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Bunuel
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IMO D

Let the side of the bigger Square be S and Smaller square be D

Diagonal length = S * Sqrt (2) therefore, diagonal of smaller square D (Sqrt 2) = S * Sqrt(2) / 3 ......implies D = S/3

Area of Bigger square = S^2 and Area of Smaller square = (S/ 3) ^ 2 >> S^2/9

Ratio = S^2 : S^2/9 >> 9:1
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Bunuel

In the figure shown, the small square divides the diagonal of the large square into three equal parts. What is the ratio of the area of the large square to the area of the small square?

A. 2:1
B. 3:1
C. 6:1
D. 9:1
E. 18:1


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Attachment:
2020-10-27_12-30-55.png

Since we are only given diagonals of square. Helpful formula is Area of Square = ½ × d2

Smaller Sq = 1/2 x x^2
Larger Sq = 1/2 x (3x)^2 = 1/2 x 9x^2

Asked ratio Larger/Smaller = 1/2 x 9x^2 / 1/2 x x^2 = 9/1 and D
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Given:
Diagonal of larger square = 3x
Diagonal of smaller square = x

Area of a square = \(\frac{1}{2 }\)* \((d^2)\)

Area of larger square = \(\frac{1}{2}\) * \(9x^2 \)

Area of smaller square = \(\frac{1}{2}\) * \(x^2\)

Ratio of area = 9:1

Ans: D
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Area of square = (1/2)*d^2, where d is the length of the diagonal

Area of large square = (1/2)*(3x)^2
Area of small square = (1/2)*x^2

Ratio = (9x^2)/(x^2)
Ratio = 9:1

Choice D
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