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The figure above shows four adjacent small squares, forming
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24 Feb 2020, 12:40
24 Feb 2020, 12:40
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#greprepclub The figure above shows four adjacent small squares, .jpg [ 19.96 KiB | Viewed 3414 times ]
The figure above shows four adjacent small squares, forming one large square. The vertices of square RSTU are midpoints of the sides of the small squares. What is the ratio of the area of RSTU to the area of the large outer square
A. \(\frac{1}{2}\)
B. \(\frac{5}{9}\)
C. \(\frac{7}{12}\)
D. \(\frac{3}{5}\)
E. \(\frac{5}{8}\)
Kudos for the right answer and explanation
Re: The figure above shows four adjacent small squares, forming
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25 Feb 2020, 00:30
25 Feb 2020, 00:30
2
Since all the adjacent small squares have equal length therefore let the side of one small square be 2 units. Side of larger outer square would be 4 units.
Area of larger square will be 16 sq units.
Now,since R,S,T and U are the mid points of the adjacent square thus it is divided further into 1 unit each.
Let outer square be ABCD. Consider triangle SAT. At will be 2+1(as T is the midpoint ) = 3 units. Apply pythagoras theorem and compute ST as the sum of the squares of other 2 sides which comes out be root 10.
area of RSTU will be 10.
HENCE ratio will be 10:16=5:8(option E)
Area of larger square will be 16 sq units.
Now,since R,S,T and U are the mid points of the adjacent square thus it is divided further into 1 unit each.
Let outer square be ABCD. Consider triangle SAT. At will be 2+1(as T is the midpoint ) = 3 units. Apply pythagoras theorem and compute ST as the sum of the squares of other 2 sides which comes out be root 10.
area of RSTU will be 10.
HENCE ratio will be 10:16=5:8(option E)
Re: The figure above shows four adjacent small squares, forming
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02 Aug 2020, 12:08
02 Aug 2020, 12:08
1
let side of square be = 4
Mid point makes small square side = 2
thus PQRS is mind point of that so the side is split again into measurements of 1 and 1.
Thus if consider TU as one side we get --> 1 side as 1 and other as 3 (one small square side(2) + half of small square(1))
hence Hyptnuse (which is the side for inner sqre) = 3 sq + 1 sq = 9 + 1 = [/square_root] 10
Area of inner sqaure = 10
Out square = 16
proportion 10/16 = 5/8
Ans E.
Kudos if you like this post!
Mid point makes small square side = 2
thus PQRS is mind point of that so the side is split again into measurements of 1 and 1.
Thus if consider TU as one side we get --> 1 side as 1 and other as 3 (one small square side(2) + half of small square(1))
hence Hyptnuse (which is the side for inner sqre) = 3 sq + 1 sq = 9 + 1 = [/square_root] 10
Area of inner sqaure = 10
Out square = 16
proportion 10/16 = 5/8
Ans E.
Kudos if you like this post!
