Question Number 69123 by TawaTawa last updated on 20/Sep/19
![](https://www.tinkutara.com/question/9409.png)
Answered by mr W last updated on 20/Sep/19
![](https://www.tinkutara.com/question/9410.png)
Commented by mr W last updated on 20/Sep/19
![A_(ABCD) =bh A_(AKD) =A A_(KLCB) =((bH)/2)−(bh−A)=A_(AKD) =A ⇒((bH)/2)−bh+A=A ⇒((bH)/2)=bh ⇒(H/2)=h ⇒(H/h)=2 (H/h)=((x+5)/x)=2 ⇒x=5](https://www.tinkutara.com/question/Q69125.png)
$${A}_{{ABCD}} ={bh} \\ $$$${A}_{{AKD}} ={A} \\ $$$${A}_{{KLCB}} =\frac{{bH}}{\mathrm{2}}−\left({bh}−{A}\right)={A}_{{AKD}} ={A} \\ $$$$\Rightarrow\frac{{bH}}{\mathrm{2}}−{bh}+{A}={A} \\ $$$$\Rightarrow\frac{{bH}}{\mathrm{2}}={bh} \\ $$$$\Rightarrow\frac{{H}}{\mathrm{2}}={h} \\ $$$$\Rightarrow\frac{{H}}{{h}}=\mathrm{2} \\ $$$$\frac{{H}}{{h}}=\frac{{x}+\mathrm{5}}{{x}}=\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{5} \\ $$
Commented by TawaTawa last updated on 20/Sep/19
![God bless you sir.](https://www.tinkutara.com/question/Q69126.png)
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$