Is It A Square?

Here is the following illustration as the start:

Because of line length symmetry for $\overline{AE}$ , we can conclude that $|AC| = 2.5$ , which is the radius of the semicircle. Thereby, $|FC| = |GC| = 2.5$ .

Because of midpoint $C$ , we see that $|BC| = |AC| - |AB| = 1.5$ and also $|CD| = |CE| - |DE| = 1.5$ . Since $\Delta BFC$ is the right triangle (as well as $\Delta CGD$ ), Pythagorean theorem shows that $\begin{array}{rl} |FB|^2 + |BC|^2 &= |FC|^2\\ |FB| &= \sqrt{|FC|^2 - |BC|^2}\\ &= 2 \end{array}$ which thus shows that the area of the rectangle is $A = |FB| \times |BD| = \boxed{6}$ .

$x=\sqrt{2.5^2-1.5^2}=2$ $units$

$area = 2(3) = 6$ $square$ $units$

The vertical side of the rectangle is essentially the geometric mean of two numbers: 1 and 3 +1 = 4 ie. the square root of 1 x 4 = 2.

It follows that the areas of the rectangle is therefore 2 x 3 = 6

diameter is 5. Hence radius is 2.5

Now from centre to one edge of the rectangle forms a right triangle with H=2.5 with B=1.5. So P=2. Area is 3*2

Add 1+3+1=5 the diameter of the circle. Thus 2.5 is the radius. Half of 3 is 1.5. Now a right triangle measuring 1.5 on one leg and 2.5 on the Hypotenuse is formed. Noticing a pattern, multiply by 10 and divide by 5 reveals a 3-4-5 right triangle. Thus 4×5/10=2. 2×3=6 the area of the rectangle.