Question
A square is inscribed in a circle and another in a semi-circle of same radius. The ratio of the area of the first square to the area of the second square is :
Answer: Option B
Let the radius of each of the circle and the semi-circle be r units
Diagonal of the first square = 2r units
Let the side of the second be a units
Then,
$$\eqalign{
& {r^2} = {a^2} + {\left( {\frac{a}{2}} \right)^2} \cr
& \Rightarrow {r^2} = \frac{{5{a^2}}}{4} \cr
& \Rightarrow {a^2} = \frac{{4{r^2}}}{5} \cr} $$
∴ Ratio of the areas of the two squares :
$$\eqalign{
& = \frac{{\frac{1}{2} \times {{\left( {2r} \right)}^2}}}{{{a^2}}} \cr
& = \frac{{2{r^2}}}{{\left( {\frac{{4{r^2}}}{5}} \right)}} = \frac{5}{2} \cr
& = \frac{5}{2} \cr
& = 5:2 \cr} $$
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Let the radius of each of the circle and the semi-circle be r units
Diagonal of the first square = 2r units
Let the side of the second be a units
Then,
$$\eqalign{
& {r^2} = {a^2} + {\left( {\frac{a}{2}} \right)^2} \cr
& \Rightarrow {r^2} = \frac{{5{a^2}}}{4} \cr
& \Rightarrow {a^2} = \frac{{4{r^2}}}{5} \cr} $$
∴ Ratio of the areas of the two squares :
$$\eqalign{
& = \frac{{\frac{1}{2} \times {{\left( {2r} \right)}^2}}}{{{a^2}}} \cr
& = \frac{{2{r^2}}}{{\left( {\frac{{4{r^2}}}{5}} \right)}} = \frac{5}{2} \cr
& = \frac{5}{2} \cr
& = 5:2 \cr} $$
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