A square is drawn inside a square. Find the area of the square as a function of the radius of the circle.
The diameter of the circle is 2*Radius
area of square= diameter^2= 4 * R^2
so that is the answer
No, the diameter of the circle is the diagonal of the square
Side of square = D/sqrt2
Area of square = D^2/2
but D = 2 R
so
area of square = 4 R^2/2 = 2 R^2
which one is the correct answer
Draw a graph. Start at x = 0 and work your way up to x = pi/6
Oh, sorry, worong question. Draw your square inside the circle and you will see that the diagonal of the square is the diameter of the circle.
The diagonal of the square divides the square into two 45,45,90 triangles
The hypotenuse of a 45,45,90 triangle id the side times sqrt 2
s^2+s^2 = diagonal^2
2 s^2 = diagonal^2
s = (1/sqrt2) diagonal
so
diagonal = side * sqrt 2 = diameter
To find the area of the smaller square as a function of the radius of the larger square, we need to establish a relationship between the two shapes.
Let's assume that the larger square has a side length of \( L \), so its area would be \( L^2 \).
Now, let's draw a circle inscribed within the larger square. Since the circle is inscribed, its diameter would be equal to the length of the square's side, which is \( L \). We can then determine that the radius of the circle is \( \frac{L}{2} \).
The diagonal of the larger square is equal to the diameter of the circle. By drawing a diagonal across the square, we create two right triangles. Using the Pythagorean theorem, we can find the value of the diagonal, which is \( \sqrt{2}L \).
Now, let's focus on the smaller square inside the larger square. Since the diagonals of a square are equal in length, the diagonal of the smaller square is also \( \sqrt{2}L \).
Recall that the diagonal of a square forms an isosceles right triangle with the sides of the square. In this case, the hypotenuse of the triangle is the diagonal of the smaller square, and one of the legs is the radius of the circle (\( \frac{L}{2} \)).
Using the Pythagorean theorem again, we can find the length of the other leg (\( s \)) of the right triangle, which represents the side length of the smaller square:
\[ s^2 = (\sqrt{2}L)^2 - \left(\frac{L}{2}\right)^2 \]
Simplifying the equation, we get:
\[ s^2 = 2L^2 - \frac{L^2}{4} \]
\[ s^2 = \frac{7L^2}{4} \]
Therefore, the area of the smaller square is given by:
\[ \text{Area of smaller square} = s^2 = \frac{7L^2}{4} \]
Since the radius of the circle is half the length of the larger square's side, we can express the area of the smaller square as a function of the radius (\( r \)) of the circle:
\[ \text{Area of smaller square} = \frac{7(2r)^2}{4} = \frac{28r^2}{4} = \frac{7r^2}{4} \]
Hence, the area of the smaller square is given by \( \frac{7r^2}{4} \) as a function of the radius of the circle.