Answer:
![94.6cm^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/xpw6r0xz5km9khxt4p0am4q8s8972n6rvk.png)
Explanation:
First we will find the area of Square PQRS.
Area of Square PQRS = Length x Width = 21 x 21
=
![441cm^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/caq2cxrpmrptx7kyuwitr1o29g9jk4i9m4.png)
Next we will found the Area of Semicircles PS and QR.
Note: Area of Semicircle PS = Area of Semicircle QR
Area of Semicircle =
![(1)/(2) \pi r^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/k3fjljgwcfy3xuyzboehsdmzdkl3kruvte.png)
Total Area of Semicircles PS and QR combined =
![2((1)/(2) \pi r^(2) )\\=\pi r^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/iybwal74xnax1105mtu9gnkjinvn8qubou.png)
We know that the diameter of PS = QR = 21 cm (due to the length of the square)
Radius = Half of Diameter = 0.5 x 21cm = 10.5cm
Total Area of Semicircles PS and QR =
![\pi (10.5)^(2) \\=110.25\pi cm^(2)](https://img.qammunity.org/2023/formulas/mathematics/high-school/7in38a49o4c9cn3n7j0l82vuboskv7og52.png)
Finally,
Area of Shaded Region = Area of Rectangle PQRS - Total Area of Semicircles PS and QR
=
![441 - 110.25\pi\\= 94.6cm^(2) (1dp)](https://img.qammunity.org/2023/formulas/mathematics/high-school/ltt6r3blcmg168o74548env5nzwizghzkl.png)
In this case , you can choose the nearest answer as there might be some rounding differences.