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Geometry Level 2

A triangle with sides k + 1 \sqrt{k+1} , k + 1 k+1 and k + 2 k+2 , where k k is a positive integer, must be

None of the others Obtuse Right angled Acute

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1 solution

Sharky Kesa
Sep 27, 2015

Since ( k + 1 ) 2 + k + 1 2 = k 2 + 3 k + 2 < ( k + 2 ) 2 (k+1)^2 + \sqrt{k+1}^2 = k^2 + 3k + 2 < (k+2)^2 , this triangle is obtuse.

How?please explain

Archiet Dev - 5 years, 1 month ago

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Pythagoras Theorem extended.

If Δ A B C \Delta ABC is acute, then

a 2 + b 2 > c 2 a^2+b^2>c^2

If Δ A B C \Delta ABC is right-angled, then

a 2 + b 2 = c 2 a^2+b^2=c^2

If Δ A B C \Delta ABC is obtuse, then

a 2 + b 2 < c 2 a^2+b^2<c^2

Sharky Kesa - 5 years, 1 month ago

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Got it ....Thnx

Archiet Dev - 5 years, 1 month ago

This could be proved by using cosine rule.

Anurag Pandey - 4 years, 4 months ago

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@Anurag Pandey Isn't it obvious from Pythagoras Theorem? Also, cosine rule is proven by Pythagoras Theorem.

Sharky Kesa - 4 years, 4 months ago

Maybe it is not like the way you think..............

Ali Imam - 11 months, 2 weeks ago

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