Question Number 211953 by efronzo1 last updated on 25/Sep/24
Commented by BHOOPENDRA last updated on 25/Sep/24
$${h}=\frac{\sqrt{\mathrm{2}}\:\left(\sqrt{{ab}}\:−{a}\:+{b}−{c}\right)}{\:\sqrt{\left(\sqrt{{ab}}\right)+{b}}} \\ $$
Commented by BHOOPENDRA last updated on 26/Sep/24
Answered by BHOOPENDRA last updated on 26/Sep/24
![△EFG∼△BCG area ratio [EFG:BCG]=a:b and altitude PG:GQ=(√a):(√b) PG=(√a) n,GQ=(√b) n PQ=((√a)+(√b) )n area of BCG=((BC.GQ)/2)⇒((((√a)+(√b) )(√b) n^2 )/2)=b n=(√((2(√b))/( ((√a)+(√b)))))⇒PQ=(√(2((√(ab))+b))) Area of ABF=ABCD−a−b−c−(√(ab)) ABF=((h(√(2((√(ab))+b))))/2) h=(((√2) ((√(ab))−a+b−c))/( (√(((√(ab))+b)))))](https://www.tinkutara.com/question/Q211984.png)
$$\bigtriangleup{EFG}\sim\bigtriangleup{BCG} \\ $$$${area}\:{ratio}\:\left[{EFG}:{BCG}\right]={a}:{b} \\ $$$${and}\:{altitude}\:{PG}:{GQ}=\sqrt{{a}}:\sqrt{{b}}\: \\ $$$${PG}=\sqrt{{a}}\:{n},{GQ}=\sqrt{{b}}\:{n} \\ $$$${PQ}=\left(\sqrt{{a}}+\sqrt{{b}}\:\right){n} \\ $$$${area}\:\:{of}\:{BCG}=\frac{{BC}.{GQ}}{\mathrm{2}}\Rightarrow\frac{\left(\sqrt{{a}}+\sqrt{{b}}\:\right)\sqrt{{b}}\:{n}^{\mathrm{2}} }{\mathrm{2}}={b} \\ $$$${n}=\sqrt{\frac{\mathrm{2}\sqrt{{b}}}{\:\left(\sqrt{{a}}+\sqrt{{b}}\right)}}\Rightarrow{PQ}=\sqrt{\mathrm{2}\left(\sqrt{{ab}}+{b}\right)} \\ $$$${Area}\:{of}\:{ABF}={ABCD}−{a}−{b}−{c}−\sqrt{{ab}} \\ $$$${ABF}=\frac{{h}\sqrt{\mathrm{2}\left(\sqrt{{ab}}+{b}\right)}}{\mathrm{2}} \\ $$$${h}=\frac{\sqrt{\mathrm{2}}\:\left(\sqrt{{ab}}−{a}+{b}−{c}\right)}{\:\sqrt{\left(\sqrt{{ab}}+{b}\right)}} \\ $$$$ \\ $$