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13 Apr 2018, 01:31
Kudos
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
61% (01:36) correct
39%
(01:35)
wrong
based on 79
sessions
History
Date
Time
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Not Attempted Yet
What is the perimeter of the triangle shown?
(1) The area of the triangle is 96 in^2.
(2) y = x + 4
Attachment:
2018-04-13_1227.png [ 5.79 KiB | Viewed 2922 times ]
13 Apr 2018, 03:21
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Bunuel
1) 1/2xy= 96
xy =192
x^2 + y^2 = 20^2 = 400
(x+y)^2 = x^2+y^2 + 2xy = 400 + 2x192 = 784
(x+y)^2 = 28^2
x+y = 28
xy = 192
using trial and error x=12/16 y=16/12
perimeter can be found
2) y=x+4
X^2 + (x+4)^2 = 20^2
2x^ + 16 + 8x = 400
2x^2 + 8x - 384 = 0
x^2 + 4x - 192 = 0
x^2 + 16 x - 12 x - 192 = 0
X= 12
y = 16
(D) imo
13 Apr 2018, 05:27
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Bunuel
Given in question
\(x^2+y^2=(20)^2\)
Question ask us to find parameter P,
\(P= 20+x+y\).
Either finding value of (x+y) or both x and y individually will suffice.
Statement 1 :The area of the triangle is 96 in^2.
Area =\(\frac{x*y}{2}\)
It gives\(xy= 192\).
Now using \((x+y)^2=x^2+y^2+2xy\),\(x^2+y^2=(20)^2\) and xy= 192, we can get
\((x+y)^2\)=784
Rejecting -ve value, we get \((x+y)=28\)
Perimeter can be calculated.
So Statement 1 alone is sufficient
Statement 2 :y = x + 4
Using \(x^2+y^2=(20)^2\) and y = x + 4, we get
\(x^2+(x+4)^2=(20)^2\)
\(x^2+4x-192=0\)
\(x\)= \(\frac{-4±\sqrt{4^2-4*1*(-192)}}{2*1}\)
\(x=12\) (Rejecting -ve value of x i.e -16)
Using y = x + 4, we get \(y =16\)
As we know value of x and y individually , Value of perimeter can be easily calculated.
So Statement 2 alone is Sufficient.
As Both Statement are alone sufficient , Answer is D
