BuilderBoi

Here: https://web2.0calc.com/questions/geometry_31898#r1

We want the term with the x to cancel out. 

Currently, we have a \(36x\), so we need \(3xy = 36x \)

This can only be hieved when the square is \(\color{brown}\boxed{12}\)

Which is \(SR\), and which is \(QR\)?

We have the following system: 

\(a+b=x\)                          (1)

\(ab=x\)                               (2)

\((a-b)(a+b)=x\)         (3)

Substituting (1) into (3), we get: \((a-b)x=x\)

Dividing both sides by x, we find \(a-b=1\)

Solving for \(a\) in terms of b, we find: \(a=b+1\)

Substituting this into (2), we get: \(b(b+1)=x\)

Simplifying, we find: \(b^2+b=x\)

Substituting this in (3), we get: \(a^2-b^2=b^2+b\)

We also know that \(a = b+1\)

Solving this, we find: \({a={ {3 + \sqrt5} \over 2}}\) and \({b={{1 + \sqrt5} \over 2}}\)

However, these are interchangeable, so b can be either of these. 

We know that \(100 + 5 = 105\)

We also know that \(50+55 = 105\)

There are 10 cases of this, so the sum is \(10 \times 105 = 1050\)

Can you find the last 2 digits from here?

We can write \(15^{15}\) as \(3^{15} \times 5^{15}\)

Thus, n can be anything from 0 up to 15, making for \({16} \) values

EDIT: Upon furthur inspection, I found that n can also be 18, making for \(\color{brown}\boxed{17}\) values. 

We can enclose the shape in a \(5 \times 4 \) grid. From here, we can subtract the areas of the regions that do not lie in the shaded shape (see graph below)

Doing so, we find the area is \(20 - 2 - 2 - 2 - 1 - 1 = \color{brown}\boxed{12}\)

Here is the graph: 

We have the following system:

\(6+b+c=24\)       (1)

\(36+b^2=c^2\)             (2)

Solve for \(c\) in the first equation: \(c=18-b\)

Squaring this gives \(b^2-36b+324\)

Substituting this in (2) gives: \(36+b^2=b^2-36b+324\)

Solving, we find that \(b= 8\)

Subsituting this in to (1), we find that \(\color{brown}\boxed{c = 10}\)

Draw the line \(y=-{1 \over 3}x\). 

It will intersect the original line at \((3,-1)\), and this point will also bisect the chord \(AB\)

The distance between the origin and the intersection point is \(\sqrt{10}\)

Using the Pythagorean Theorem, the radius of the circle is \(\sqrt{209}\).

Now can you find the area?

Draw altitude \(AM\). 

Because the trapezoid is isoceles (\(AD = BC\)), \(MD =1\)

Applying the Pythagorean Theorem, \(x= \color{brown}\boxed{\sqrt{5}}\)