Question

A tape is used to form a perfect circle of diameter $D$. Later, the same length of tape is used to form a rectangle inside the circle such that the corners of the rectangle lie on the circumference of the circle. What is the area of rectangle in square inches, if the diameter of circle is 4 in?
$(Take\pi \approx 3)$

Answer 1

$Diagonaloftherectangle=dDiameterofthecircle=D$

As the rectangle is inscribed in the circle, we have:

$D=d.....Equation1$


Now, inspecting the dimensions of the rectangle, we find:

$Lengthoftherectangle=lWidthoftherectangle=wDiagonaloftherectangle=d$


Using the Pythagorean theorem, we get:

$d^2=l^2+w^2$
Using Equation 1
$⟹D^2=l^2+w^2......Equation2$

Now, we know that the length of the tape used to make the rectangle is the same as that used to make the circle.

$⟹Circumferenceofthecircle=Perimeteroftherectangle⟹2\pi \left(radius\right)=2\times (l+w)⟹\pi \times \left(diameter\right)=2(l+w)⟹\frac{\pi D}{2}=(l+w)$
$⟹\frac{\pi D}{2}=l+w$

Squaring both sides, we get:

$[\frac{\pi D}{2}{]}^2=(l+w{)}^2⟹\frac{{\pi }^2{D}^2}{4}=l^2+w^2+2\times l\times w$

Now, using Equation 2, we get:

$\frac{{\pi }^2{D}^2}{4}=D^2+2lw⟹\frac{{\pi }^2{D}^2}{4}-D^2=2lw⟹D^2[\frac{{\pi }^2}{4}-1]=2lw⟹\frac{D^2}{2}[\frac{{\pi }^2}{4}-1]=l\times w⟹\frac{D^2}{2}[\frac{{\pi }^2}{4}-1]=length\times width⟹\frac{D^2}{2}[\frac{{\pi }^2}{4}-1]=Rectangl{e}^{\prime }sarea$

$\text{Putting D = 4 in and}\pi =3$
$⟹Rectangl{e}^{\prime }sarea=\frac{4^2}{2}[\frac{3^2}{4}-1]$
$⟹Rectangl{e}^{\prime }sarea=8[\frac{9}{4}-1]$
$⟹Rectangl{e}^{\prime }sarea=8\left[\frac{5}{4}\right]$
$⟹Rectangl{e}^{\prime }sarea=10\text{sq in}$