the cone. (It will help find the radius or the circle)
3. What does this mean?

bump!

bumperino

3: Isn't it a cone?

1. 12 meter