In triangle △abc, ∠abc=90°, bh is an altitude. find the missing lengths. ab=4 and bc=3, find ah, ch and bh.

AH = 3.2

CH = 1.8

BH = 2.4

I hope this helps! Brainliest please!

Step-by-step explanation:  

It can be convenient to compute the length of the hypotenuse of this triangle (AC). The Pythagorean theorem tells you ...

  AC^2 = AB^2 + CB^2

  AC^2 = 4^2 + 3^2 = 16 + 9 = 25

  AC = √25 = 5  

The altitude divides ∆ABC into similar triangles ∆AHB and ∆BHC. The scale factor for ∆AHB is ...

 scale factor ∆ABC to ∆AHB = AB/AC = 4/5 = 0.8  

And the scale factor to ∆BHC is ...  

 scale factor ∆ABC to ∆BHC = BC/AC = 3/5 = 0.6  

Then the side AH is 0.8·AB = 0.8·4 = 3.2

And the side CH is 0.6·BC = 0.6·3 = 1.8  

These two side lengths should add to the length AC = 5, and they do.  

The remaining side BH can be found from either scale factor:    

 BH = AB·0.6 = BC·0.8 = 4·0.6 = 3·0.8 = 2.4

The sides of interest are ...  

 AH = 3.2

 CH = 1.8

 BH = 2.4

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call the larger triangle abc

then the smaller triangle cde is similar to abc (all angles are equal, aaa).  

ab is the side that is 36, and we wish to find de, the side corresponding to ab

area_cde = area_abc - area_cde --> area_abc / area_cde = 2

let x=length of de

ratio of areas = (ratio of sides)^2

+2+=+%2836%2fx%29%5e2+

+2+=+1296%2fx%5e2+

++x%5e2+=+1296%2f2+=+648+

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