Inside an equi-angular triangular park there is a flower bed forming a similar triangle. Around the flower bed, runs a uniform path of such a width that the sides of the park are exactly double the corresponding sides of the flower bed. The ratio of the areas of the path to the park is

1 : 1

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1 : 2

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1 : 3

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3 : 4

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Solution

The correct option is D $3 : 4$
$\frac{a r e a o f p a r k - a r e a o f f l o w e r b e d}{a r e a o f p a r k} = \frac{\frac{\sqrt{3}}{4} \times [(2 x)^{2} - x^{2}]}{\frac{\sqrt{3}}{4} \times {(2 x)}^{2}} = \frac{3 x^{2}}{4 x^{2}} = 3 : 4$

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The dimensions of a rectangular field are 50 m by 40 m. A flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and gravelling the path at Rs. 30 and Rs. 20 per square meter, respectively, is Rs. 52,000. Find width of the gravel path.

Q. The dimensions of a rectangular field are

50 m

40 m

. A flower bed is prepared inside this field leaving a gravel path of uniform width all round the flower bed. The total cost of laying the flower bed and gravelling the path at

R s 30

and

R s 20

per square metre, respectively is

R s 52, 000

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Inside an equi-angular triangular park there is a flower bed forming a similar triangle. Around the flower bed, runs a uniform path of such a width that the sides of the park are exactly double the corresponding sides of the flower bed. The ratio of the areas of the path to the park is

The correct option is D 3:4areaofpark−areaofflowerbedareaofpark=√34×[(2x)2−x2]√34×(2x)2=3x24x2=3:4

The correct option is D $3 : 4$
$\frac{a r e a o f p a r k - a r e a o f f l o w e r b e d}{a r e a o f p a r k} = \frac{\frac{\sqrt{3}}{4} \times [(2 x)^{2} - x^{2}]}{\frac{\sqrt{3}}{4} \times {(2 x)}^{2}} = \frac{3 x^{2}}{4 x^{2}} = 3 : 4$