If you draw the a square inside the inscribed circle, the circle's diameter is also the diagonal of the square inside it.
The side of the square is equal to the diameter of its inscribed circle.
If 4cm is the side of the bigger square, then the diameter of the inscribed circle is also 4 cm, and the diagonal of the smaller circle is also 4 cm.
When a diagonal is drawn inside the square, it divides the square i two equal isosceles triangles. In this problem, the hypotenuse (diagonal) is equal to 4 cm.
To solve for the area of the smaller square, find the side of the isosceles triangle.
Let x be the side:
[tex]4 ^{2} = x^{2} + x^{2} [/tex]
[tex]4 ^{2} = 2 x^{2} [/tex]
[tex] \sqrt{ 4^{2} } = \sqrt{2 x^{2} } [/tex]
[tex]4 = x \sqrt{2} [/tex]
[tex] (\frac{4}{ \sqrt{2} } )( \frac{ \sqrt{2} }{ \sqrt{2} }) =x[/tex]
[tex]x= \frac{4}{2} \sqrt{2} [/tex]
[tex]x = 2 \sqrt{2} [/tex]
Area of big square:
A = (4 cm)²
A = 16 cm²
Area of small square:
=[tex](2 \sqrt{2} ) ^{2} [/tex]
=[tex](4) ( \sqrt{4} ) = (4) (2cm)[/tex]
A = 8 cm²
The area of the small (er) square is 8 cm²
