Re: In right triangle, ABC, the ratio of the longest side to the
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15 Nov 2018, 09:14
This triangle is a \(3-4-5\) pythagoras triplet.
Let us assume the sides are \(3x:4x:5x\)
In a right triangle the two sides apart from the hypotenuse are at right angle to one another
we have the area of this triangle as \(\frac{1}{2} * 3x * 4x\)
Now, we have been given that the area is between 50 and 150 not inclusive let us take one case at a time
\(\frac{1}{2}\) * \(3x * 4x = 50\)
or, \(1/2 * 12x^2 = 50\)
or, \(12x^2 = 100\)
or, \(x^2 = 100/12\)
or, \(x = 10/3.46\)
or,\(x = 2.89\)
since the smallest side is \(3x\) we have the smallest side as approximately \(8.67\)
Now if we take the area as \(150\)
we get,
\(1/2 * 3x * 4x = 150\)
\(6x^2 = 150\)
or, \(x = 5\)
therefore \(3x = 15\)
therefore if smallest side = y
we have \(8.67 < y < 15\)