Re: The length of sides of triangle PQR are p, q and r units. The length
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20 Feb 2020, 07:15
We got 3 types of triangles:
—> acute—\(s^{2} + t^{2} < u^{2}\)
—> obtuse — \(s^{2} + t^{2} > u^{2} \)
—right —\(s^{2} + t^{2} = u^{2}\)
Let’s say that triangle STU is right angled triangle:
\(s^{2} + t^{2} = u^{2}\)
—> In order STU to be a right angled triangle, we have to prove that triangle PQR is a triangle (p+q > r, p+r > q or r+q > p)
—> we got \((p —q)^{2} = r^{2}\)
—> p—q = r ( it shouldn’t be a triangle) —> STU is not a right-angled triangle.
Let’s say that it’s an obtuse angled triangle. —\(s^{2} + t^{2} < u^{2}\)
—> we got \((p—q)^{2} < r^{2}\)
(p—q—r)(p—q+r) < 0
(p+r—q ) — always greater than zero
—> p—q—r < 0
—> p< q+r ( it is one of the features of a triangle)
—> STU is an obtuse-angled triangle.
Let’s say that it is an acute-angled triangle \(s^{2} + t^{2} > u^{2}\)
—> we got \((p—q)^{2} > r^{2}\)
(p—q—r)( p—q+r) > 0
—> p+r—q is always greater than zero.
But —> p—q—r > 0
p > q+r ( it is WRONG)
Triangle STU can be Obtuse—angled triangle
The answer is C
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