QUANTITATIVE APTITUDE

The largest square that can be inscribed in a right angled triangle ABC when one of its vertices lies on the hypotenuse of the triangle

Solution 1:

From the above diagram,  Δ$\mathrm{\Delta }$ABC and Δ$\mathrm{\Delta }$AFD are similar.
⇒Tanα=ADFD=ABBC$\Rightarrow Tan\alpha ={\displaystyle \frac{AD}{FD}}={\displaystyle \frac{AB}{BC}}$
⇒ADx=ab$\Rightarrow {\displaystyle \frac{AD}{x}}={\displaystyle \frac{a}{b}}$
⇒AD=xab$\Rightarrow AD={\displaystyle \frac{xa}{b}}$ - - - - - (1)
Also, Δ$\mathrm{\Delta }$ABC and Δ$\mathrm{\Delta }$EGC are similar.
⇒Tanα=GEEC=ab$\Rightarrow Tan\alpha ={\displaystyle \frac{GE}{EC}}={\displaystyle \frac{a}{b}}$
⇒xEC=ab$\Rightarrow {\displaystyle \frac{x}{EC}}={\displaystyle \frac{a}{b}}$
⇒EC=xba$\Rightarrow EC={\displaystyle \frac{xb}{a}}$ - - - - - (2)
We know that c= AD + x + EC
⇒c=xab+x+xba$\Rightarrow c={\displaystyle \frac{xa}{b}}+x+{\displaystyle \frac{xb}{a}}$
⇒c=x(ab+1+ba)$\Rightarrow c=x\left({\displaystyle \frac{a}{b}}+1+{\displaystyle \frac{b}{a}}\right)$
⇒c=x(a2+ab+b2ab)$\Rightarrow c=x\left({\displaystyle \frac{a^2+ab+b^2}{ab}}\right)$
⇒x=(abca2+ab+b2)$\Rightarrow x=\left({\displaystyle \frac{abc}{a^2+ab+b^2}}\right)$
Side of the square = abca2+b2+ab${\displaystyle \frac{abc}{a^2+b^2+ab}}$

Solution 2:

From the above diagram, drop a perpendicular to AC from vertex B. 

Area of Δ$\mathrm{\Delta }$ABC = 12×a×b=12×BF×c${\displaystyle \frac{1}{2}}\times a\times b={\displaystyle \frac{1}{2}}\times BF\times c$

∴BF=abc$\therefore BF={\displaystyle \frac{ab}{c}}$ - - - - - (1)

Now Δ$\mathrm{\Delta }$BDE and Δ$\mathrm{\Delta }$ABC are similar. 

⇒BGBF=DEAC$\Rightarrow {\displaystyle \frac{BG}{BF}}={\displaystyle \frac{DE}{AC}}$

⇒abc−xabc=xc$\Rightarrow {\displaystyle \frac{{\displaystyle \frac{ab}{c}}-x}{{\displaystyle \frac{ab}{c}}}}={\displaystyle \frac{x}{c}}$

⇒1−xcab=xc$\Rightarrow 1-{\displaystyle \frac{xc}{ab}}={\displaystyle \frac{x}{c}}$

⇒1=xcab+xc$\Rightarrow 1={\displaystyle \frac{xc}{ab}}+{\displaystyle \frac{x}{c}}$

⇒1=x(cab+1c)$\Rightarrow 1=x\left({\displaystyle \frac{c}{ab}}+{\displaystyle \frac{1}{c}}\right)$

⇒1=x(c2+ababc)$\Rightarrow 1=x\left({\displaystyle \frac{c^2+ab}{abc}}\right)$

⇒x=abcc2+ab$\Rightarrow x={\displaystyle \frac{abc}{c^2+ab}}$ = abca2+b2+ab