Hello, my dear brilliant members. Well, I wanted to give you some simple challenges. These are 2 of my new formulas which are (A)Finding the height of any triangle with sides a,b,c in terms of a, b and c. Obviously, there can be 3 heights but you find for base b. (B) Finding the length of median of any triangle of side a, b, c in terms of a, b and c and the median should be bisecting side b. Well, it is extremely easy and the challenge is that you find these formulas with proper proofs. Don't Google it as I am pretty sure you all can do it. Thank you and everyone comment on this, try and find the formulas yourself and post it in the comments.

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TopNewestu suck – Ketan Bansal · 2 years ago

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Proofs:(A) Take \(\triangle ABC\) with an altitude \(BD\) of length \(h\) perpendicular to \(b\). Let \(DC=x\), and thus, \(AD=b-x\). Now, by Pythagoras Theorem, \(a^{2}-x^{2}=c^{2}-(b-x)^{2}\) \(\Rightarrow\) \(a^{2}+b^{2}-c^{2}=2bx\) \(\Rightarrow\) \(x=\frac{a^{2}+b^{2}-c^{2}}{2b}\)Also, \(x^{2}+h^{2}=a^{2}\) \(\Rightarrow\) \(h^{2}=a^{2}+x^{2}\)

Substituting for \(x\) into the first equation, \(h^{2}=c^{2}-(b-\frac{a^{2}+b^{2}+c^{2}}{2b})^{2}\)

By simplifying and bringing to the same denominator, we get \(h=\sqrt{\frac{2b^2c^2 + 2c^2b^2 + 2a^2b^2-a^4-b^4-c^4}{4b^2}}\) \(\Rightarrow\) \(h=\frac{\sqrt{2b^2c^2+2c^2a^2+2a^2b^2-a^4-b^4-c^4}}{2b}\), which can be simplified, as shown in my other comment. – Nanayaranaraknas Vahdam · 3 years, 2 months ago

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– Kushagra Sahni · 3 years, 2 months ago

It still can be simplified. Apply the identity a^2-b^2Log in to reply

– Kushagra Sahni · 3 years, 2 months ago

And yes. Post a proof for median one as wellLog in to reply

I have gotten both formulae.

(A) \(h_b = \frac{\sqrt{2a^{2}b^{2}+2b^{2}c^{2}+2c^{2}a^{2}-a^{4}-b^{4}-c^{4}}}{2b}\), where \(h_b\) denotes the altitude incident on \(b\)

(B) \(x_b = \frac{\sqrt{2a^{2}+2c^{2}-b^{2}}}{2}\), where \(x_b\) denotes the median incident on \(b\)

Should I post a proof? – Nanayaranaraknas Vahdam · 3 years, 2 months ago

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– Kushagra Sahni · 3 years, 2 months ago

COME ON FRIEND, REPLY. AND PLEASE RESHARE THIS NOTELog in to reply

– Nanayaranaraknas Vahdam · 3 years, 2 months ago

Sorry, been caught up with IIT classesLog in to reply

– Kushagra Sahni · 3 years, 2 months ago

No problem friend, please reshare this note and simplify your altitude formula and post a proof for both.Log in to reply

– Kushagra Sahni · 3 years, 2 months ago

AND YES, ALSO TELL ME THAT ARE THESE GOOD DISCOVERIES OR VERY EASY ONES ?Log in to reply

– Kushagra Sahni · 3 years, 2 months ago

YOUR MEDIAN ONE IS CORRECT, BUT THE ALTITUDE ONE CAN BE SIMPLIFIED FURTHER. SIMPLIFY IT USING SUITABLE IDENTITIES AND THEN POST A PROOF FOR BOTHLog in to reply

– Nanayaranaraknas Vahdam · 3 years, 2 months ago

The altitude formula can be simplified to \(h_b=\frac{\sqrt{(a^{2}+b^{2}+c^{2})^{2}-4(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})}}{2b}\)Log in to reply