Tag: solution

Solution to Related Functions

let:f(x) = g(x+1)-x, g(x) = f(2x)-2

Now, we want to find the derivative of f( g(x) ). It is important to remember that these functions are related, entangled, if you will. With that in mind, we must not forget that derivates do not require precise functions. So. f(x) = g(x+1) – x, and that equals f(2x+2) – 2. That makes f(x) = f(2x+2) – x – 2. That means when we add x+2 to x, f(x) increases by x+2. Therefore the derivative of f(x) is 1. Now, g(x) is 2 less than twice f(x), and thus its derivative is the same as that of 2f(x). That makes g(x)’s derivative 2. Because the derivative of f( g(x) ) equals their separate derivates multiplied. That makes our answer 2.

Solution to Altitude Triangle

Triangle for this challenge.

In this challenge, we’re trying to find the area of this triangle, with perimeter 20. Now, it is important to recognize that this is an isosceles triangle. That means that 2 sides are equal, and the third side is bisected. That means that if we let x and y be the side lengths, x+2y=20. Also, using the pythagorean theorem, x2 /2+25 = y2. The only solution to this is x=7.5, and y=6.25. That makes the area 5x/2, or 18.75 u2

Solution to Perfect Square

The Square!

We wanted to find the area of a square! Now, the first thing we should take note of is that the small triangle is similar to the larger triangles because all of it’s angles are the same. If we let x and y be the 2 segments of the top line, x on left and y, right, then we can make an equation. specifically, find the area of the small triangle via hypotenuse comparison. The hypotenuse of the small triangle is y, and the hypotenuse of the big one is √(y2+(x+y)2). This makes y/√(y2+(x+y)2) equal one third, or (x+y)2=8y. To get this we simply raise both sides to the power of -2, subtract 1 from both sides, and then multiply by y. We also know that the area of the whole square is (x+y)^2, and that, since each outer region is a third of y(x+y), the small square is (x+y)2 – 8/3*y(x+y). That gives us the equation (x-5/3*y)*(x+y)=1. We can combine our 2 equations to solve for x and y. The positive, real number solution is x ≈ 1.120, y ≈ 0.2269, so the side of the square is ≈1.347, making it’s area 1.815 square units. And there’s our answer.

Solution to Divided Rectangle

Squared rectangle for this problem.

In this challenge, we’re trying to find the area of a rectangle. The first square has side length 1. It shares a side with another, which must also have side 1. This makes the 2 big squares above them both have side length 2. This makes the square to the right have side length 5 units, and the one below, 7. Then we must do some algebra and find that 4x=12, x=3. This makes the very large square have side length 9, and thus the rectangle is 12 by 16. This gives it an area of 192.

Solution to Shaded Square

Square for this challenge.

Previously, we were trying to find the area of the non-shaded region. You will probably have found that the upper line must be parallel to the horizontal sides, and situated 2/3 of the way up the vertical ones. We know this because the steep diagonal lines clearly end 4/3 of the way across the top. This makes the slope of the steep diagonal line 1.5. The shallow diagonals must have slopes of 2/3. It follows that the lower intersections are 1/6 of the way through. This makes the lower shaded triangles both have area (1/3 * 1/2 / 2) – (1/3 * 1/6) = 1/18 each (A=B*H/2.) The upper shaded triangle clearly has base 1/2; and, height 1/3 as we showed earlier. This makes it’s area 1/12. This means the total shaded area 1/18 + 1/18 + 1/12 = 7/36, and the non-shaded, 1 – 7/36 = 29/36. The answer is therefore 29/36.

Solution