Question

Given: ΔABC, CM ⊥ AB AC = 10, CM = 4 AM:BM = 2:5 Find: AB, CB

Answer 1

Answer:

 $AB = 7\sqrt{21}\\CB= \sqrt{541}$

Step-by-step explanation:

 Given:  Δ ABC, CM⊥ AB

    $AC = 10, CM = 4 AM:BM = 2:5$

Now, consider  $AM:BM = 2:5 = 2x:5x$

Let In ΔCMA  H=- 10 , P= 4 , B= 2x

By, Pythagoras theorem,  $H^2=P^2+B^2$

putting values we get, $10^2=4^2+(2x)^2$

⇒ $100=16+4x^2$

⇒$x^2= 21$

⇒$x= \sqrt{21}$

which gives us $AM = 2x= 2\sqrt{21}$ and  $MB = 5x= 5\sqrt{21}$

⇒$AB= 2\sqrt{21}+5\sqrt{21}$

⇒$AB= 7\sqrt{21}$

Now, Let In ΔCMB  H=- ? , P= 4 , B= 5√21

By, Pythagoras theorem,  $H^2=P^2+B^2$

putting values we get, $H^2=4^2+(5\sqrt{21})^2$

⇒ $H^2=16+525$

⇒$H^2=541$

⇒$H= \sqrt{541}$

⇒ $CB= \sqrt{541}$

Therefore,  $AB = 7\sqrt{21}\\CB= \sqrt{541}$