Chapter 11.5, Problem 24PSC

a

To determine

To calculate: The ratio of a side of a square to the side of the hexagon

Answer to Problem 24PSC

The ratio of a side of square to the side of hexagon is $\sqrt{2}$ .

Explanation of Solution

Given information : The square and hexagon are inscribed in the circle,

Formula used: Diameter of a circle is $2r$

Diagonal of square = $\sqrt{2}a$ (where a is the side of square)

Calculation: As, square is inscribed inside the circle,

  $\Rightarrow$ Diagonal of the square will be equal to the diameter of circle,

  $\Rightarrow$ Diagonal = $2r$

Now, diagonal of square = $\sqrt{2}a$

  $\begin{array}{l}\Rightarrow \sqrt{2}a=2r\\ \Rightarrow a=\frac{2r}{\sqrt{2}}\\ \Rightarrow a=\frac{2r}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}\\ \Rightarrow a=\frac{2\sqrt{2}r}{2}\\ \Rightarrow a=r\sqrt{2}\end{array}$

Now, as hexagon is inscribed inside the circle, the side of hexagon is always equal to the radius of the circle,

  $\Rightarrow$ Side of hexagon = r

Now, ratio of side of square to the side of hexagon = $r\sqrt{2}:r$

  $\begin{array}{l}\Rightarrow \frac{r\sqrt{2}}{r}\\ \Rightarrow \sqrt{2}\end{array}$

b

To determine

To calculate: The ratio of the area of the square to the area of the hexagon

Answer to Problem 24PSC

The ratio of the area of the square to the area of the hexagon is $\frac{4}{3\sqrt{3}}$ .

Explanation of Solution

Given information : The square and hexagon are inscribed in the circle,

Formula used:

Area of square = $a^2$

Where “ a ” is side of square

Area of hexagon = $\frac{3\sqrt{3}}{2}{a}^2$

Calculation: As, side of square inscribed in a circle is $(a)=r\sqrt{2}$ ,

Area = ${(r\sqrt{2})}^2$

  $Area=2r^2$

Now, a side of the hexagon inscribed in a circle is $(a)=r$ ,

  $\Rightarrow$ Area = $\frac{3\sqrt{3}}{2}{r}^2$

Now, ratios of area of the square and the hexagon = $2r^2:\frac{3\sqrt{3}}{2}{r}^2$

  $\begin{array}{l}\Rightarrow \frac{2r^2}{\frac{3\sqrt{3}}{2}{r}^2}\\ \Rightarrow \frac{2(2)}{3\sqrt{3}}\\ \Rightarrow \frac{4}{3\sqrt{3}}\end{array}$