Question

I need help asap pls and thank you ;)

Answer 1

Answer:

$\text{Length of AB is }\frac{ah}{a+h}$

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

$\frac{ND}{NP}=\frac{CD}{MP} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{AB}{PM} \\\\\frac{ND}{NP}=\frac{x}{h} \thinspace\thinspace and\thinspace\thinspace \frac{KA}{KP}=\frac{x}{h}\\\\\frac{NP}{ND}=\frac{h}{x} \thinspace\thinspace and\thinspace\thinspace \frac{KP}{KA}=\frac{h}{x}\\\\\frac{PD}{ND}=\frac{h}{x}-1 \thinspace\thinspace and\thinspace\thinspace \frac{AP}{KA}=\frac{h}{x}-1$

$KA(\frac{h}{x}-1)=AP$

$ND(\frac{h}{x}-1)=PD$

Adding above two, we get

$(KA+ND)(\frac{h}{x}-1)=(AP+PD)$

⇒ $(KN-AD)=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x}{(\frac{h}{x}-1)}$

⇒ $a-x=\frac{x^2}{h-x}$

⇒ $x^2=ah-ax-xh+x^2$

⇒ $x(h+a)=ah$

⇒ $x=\frac{ah}{a+h}$