Question

the length and breadth of a rectangular piece of paper are 28cm and 14cm.A semicircular portion is cut off from the breadth side and a semicircular portion is added on the length side
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Answer 1

dude, ur question is incomplete

Answer 2

Step-by-step explanation:

$length \: of \: the \: rectangular \: piece \: of \: paper \: = 28cm \\ breadth \: of \: the \: rectangular \: piece \: of \: paper \: = 14cm \\ \\ area \: of \: rectangular \: paper \: a_{1} = ({28.14}) {cm}^{2} = 392 {cm}^{2} \\ radius(r) \: of \: the \: removed \: semicircular \: portion \: = \frac{1}{2} \pi {r}^{2} \\ \\ = \frac{1}{2} \times \frac{22}{7} \times ( {14})^{2} {cm}^{2} \\ \\ = \frac{11}{7} \times 49 {cm}^{2} \\ \\ = 77 {cm}^{2} \\ \\ \\ radius \: of \: the \: semicircular \: portion \: added = \frac{28}{2} = 14cm \\ area \: (a_{3}) \: of \: the \: added \: semicircular \: potion \: = \frac{1}{2} \pi {r}^{2} \\ = \frac{1}{2} \times \frac{22}{7} \times( {14})^{2} {cm}^{2} \\ = 308 {cm}^{2} \\ \\ area \: of \: the \: shaded \: region \: = a_{1} - a_{2} + a_{3} \\ = (392 - 77 + 308) {cm}^{2} \\ = 623 {cm}^{2}$