The area of a circle with radius a is less than the area of

Question

The area of a circle with radius  a  is less than the area of a square with sides of length  ka . Which of the following could be the value of k?

I.  \frac{5}{3}

II.  \frac{6}{3}

III.  \frac{7}{3}

A. I only
B. II only
C. III only
D. II and III only
E. I, II, and III

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zhanbo · Answer

pi*a^2 < (ka)^2 => (pi - k^2)*a^2<0 Because a^2 is not negative, pi - k^2 < 0 k^2 > pi pi is approximately 3.14 I. (5/3)^2=25/9 < 3 II 2^2 = 4 > pi III (7/3)^2 > 2^2 > pi The answer is thus (D) .

paragamrutkar · Answer

\pi * a^2 < k^2 * a^2 k^2 > \pi \approx 3.14 I. k = \frac{5}{3} ; k^2 = \frac{25}{9} < 3 < 3.14 . Therefore, k eq \frac{5}{3} II. k = \frac{6}{3} = 2 ; k^2 = 4 > 3.14 . Therefore, k could be \frac{6}{3} . III. k = \frac{7}{3} ; k could be \frac{7}{3} because \frac{7}{3} > \frac{6}{3} . D

gmatcafe · Answer

The area of the circle is pi * a^2
The area of the square is k^2 * a^2
pi * a^2 < k^2 * a^2
3,14 < k^2
since k takes only positive values, k > 1,8 (approximately)

Out of 3 choices, II and III has value greater than 1,8. So IMO the correct answer is D.

ThatDudeKnows · Answer

6/3 = 2
For II, we've got a circle with a diameter equal to the side of the square; the circle is inscribed inside the square. Of course it is smaller. II works. A and C are out.

For III, if we make the square even larger, that'll surely work, too, so we need III. B is out.

For I, make a=1. Area of the circle is  \pi  and area of the square is  \frac{5}{3}^2 = \frac{25}{9} , which is less than 3, so the circle is larger. I doesn't work. E is out.

Answer choice D.

The area of a circle with radius a is less than the area of

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The area of a circle with radius a is less than the area of

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