How the heck do you do this?

Question

http://hometown.aol.com/amahol777/myhomepage/index.html

Answer 1

If you divide the triangle down the middle you get two identical triangles each with an angle of 55 degrees at the top. You can use this angle with the radius to calculate the height and therefore the area of the triangle (make sure you calculate the whole triangle area and not just one of the half triangles though).

To calculate the area of the blue shaded section you need the area of the segment of the circle. This is easy, just find the area of the whole circle then the area of the segment is just 110/360 times the whole circle.

Subtract the area of the triangle from the aread of the segment and you have the area of the blue shaded section.

Answer 2

Difficult to say how the teacher wants you to do it, because the information in the diagram is inconsistent (I'm assuming that the 110 means that the central angle is 110°). For a central angle of 110° the base of the triangle would be 32.77 units to the nearest hundredth.

Anyway, the method I think you're most likely expected to use is this:
Area of the whole segment (triangle plus shaded area) = 110/360 * area of circle = 110/360 * 3.14 * 20^2 = 383.78 square units.
Now draw a line perpendicular to the base of the triangle and going from the base up to the centre of the circle. This splits the triangle into two congruent halves (you should be able to show two matching sides and all matching angles fairly easily); so you have two right triangles with hypotenuse 20 and base 12, hence height 16 (a 3-4-5 triangle). So the area of the triangle is 2 * (bh/2) = 2 * (12*16/2) = 192 square units.

Hence the shaded area is 383.78 - 192 = 191.78 square units.

Unfortunately, as I said before, the values given are inconsistent. I'm assuming you haven't gotten up to trigonometry yet; if you have, you should be able to see why the values are wrong. Drawing the perpendicular bisector as before, and noting that the small angles of the triangle are 35°, we get cos 35° = 12/20 = 0.6, but actually cos 35° = 0.819 to 3 decimal places.

Answer 3

The area of the whole circle is (pi)r^2. The area of the whole segment containing the triangle and the section whose area is required is [(pi)r^2]110/360. Find the area of the triangle as half of base*height.Substract it from the area of the sector and hence obtain area of the section