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Farmer Pythagoras has a field in the shape of a right triangle. The ri

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New post 17 Nov 2018, 10:09
2
2
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

18% (01:53) correct 82% (04:15) wrong based on 15 sessions

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GMATbuster's Weekly Quant Quiz#9 Ques 6


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Farmer Pythagoras has a field in the shape of a right triangle. The right triangle's legs have lengths 3 and 4 units. In the corner where those sides meet at a right angle, he leaves a small unplanted square S so that from the air it looks like the right angle symbol. The rest of the field is planted. The shortest distance from S to the hypotenuse is 2 units. What fraction of the field is planted?

A) 25/27
B) 26/27
C) 73/75
D) 145/147
E) 74/75

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New post 26 Nov 2018, 06:51
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New post 17 Nov 2018, 11:35
Since the shortest distance is the perpendicular to hypotenuse say x,
1/2(5)(x)= 1/2(3)(4)
Thus x= 12/5= 2.4
Thus length of diagonal of square= 2.4-2= 0.4
Thus area of square= 0.4x0.4= 0.16
Thus area left for farming= 6-0.16= 5.84
Fraction= 5.84/6= 73/75
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Re: Farmer Pythagoras has a field in the shape of a right triangle. The ri  [#permalink]

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New post 17 Nov 2018, 12:00
From the image attached, we have found diagonal of square=(2/5)
2(side^2)=(diagonal^2) (In a square, relation between diagonal and side)
side^2= area of square S =2/25
total area of the field=(1/2)*4*3=6 ((1/2*base*height))
area planted=6-(2/25)=148/25
fraction of the field planed=(148/25)/6=74/75
IMO:OPTION E
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Re: Farmer Pythagoras has a field in the shape of a right triangle. The ri  [#permalink]

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New post 17 Nov 2018, 12:03
Let the side of square S be x. Hypotenuse is 5 from Pythagoras theorem. Draw lines from top right upper vertex of S to two end points of the hypotenuse of the triangle. So, now we have divided this into square S + 1 thin triangle with base x and height (3-x) + one thin triangle with base x and height (4-x) + one triangle with base 5 and height 2. All this must add up to total area of 0.5*3*4 = 6

Hence, \(x^2 + 0.5*x*(3-x) + 0.5*x*(4-x) + 0.5*2*5 = 6\), which gives

\(x^2 + 1.5*x - 0.5*x^2 + 2*x - 0.5*x^2 + 5 = 6\)

\(3.5*x = 1\) or \((7/2)*x = 1\) or x = 2/7

Area of S = \((2/7)^2\) = 4/49

Area of planted region = 6 - (4/49)

Fraction needed = [6 - (4/49)]/6 = 1 - \(\frac{4}{(49*6)}\) = 1 - \(\frac{2}{147}\) = 145/147

Hence Option D
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Re: Farmer Pythagoras has a field in the shape of a right triangle. The ri   [#permalink] 17 Nov 2018, 12:03
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