#Geometry Geometry and Progressions! :) · Level pending (100 points)
In an isosceles right triangle, ABC,with the right angle at C, the length of a leg is 20 inches.Perpendiculars are dropped from C to the hypotenuse,from the foot of this perpendicular o leg CB,from the foot of this perpendicular to AB, then back to leg CB,etc. without end.Find the sum of the lengths of all the perpendiculars.
NOTE: write your answer in two four decimal places
The answer is 48.28427.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Firstly, observe that the triangle is RIGHT isosceles, hence the length of perpendicular from C on AB is same as the half of the length of AB. Property used is, in any isosceles triangle, the perpendicular from the vertex cuts the opposite side in half and is also the angle bisector of the vertex angle.
Say the point was D, hence CD is the perpendicular ( and length CD=10*sqrt(2)). Now from D, another perpendicular is dropped on CB, and using the same method as above, we can have the length as 10.
Now drop on more perpendicular, and get that the length is 5*sqrt(2).
See that the length of the perpendiculars make a GP with a = 10*sqrt(2), and r = 1/sqrt(2). Also that the GP is infinite.
Hence the sum required is sum S of GP.
S = a/(1-r)
This wields the answer as 48.28427.