Question

Write a Pythagorean triplet whose one member is 50

Answer 1

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{The \ required \ pythagorean \ triplet \ is}$

$\sf{50, \ 2\sqrt{373}, \ 14\sqrt7.}$

$\sf\pink{To \ find:}$

$\sf{Pythagorean \ triplet \ whose \ one \ member \ is \ 50.}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{By \ identity}$

$\sf{(a+b)^{2}=a^{2}+b^{2}+2ab}$

$\sf{i.e. (a+b)^{2}=(a^{2}+b^{2})+2ab}$

$\sf{Also,}$

$\sf{Hypotenuse^{2}=(Side \ 1)^{2}+(Side \ 2)^{2}}$

$\sf{On \ comparing \ both, \ we \ get}$

$\sf{Hypotenuse^{2}=(a+b)^{2}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\boxed{\sf{Hypotenuse=a+b}}$

$\sf{(Side \ 1)^{2}=a^{2}+b^{2}}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\boxed{\sf{Side \ 1=\sqrt{a^{2}+b^{2}}}}$

$\sf{(Side \ 2)^{2}=2ab}$

$\sf{On \ taking \ square \ root \ of \ both \ sides}$

$\boxed{\sf{Side \ 2=\sqrt{2ab}}}$

$\sf{Here, \ a \ and \ b \ can \ be \ any \ integer.}$

$\sf{Here \ we \ need \ any \ one \ term \ 50.}$

$\sf{\therefore{Let \ a=14 \ and \ b=36}}$

$\sf{\therefore{a+b=14+36=50,}}$

$\sf{\sqrt{a^{2}+b^{2}}=\sqrt{14^{2}+36^{2}}}$

$\sf{=\sqrt{196+1296}}$

$\sf{=\sqrt{1492}}$

$\sf{\therefore{\sqrt{a^{2}+b^{2}}=2\sqrt{373},}}$

$\sf{\sqrt{2ab}=\sqrt{2\times14\times36}}$

$\sf{=\sqrt{2\times2\times7\times9\times2\times2}}$

$\sf{\therefore{\sqrt{2ab}=12\sqrt7}}$

$\sf\purple{\tt{\therefore{The \ required \ pythagorean \ triplet \ is}}}$

$\sf\purple{\tt{50, \ 2\sqrt{373}, \ 14\sqrt7.}}$

______________________________________

$\sf\blue{\underline{\underline{Verification:}}}$

$\sf{50^{2}=2500...(1)}$

$\sf{(2\sqrt{373})^{2}+(12\sqrt7)^{2}=1492+1008}$

$\sf{\therefore{(2\sqrt{373})^{2}+(12\sqrt7)^{2}=2500...(2)}}$

$\sf{...From \ (1) \ and \ (2), \ we \ get}$

$\sf{50^{2}=(2\sqrt{373})^{2}+(12\sqrt7)^{2}}$

$\sf{Hence, \ the \ required \ pythagorean \ triplet \ is}$

$\sf{50, \ 2\sqrt{373}, \ 12\sqrt7.}$