Ratio of inscribed squares

Geometry Level pending

  • Let the area of the inscribed square ABDC in the circle be A1
  • Let the area of the inscribed square MNOP in the semi-circle be A2

Determine A 2 A 1 \frac{A2}{A1}

Give your answer as a decimal with 3 digits (if necessary)


The answer is 0.4.

Magnus Seidenfaden by Magnus Seidenfaden , 19, Denmark

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1 solution

A1 - If we draw a line from A to D through the center of the circle we can find A1 in terms of the radius. - If we let r be the radius and x the side length of the square then we can use the pythagorean theorem to get an expression for A1:

  • x 2 + x 2 = ( 2 r ) 2 x^2+x^2=(2r)^2
  • 2 x 2 = 4 r 2 2x^2=4r^2

  • x 2 = 2 r 2 x^2=2r^2

  • As x 2 x^2 is the area of ABDC we have found an expression for A1

A2 - The line ED is the same as the radius of the circle so we call that length r. - The sidelength of the square will be called x - If we observe the triangle ENO we can see that the sidelength x is double the length of EO

  • Using the pythagorean theorem:
  • ( 1 / 2 ∗ x ) 2 (1/2*x)^2 + x 2 x^2 = r 2 r^2
  • 5 / 4 ∗ x 2 5/4*x^2 = r 2 r^2
  • x 2 x^2 = 4 / 5 ∗ r 2 4/5*r^2

  • A2= 4 / 5 ∗ r 2 4/5*r^2

  • Now we can calculate A 2 A 1 \frac{A2}{A1} :

  • 4 / 5 ∗ r 2 2 ∗ r 2 \frac{4/5*r^2}{2*r^2}
  • 4 / 5 2 \frac{4/5}{2}
  • 2/5=0.4
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