Question

The red rectangle in the figure is the maximum possible rectangle in the semicircle. What is its area (in cm)?
22 cm​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

121 cm^2

Step-by-step explanation:

The maximum possible rectangle ABCD (say) is inscribed inside semicircle of diameter of 22 cm. Let the Center of semi circle  is O which lies on line AB. Let the length of the rectangle is 2x cm.

Then,

AO = x cm;

OD = 11 cm (radius)

By pythogorous theorem,

$AD = \sqrt{121-x^2}$

Let the area function A(x) denotes the area of the rectangle, for maximum value its derivative must be zero

$A(x) = 2x\sqrt{121-x^2}\\ A'(x) = 2\sqrt{121-x^2} - \frac{2x^2}{\sqrt{121-x^2} } \\A'(x) =0 \implies x = 11/\sqrt{2} \\A(11/\sqrt{2}) = 2 \times 11/\sqrt{2} \times 11/\sqrt{2} = 121 cm^2$

#SPJ2