Inscribed and Circumscribed Circles

Suppose the length of each side of $ABCD$ is $2s$

By drawing a line from center point to point $A$ and a line from center point perpendicularly to $AB$ , we have a right triangle and get the radius of $Γ_1$ is $s\sqrt{2}$ and $Γ_2$ is $s$

Here we have

$\pi (s\sqrt{2})^2 = 100$

$2\pi s^2 = 100$

and get

$\pi s^2 = 50$ which is the area of $Γ_2$

Let the side of the square be a. Radius of the circumscribed circle =half the diagonal of the square=√2a/2. Radius of the inscribed circle=half the side of the square=a/2. Area of the circumscribed circle=2a^2/4 *π=100 ⇒a^2/4=50/π. Area of the inscribed circle=a^2/4 * π=50/π * π=50

Both circle have their centers on the center of the square. Thus the radius of the inscribed circle is equal to the side length of the square ABCD (call it $a$ ). The circumscribing circle has a radius equal to $a\sqrt{2}$ . Thus the ratio of the areas of the circumscribing circle to that of the inscribed circle = $\frac{1}{2}$ . The area of the inscribed circle is therefore $= \frac{1}{2} \times 100 = 50$

• AC cut BD at O. => O is the center of T1 and also the center of T2.
• The area is OA^2 x pi = 100, so OA=R1=square root of (100/pi).
• As ABO is an isosceles and right-angled at O, according to Pytago theorem, we have OA^2 + OB^2 = AB^2
  => (100/pi) +(100/pi) = AB^2 => AB^2 = (200/pi).
  
• Let H be the foot of the perpendicular from O to BC, the triangle OBC is a right-angled triangle at H. And BH^2 = (BC^2)/4 = (AB^2)/4 = (50/pi) => OB^2 - BH^2 = OH^2 => (100/pi) - (50/pi) = OH^2 => OH^2 = 50/pi
• We have OH is the radius of T2 - the circle inscribed in ABCD, so the area of T2 is OH^2 x pi = (50/pi) x pi = 50.

Let side of the square be a. The circumscribed circle Γ1 has an area of 100 and clearly it's radius is half of the diagonal of the square i.e. x=a/2^0.5

Therefore 100= pi * x * x (i)

Let area of the inscribed circle be y. Clearly the radius of this circle is half the side of the square.

z=a/2

y= pi * z * z (ii)

Dividing (i) and (ii)

we get y=50 sq. unit

let $ABCD$ has side length of $x$ , Then $Γ_{1}$ will have area $\pi(\frac{\sqrt{2}\ x}{2})^{2}$ $=$ $\frac{\pi\ x^{2}}{2}$ ,and $Γ_{2}$ will have area $\pi(\frac{x}{2})^{2}$ $=$ $\frac{\pi\ x^{2}}{4}$ .Now, $\frac{Γ_{1}}{ Γ_{2}}$ $=$ $2$ . Thus the area of $Γ_{2}$ is $\frac{100}{2}$ $=$ $50$

Let the side of the square be a units. Radius of outer circle = a/(sqrt. 2) Equating to area = 100, we get pi*(a^2) = 200

Radius of inner circle = a/2 Area = pi (a^2)/4 Substituting the value of pi (a^2); we get area = 50 units

$\Gamma_1$ and $\Gamma_2$ have the same centre. assume it's $O$ and the radius $\Gamma_2$ is $r_2$ , radius of $\Gamma_1$ is $r_1$ . The radius of the $\Gamma_1$ is a half of square $\text{ABCD}$ sides.The radius of $\Gamma_2$ is a half of square $\text{ABCD}$ diagonal. then $r_2=\frac{\sqrt{2}r_1}{2}\Leftrightarrow r_2^2=\frac{r_1^2}{2}$ . We know area of $\Gamma_1$ is $\pi r_1^2=100$ . Hence, area of $\Gamma_2$ is $\pi r_2^2=\frac{\pi r_1^2}{2}=\frac{100}{2}=50$ . *p.s. sorry my english realy bad

Let $O$ be the center of the circles, and $E$ be the perpendicular from $O$ to $AB$ . $\Gamma_1$ and $\Gamma_2$ are similar by a ratio of $\frac {OA}{OE} = \sqrt{2}$ . Hence, the area of $\Gamma_2$ is $\frac {100}{\left(\sqrt{2}\right) ^2} = 50$ .

Let side of the square be a. The circumscribed circle Γ1 has an area of 100 and clearly it's radius is half of the diagonal of the square i.e. x=a/2^0.5 Therefore 100= pi * x * x (i) Let area of the inscribed circle be y. Clearly the radius of this circle is half the side of the square. z=a/2 y= pi * z * z (ii) Dividing (i) and (ii) we get y=50 sq. un