Multiple choice

$ABCD$ is a rectangle inscribed in a quadrant of a circle having radius $10cm$. If $AD=2\sqrt { 5 } cm$, find the area of $ABCD$.

  1. $30{ cm }^{ 2 }$
  2. $50{ cm }^{ 2 }$
  3. $40{ cm }^{ 2 }$
  4. $35{ cm }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let the rectangle have sides x and y. Since it is inscribed in a quadrant of radius 10, the diagonal from the origin to the opposite corner is 10. Thus, x^2 + y^2 = 10^2 = 100. Given one side is 2*sqrt(5), (2*sqrt(5))^2 + y^2 = 100 => 20 + y^2 = 100 => y^2 = 80 => y = sqrt(80) = 4*sqrt(5). Area = x * y = 2*sqrt(5) * 4*sqrt(5) = 8 * 5 = 40.

AI explanation

Because the rectangle ABCD is inscribed in the quadrant, vertex C lies on the circumference, making the diagonal AC equal to the radius of 10 cm. Using Pythagoras theorem in triangle ADC, the length of CD squared plus AD squared equals AC squared. This gives CD squared plus (2 root 5) squared equals 10 squared, meaning CD squared plus 20 equals 100, so CD is the square root of 80, which is 4 root 5 cm. The area of the rectangle is length AD times length CD, calculated as 2 root 5 times 4 root 5, giving 40 square cm.

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