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07 Mar 2019, 00:24
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The hypotenuse of a right triangle is 16 ft longer than the length of the shorter leg. If the area of this triangle is exactly 120 ft², what is the length of the hypotenuse in feet?
A. 26
B. 32
C. 40
D. 64
E. 80
A. 26
B. 32
C. 40
D. 64
E. 80
07 Mar 2019, 01:25
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MartinTao
Two ways..
(1) I would recommend use of choices as a priority in such a question.
Let the hypotenuse be option A, that is 26, so one leg = 26-16=10....
So the other side = \(\sqrt{26^2-10^2}=24\)
Thus, the area = (24*10)/2=120... TRUE
so answer A
(2) Algebraic method
Let the legs be x and y, so hypotenuse = x+16 and area xy/2=120 or xy=240
by Pythagorean theorem \((x+16)^2=x^2+y^2.....x^2+32x+256=x62+y^2.......y^2=32x+256\) substitute x= 240/y from xy=240
Thus, \(y^2=32*\frac{240}{y}+256.......y^3-256y=32*240\)
We are not asked cubic functions, that is power of 3, but we can use \(y(y^2-256)=32*240.......(y-16)(y+16)y=32*240=8*24*40=(24-16)*24*(24+16)\)
so y = 24, x= 240/24=10 and hypotenuse = \(\sqrt{24^2+10^2}=26\)
A
11 Mar 2019, 06:46
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MartinTao
Another way is working with triangle properties.
We know that the sum of two sides must be greater than the third side. Let's work with this:
short leg: a
long leg: b
hypotenuse: c=a+16
This means, the long leg has to be longer than 16 ft.
Now, we factorize 240 (which is a*b).
\(240=2^4*3*5\)
\(2^4\) is \(16\), but we need greater than \(16\), so lets try \(2^3*3=24=b\).
In this case \(a=10, b=24, c=26\).
\(a+b>c...10+24=34>26\)
\(a+c>b...10+26=36>24\)
\(b+c>a...24+26=50>10\)
Answer (A)
