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20 Feb 2020, 05:14
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
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Question Stats:
22% (02:41) correct
78%
(02:31)
wrong
based on 221
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The length of sides of triangle PQR are p, q and r units. The length of sides of triangle STU are s, t and u units. The following relations hold true:
\(p(q+r-p)=s^2\)
\(q(p+r-q)=t^2\)
\(r(p+q-r)=u^2\)
The type of triangle STU is
(A) Acute angled
(B) Right angled
(C) Obtuse angled
(D) Either A or B
(E) Either A or C
Are You Up For the Challenge: 700 Level Questions
\(p(q+r-p)=s^2\)
\(q(p+r-q)=t^2\)
\(r(p+q-r)=u^2\)
The type of triangle STU is
(A) Acute angled
(B) Right angled
(C) Obtuse angled
(D) Either A or B
(E) Either A or C
Are You Up For the Challenge: 700 Level Questions
20 Feb 2020, 08:15
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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
Posted from my mobile device
—> 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
Posted from my mobile device
27 Aug 2021, 07:06
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lacktutor
Explanation above is good but with a minor correction in conditions for both acute and obtuse angled triangles. The inequalities needs to be reversed as below:
—> acute - \(s^{2} + t^{2}> u^{2}\)
—> obtuse - \(s^2 + t^2 < u^2\)
Everything remains same in the explanation, just that obtuse must be replaced with acute and vice-versa.
This will change your posted answer from "C" to "A" (which also matches official answer)
