Answer :
Answer:
[tex]\text{Length of AB is }\frac{ah}{a+h}[/tex]
Step-by-step explanation:
Given △KMN, ABCD is a square where KN=a, MP⊥KN, MP=h.
we have to find the length of AB.
Let the side of square i.e AB is x units.
As ADCB is a square ⇒ ∠CDN=90°⇒∠CDP=90°
⇒ CP||MP||AB
In ΔMNP and ΔCND
∠NCD=∠NMP (∵ corresponding angles)
∠NDC=∠NPM (∵ corresponding angles)
By AA similarity rule, ΔMNP~ΔCND
Also, ΔKAP~ΔKPM by similarity rule as above.
Hence, corresponding sides are in proportion
[tex]\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1\\[/tex]
[tex]KA(\frac{h}{x}-1)=AP[/tex]
[tex]ND(\frac{h}{x}-1)=PD[/tex]
Adding above two, we get
[tex](KA+ND)(\frac{h}{x}-1)=(AP+PD)[/tex]
⇒ [tex](KN-AD)=\frac{x}{(\frac{h}{x}-1)}[/tex]
⇒ [tex]a-x=\frac{x}{(\frac{h}{x}-1)}[/tex]
⇒ [tex]a-x=\frac{x^2}{h-x}[/tex]
⇒ [tex]x^2=ah-ax-xh+x^2[/tex]
⇒ [tex]x(h+a)=ah[/tex]
⇒ [tex]x=\frac{ah}{a+h}[/tex]