<p>Ok, I don't know how to find the answer for these 2 problems:</p>
<p>Problem 1:
Figure 8 shows a triangle inscribed in a semicircle. What is the area of the triangle in terms of theta ?
A) thetapi/2
B) theta/2
C) tan theta
D) sin theta
E 2 sin theta cos theta
Figure 8 is a semicircle with diameter 2. One of the triangle's sides is the diameter, ? is an acute angle, another of the triangle's sides hits the other side of the semicircle, but not right above the midpoint of the diameter--that is, if you were to draw the radius of the circle to the corner of the triangle, it would not make 90 degrees. Yeahh, it's kind of hard to picture.</p>
<p>Problem 2: The radius of the base of a right circular cone is 6 and the radius of a parallel cross section is 4. If the distance between the base and the cross section is 8, what is the height of the cone?
A) 11
B) 13 1/3
C) 16
D) 20
E) 24</p>
<p>I know the answers to these, just not how to get the answer. I'd appreciate someone's help in understanding how to solve them.</p>
<p>For the first question, who is theta? Anyway, the triangle is right (any triangle with the 3 vertices on a circle and one of the sides as the diameter is right). If theta is the angle between the diameter and one of the other sides, then the answer is E (product of legs/hypotenuse).
For the second question, imagine the intersection of a plane that contains the vertex of the cone and the center of the base with the cone. You get an isosceles triangle where the base is a diameter and the height is the height of the cone. This triangle and the one who has the diameter of the cross section 4 form two similar triangles. If you consider just half of the 2 triangles, you have 2 similar right triangles, one with the legs x and 3, the other with the legs (x-8) and 2. Solve for x the equation 2/3=(x-8)/x and you get the solution (E)
Are these the answers?</p>
<p>Yeah, you're right on problem 2, thanks.
But problem 1, the triangle cannot be a right triangle. It is impossible to have a triangle inscribed in a semicircle with one of the sides as the diameter be right. In fact, the only way I could see it being some sort of special triangle would be if it were 2 45-45-90 triangles if split into two, but then a line drawn from the diameter to the opposite angle would have to be at 90 degrees to the angle, which it isn't. E is the right answer though, but not because it's a right triangle...</p>
<p>I am sure that the triangle is right. I've learned a theorem about that when I was studying plane geometry. If you don't believe me, try an exact draw: Use a compass to draw a circle, draw a diameter and choose any point from the circle. Unite it with the endpoints of the diameter and measure the angle. You'll find it's right, even if it doesn't seem so because the legs aren't horizontal and vertical.</p>
<p>ahhh ok I thought you meant theta was right. Thanks!!</p>
<p>the first problem is E b/c of the formula A = .5 ab sin theta,
where theta is the included angle between the sides</p>