Floor of a room is of dimensions 5 m $\times$ 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Fig. 11.18. The area of floor that remains uncovered with tiles is

4.3 m^{2}

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1.57 m^{2}

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3.14 m^{2}

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2.52 m^{2}

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Solution

The correct option is C $4.3 m^{2}$
$G i v e n - A r o o m h a s d i m e n s i o n s 5 m \times 4 m . 8 \times 10 = 80 i d e n t i c a l c i r c i l a r t i l e s o f d i a m e t e r d = 50 c m = 0.5 m a r e p l a c e d i n s u c h a w a y t h a t e a c h c i r c l e t o u c h e s f o u r o t h e r s a n d t h e w a l l s t o u c h t h e t i l e s . T o f i n d o u t - T h e a r e a o n t h e r o o m n o t c o v e r e d b y t h e t i l e s . S o l u t i o n - T h e a r . r o o m = 4 \times {5 m}^{2} = 20 m^{2} a n d a r . c i r c l e s = 80 \times π \times \frac{d^{2}}{4} = 80 \times \frac{22}{7} \times \frac{0 {.5}^{2}}{4} m^{2} = 15.7 m^{2.} N o w a r . n o t c o v e r e d b y t h e c i r c l e s = a r . s q u a r e - a r .80 c i r c l e s . = (20 - 15.7) m^{2} = 4.3 m^{2} . A n s - O p t i o n A$

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Floor of a room is of dimensions 5 m × 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Fig. 11.18. The area of floor that remains uncovered with tiles is

Floor of a room is of dimensions 5 m $\times$ 4 m and it is covered with circular tiles of diameters 50 cm each as shown in Fig. 11.18. The area of floor that remains uncovered with tiles is