Have Questions

[cwilli9000]

May I have help on some or all the questions please?
Thanks
06A-50, 06A58, 06A-59, 06A-60
06E-40, 06E-49, 06E-50, 06E-59, 06E-60

[AuSmith]

I'll give a sort of guide, then you can try again. If you still don't get it, I can give more detail.[unparseable or potentially dangerous latex formula]

06A-50
What if you were given the side length? Find the surface area in terms of the side length, set it equal to 2150 and solve.

Alt soln: Find the area of the four triangles. Cut two of the opposite triangles down the middle (into four congruent right triangles). Fill in the gaps to make a wedge looking thing inside the cube. Then, reflect half of the wedge so that it lies in the same plane as the other half. So, the area of the four triangles is [unparseable or potentially dangerous latex formula].

06A-60
I guess regular trapezoid means isosceles trapezoid. If you have 3 equal lengths emanating from a point, it suggests a circle, but four lengths beg. Again, find the area of the trapezoid in terms of the radius (length AM), then solve. Use the property of circles that an inscribed angle intercepts an arc twice as big (ask if you're not familiar) to get a central angle... use trigonometry to get the height and then the area of the trapezoid in terms of R.

06E-40
Law of cosines

06E-49
Really, the frustum plays no part in this problem. It gives you the height (2h) of a right isosceles and asks for half the hypotenuse.

06E-50
Be careful - you might mistake the isosceles triangle for an equilateral. Solve for [unparseable or potentially dangerous latex formula] with Pythagorean: [unparseable or potentially dangerous latex formula]. Then, find the area of the pentagon.

06E-59
Find the point of intersection ([unparseable or potentially dangerous latex formula]). Integrate [unparseable or potentially dangerous latex formula]. Actually, for shaded area, you should integrate over the absolute value of the difference, but this difference is always positive, so we can forget the abs.

06-E60
Note: Given any three significant (not all three angles) measurements of a triangle, we can determine the measures of all 3 sides and 3 angles using law of sines and law of cosines. This problem gives you three values on the rightmost triangle and wants you to work back to the leftmost triangle.

Find the third side and the other bottom angle of the rightmost triangle with law of cosines then law of sines. Now, the bottom angle of the middle triangle is within reach and the middle triangle also shares a common side with the rightmost. Find the next side with law of sines. Finally, law of cosines gets the desired side.