# I need help on a similarity question.The triangle MNO is right-angled at N. If MN=3.4cm and NO=1.8cm, find NP, the length of the perpendicular to the hypotenuse and also the lenghts of MP and PO.

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(1) First, draw right triangle MNO, with an altitude drawn from right angle N. Note that `MO=sqrt(74/5)` by the pythagorean theorem.

(2) `Delta MNO` ~ `Delta MPN` ~ `Delta NPO` (In each case by AA~; they each have a right angle, and they either share `/_ M` or `/_N` )

(3) Let MP=x, PO=y, NP=h. Then by proportionality of sides we have:

(a):`(MN)/(MP)=(MO)/(MN)=(NO)/(PN)` or `3.4/x=(sqrt(74/5))/3.4=1.8/h`

(b): `(MP)/(NP)=(MN)/(NO)=(PN)/(PO)` or `x/h=3.4/1.8=h/y`

(c): `(MN)/(NP)=(MO)/(NO)=(NO)/(PO)` or `3.4/h=(sqrt(74/5))/1.8=1.8/y`

(4) from (a) we have `3.4/x=(sqrt(74/5))/3.4`

Then `3.4^2=sqrt(74/5)x=>x~~3.004878916`

from (c) `1.8^2=sqrt(74/5)y=>y~~.8421978968`

Then from (b) `h^2=xy => h~~sqrt(3*.8)~~ 1.59081825`

**(5) So** `NP~~1.6,MP~~3.0,PO~~.8` each length in cm.

** In all cases of a right triangle with the altitude drawn to the hypotenuse, you will have the following:

Each leg is the geometric mean between the segment cut on the hypotenuse adjacent to the leg and the hypotenuse.

The length of the altitude is the geometric mean between the segments cut on the hypotenuse.