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30 Oct 2020, 00:15
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C
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Difficulty:
15%
(low)
Question Stats:
89% (02:32) correct
11%
(03:06)
wrong
based on 71
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In the figure shown, squares A and B are in rectangle R. The area of rectangle R is 216. The width (vertical dimension) of rectangle R is 12. The area of square A is 4 times the area of square B. The sum of the areas of squares A and B is 80. What is the value of x?
A. 1
B. 3/2
C. 2
D. 5/2
E. 3
Attachment:
2020-10-27_12-34-44.png [ 16.95 KiB | Viewed 2491 times ]
30 Oct 2020, 01:13
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Bunuel
Area of the rectangle l * b = 216 and b = 12.
The length (horizontal) = 216/12 = 18
Given, that \(A^2 = 4(B)^2\) and \(A^2 + B^2 = 80\)
Substituting for \(A^2\), we get \(4(B)^2 + B^2 = 5B^2 = 80\)
\(B^2 = 16\). So B has a side length of 4
\(A^2 = 4 * 16 = 64\). So A has a side length of 8
From the figure Length of the rectangle = Length of A + Length of B + 3x
3x = 18 - 4 - 8 = 6
x = 2
Option C
Arun Kumar
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30 Oct 2020, 07:38
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The length of rectangle = 216/12 = 18 ( Since area = 216 and one width is 12)
Given, that a^2 = 4 * b------1
and a^2+b^2 = 80 -----2
solving 1 and 2 we can get
b= 4 ( smaller square side)
hence a=8 units.
now, if see carefully, a+b+3x =18
4+8+3X = 18
X= 2
Hence IMO C
Given, that a^2 = 4 * b------1
and a^2+b^2 = 80 -----2
solving 1 and 2 we can get
b= 4 ( smaller square side)
hence a=8 units.
now, if see carefully, a+b+3x =18
4+8+3X = 18
X= 2
Hence IMO C
Bunuel
