| (i) Area of inscribed circle |
| (ii) Area of circumscribed circle |
| (iii) Area of shaded region |
A)
(i) (ii) (iii) \[\frac{1100}{7}c{{m}^{2}}\] \[\frac{550}{7}c{{m}^{2}}\] \[\frac{550}{7}c{{m}^{2}}\]
B)
(i) (ii) (iii) \[\frac{550}{7}c{{m}^{2}}\] \[\frac{1100}{7}c{{m}^{2}}\] \[\frac{550}{7}c{{m}^{2}}\]
C)
(i) (ii) (iii) \[\frac{550}{7}c{{m}^{2}}\] \[\frac{550}{7}c{{m}^{2}}\] \[\frac{1100}{7}c{{m}^{2}}\]
D)
(i) (ii) (iii) \[\frac{1100}{7}c{{m}^{2}}\] \[\frac{1100}{7}c{{m}^{2}}\] \[\frac{550}{7}c{{m}^{2}}\]
Correct Answer: B
Solution :
Side of the square \[ABCD=10\text{ }cm\] Diameter of the inscribed circle = 10 cm (i) Area of inscribed circle \[=\pi \times {{\left( \frac{10}{2} \right)}^{2}}\] \[=\frac{22}{7}\times 5\times 5=\frac{550}{7}\,c{{m}^{2}}\] Diameter of circumscribed circle = Diameter square \[ABCD=\sqrt{{{10}^{2}}+{{10}^{2}}}=10\sqrt{2}\] (ii) Area of circumscribed circle \[=\pi \times {{\left( \frac{10}{2}\sqrt{2} \right)}^{2}}\] \[=\frac{22}{7}\times 5\sqrt{2}\times 5\sqrt{2}=\frac{1100}{7}c{{m}^{2}}\] (iii) Area of shaded region = Area of circumscribed - Area of inscribed circle \[=\frac{1100}{7}-\frac{550}{7}=\frac{550}{7}\,c{{m}^{2}}\]
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