question_answer

In figure 5, PQRS is a square lawn with side \[PQ=42\text{ }m\]. Two circular flower beds are there on the sides PS and QR with center at O, the intersection of its diagonals. Find the total area of the two flower beds (shaded parts).

Answer:

Given, PQRS is a square with side \[42\text{ }m\].

Let its diagonals intersect at O.

Then,                \[OP=OQ=OR=OS\]

and                   \[\angle POS=\angle QOR=90{}^\circ \]

\[P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}\]

\[PR=(\sqrt{2}\times 42)m\]

Now,                 \[OP=\frac{1}{2}\times (\text{diagonal})=21\sqrt{2}\,m\]

\[\because \] Area of flower bed PAS = Area of flower bed QBR

\[\therefore \] Total area of the two flower beds = Area of flower bed PAS + Area of flower bed QBR

\[=2\times [Area\,\,of\,\,sector\,\,OPAS-Area\,\,of\,\,\Delta POS]\]

\[=2\times \left[ \pi {{r}^{2}}\frac{\theta }{360{}^\circ }-\frac{1}{2}{{r}^{2}}\sin \theta  \right]\]

[Where, \[\theta =90{}^\circ \]]

\[=2\times \left[ \frac{22}{7}\times {{(21\sqrt{2})}^{2}}\frac{90{}^\circ }{360{}^\circ }-\frac{1}{2}\times 21\sqrt{2}\times 21\sqrt{2} \right]\]

\[[\because \,\,sin\,\,90{}^\circ =1]\]

\[=2\times \left[ \frac{22}{7}\times 21\times 21\times 2\times \frac{1}{4}-\frac{1}{2}\times 21\times 21\times 2 \right]\]

\[=2[33\times 21-441]\]

\[=2[693-441]\]

\[=504\,\,{{m}^{2}}\]

Hence area of flower beds is \[504\text{ }{{m}^{2}}\].