Second moment of area about an axis along any diameter in the plane of the cross section (i.e. x-x, y-y) is each equal to (1/4)pi r^4.
The second moment of area about the zz-axis (along the axis of the cylinder) is the sum of the two, namely (1/2)pi r^4.
The derivation is by integration of the following:
int int y^2 dA
over the area of the cross section, and can be found in any book on mechanics of materials.
8 is the angle DHL with H the vertex
Answer:
1. QS and RT= 15
2. Im not positive but I think it's TQ=18
Step-by-step explanation:
For question 1, you would do 6x+3=8x-1 and solve to get the result of x=2
Then you'd plug in 6 and get QS and RT=15
For the second one I am unsure because I wasn't positive how to solve but it seems like the side RQ might be 1/2 of TQ. Sorry if thats wrong I can try and help more if you need
Answer:
The graph in the bottom right corner
