In the given diagram, ABCD is a square and semi-circular regions have been added to it by drawing two semi-circles with AB and CD as diameters. If the total area of the three regions is 350 sq.cm, then the length of the side of the square is equal to : <img src="/images/question-image/arithmetic-ability/area/1577193436-in-the-given-diagram-abcd-is-a-square-and-semi-circular-regions.png" title="Area mcq question image" alt="Area mcq question image">
Correct Answer: 14 cm
Let the length of the side of the square be x cm
Then, radius of each semi-circle = $$\left( {\frac{x}{2}} \right)$$ cm
Total area :
$$\eqalign{ & = \leftc{m^2} \cr & = \left( {{x^2} + \frac{\pi }{4} \times {x^2}} \right)c{m^2} \cr} $$
$$\eqalign{ & \therefore {x^2} + \frac{\pi }{4} \times {x^2} = 350 \cr & \Rightarrow {x^2} + \frac{{22{x^2}}}{{28}} = 350 \cr & \Rightarrow {x^2} + \frac{{11{x^2}}}{{14}} = 350 \cr & \Rightarrow \frac{{25{x^2}}}{{14}} = 350 \cr & \Rightarrow {x^2} = \left( {\frac{{350 \times 14}}{{25}}} \right) \cr & \Rightarrow {x^2} = 196 \cr & \Rightarrow x = 14\,cm \cr} $$
Then, radius of each semi-circle = $$\left( {\frac{x}{2}} \right)$$ cm
Total area :
$$\eqalign{ & = \leftc{m^2} \cr & = \left( {{x^2} + \frac{\pi }{4} \times {x^2}} \right)c{m^2} \cr} $$
$$\eqalign{ & \therefore {x^2} + \frac{\pi }{4} \times {x^2} = 350 \cr & \Rightarrow {x^2} + \frac{{22{x^2}}}{{28}} = 350 \cr & \Rightarrow {x^2} + \frac{{11{x^2}}}{{14}} = 350 \cr & \Rightarrow \frac{{25{x^2}}}{{14}} = 350 \cr & \Rightarrow {x^2} = \left( {\frac{{350 \times 14}}{{25}}} \right) \cr & \Rightarrow {x^2} = 196 \cr & \Rightarrow x = 14\,cm \cr} $$
