Radius of a quarter circle

Consider my diagram. Apply pythagorean theorem on the right triangle. We have,

$R^2=(R-4)^2+(R-2)^2$

$R^2=R^2-8R+16+R^2-4R+4$

$0=R^2-12R+20$

Factor.

$(R-2)(R-10)=0$

$R-2=0$ $\implies$ $R=2$

$R-10=0$ $\implies$ $R=10$

The value of $R$ is $\large{\color{#D61F06}\boxed{10}}$ since $2$ is not possible.

The sides of the rectangle are R-2 and R-4. The diagonal of the rectangle is the same as the radius of the circle. But it can also be written with the Pythagorean theorem as of R^2=(R-2)^2+(R-4)^2. Solving this quadratic equation we get two answers: 2 and 10. 2 is invalid because the length has to be greater than 4. This means the answer is 10.

A trigonometric solution: Begin by considering the radius $r$ along the diagonal of the inscribed rectangle, beginning from the origin and ending at the vertex. With the angle $\theta$ formed between the base of the rectangle and $r$ we can express the ratio of the lengths of the height and base of the rectangle as $\dfrac{\sin(\theta)}{\cos(\theta)}$ . From the diagram we know that $\dfrac{\sin(\theta)}{\cos(\theta)} = \dfrac{r-4}{r-2} \Rightarrow \tan(\theta) = \dfrac{3}{4}$ .

Using this ratio we can compute the length of $r$ as follows: $\sin(\arctan(\dfrac{3}{4})) = \dfrac{3}{5}$

The height of the inscribed rectangle is therefore $\dfrac{3}{5}$ of the radius $r$ .

From the diagram we know that $4 = \dfrac{2}{5}r$

Solving for $r$ we get $r = 10$

We know that the radius can be expressed in any of the following ways, keeping in mind that it is also equal to the diagonal of the rectangle, as its two opposed vertices lie in the center and on the circle respectively:

$r=4+l,\space r=2+L,\space r=\sqrt{l^2+L^2}$

From these relations we can substitue $L$ with $l+2$ and proceed by solving for $l$ :

$(4+l)^2=(\sqrt{2l^2+4l+4})^2$

The solutions of the quadratic are:

$l^2-4l-12=0\implies l_{1,2}=\dfrac{4\pm\sqrt{16+4\cdot 12}}{2}$

And therefore $l\in\{6, -2\}$ , and because $-2$ is impossible geomterically, we're left with $\boxed{r=6+4=10}$ .

I printed out the problem. I placed a compass on the "4" segment & then spun the compass around to continue down the vertical axis, marking a new segment that = "8"... (2*4=8). I reset my compass to "8" & and placed one end at the center intersect & the other end along the horizontal axis. It was equal to the side of the rectangle. The radius was then the sum of the two segments 8 +2= 10.

You can solve it from the options. 10-2=8 and 10-4=6 which are the sides of the rectangle. Now 6, 8 and 10 make up a right angled triangle as 6^2 + 8^2 =10^2. So, 10 is the right answer.

A very simple logic can be applied to do from the given options, we can easily find that if one of the side of formed rectangle is taken as L and other as B then R will be L+2 and B+4 simultaneously and also it will be √(L² + B²). Now simply 6,8,10 a famous Pythagorean Triplet comes handy and yeah we found the answer without any calculation.

Of all the given solutions only 10 forms a Pythagorean triplet with the other two being 2 and 4 less than 10. i.e. $10^{2}=(10-2)^{2}+(10-4)^{2}$

Of the four options, 10 is the only one that yields sides of the rectangle which, by Pythagoras, yield an integer (10) for its diagonal.

Doesn't anybody just look at the vertical line, covers the other parts to avoid optical illusions, see that the part of the line that is not labeled is a bit longer than the labeled part, checks the available answers and thus just knows it must be 10?

My way of dealing with problems is to just estimate.

This is just a way to solve the quadratic equation you get from the Pythagorean theorem without expanding two squared binomials:

$r^2 = (r-4)^2+(r-2)^2$

$r^2-(r-4)^2 = (r-2)^2$

$(r-(r-4))(r+r-4) = (r-2)^2$

$4(2r-4) = (r-2)^2$

$8 = r-2$

$r=10$

We can divide by $r-2$ to get from the 4th to the 5th line because the radius cannot be 2

The first thing to notice is that you can divide the rectangle into two right-angled triangles by drawing a diagonal line. It can later be stated that among the natural numbers up to twelve, there are only two Pythagorean triples: G1 = (3, 4, 5) and G2 = (6, 8, 10). Now you have to add the given numbers (2 and 4) with the cathetes ((3, 4); (6, 8)) to get the radius out. If you compare the results, the same sum should come out, since we have an equation. This only applies in the case of numbers 6 and 8. The sums are equal to ten, which means that only answer C can be correct. Here is a mathematical representation:

Initial conditions:

• x, y ∈ IN (since the anwers (z) are among the natural numbers, too)
• $x^2+y^2=z^2$
• $x+2=y+4$
• $x=(4;8);y=(3;6)$
• $x-y=4-2$ (the difference is the same => true)

Solution:

• G1: $x=4;y=3$

• $4-3≠2$ -> wrong

• G2: $x=8;y=6$

• $8-6=2$ -> true

I hope that you can understand it.

Have a nice day! ;)