Junior Math Questions?

1. The window is an equilateral triangle, meaning all sides are the same length, 7.5 feet. To find the height of the window, use the Pythagorean Theorem, a²+b²=c². Note that to solve for the height, we need to know the base and hypotenuse. The base for the right triangle is ½ of 7.5 (because it's only half of the equilateral triangle), and the hypotenuse is 7.5, per the problem. So do c² − a² = b² and you'll have your answer.

2. This is also a right triangle. Even better, it's a 45-45-90 triangle. That means the triangle is also isosceles, or has two sides of the same length. I hope you remember your triangle properties; if in a 45-45-90 you have a side length s, then the hypotenuse has a length of s√2. So, we know the hypotenuse is 118 yards. How do you solve for s?

3. Very similar to #2. Except this time, it's a 30-60-90 triangle. Remember, if the shortest side has length s, then the hypotenuse has length 2s, and the other side has length s√3. Using that info, you can find out the distance (one of the sides).

4. The volume of a cylinder is V = πr²h.

5. First calculate the volume of the chocolate mold and the volume of a chocolate bar. Multiply the chocolate bar volume by 12. Simply divide the volume of the chocolate bars by the volume of the mold.

6. The surface area of a cone is SA = πr(r+s), where r is the radius and s is the slant height (which you can figure out with the Pythagorean Theorem).

7. For simple problems like the above, you need to have at least one of the angles (like a 90° one). However, if you really need to find an angle with only 3 sides, you need to use the Law of Cosines:

c² = a² + b² − 2ab×cos C

b² = a² + c² − 2ac×cos B

a² = b² + c² − 2bc×cos A

From there you can use the Law of Cosines again or the Law of Sines to find another angle. And to find the last angle just use the triangle property that all angles measure to 180°.

8. Set up a proportion and cross multiply. All triangles with a scale factor should be similar.

9. Unless you know some angles that can determine the side lengths, you can't.

At first I inspiration 36 additionally, however I see it is soliciting for the smallest quantity of cubes now not the largest. 36 is the largest quantity of cubes he would get. Well, the largest dice he could make is 2x2x2, when you consider that the smallest facet is two inches. He can get 3 of the ones out of the tremendous piece. Now you've a 1x2x6 block of fudge that needs to be reduce into 12 1x1x1 portions. 12 + three is 15. In order to get the largest dice out of a rectangle, you have to make it the dimensions of the smallest facet.