1)
The diagram is not drawn to scale, but the measurements of the line segments and the right angles are correctly labeled. Find the length of 
if 
.
![[asy] size(6cm);pair A,B,C,D,EE;B=(0,0);A=(0,6);EE=(3.4,0);C=(6.8,0);D=(6.8,1);draw(A--B--C--D--EE--A);draw(rightanglemark(C,B,A,10));draw(rightanglemark(D,C,B,10));draw(rightanglemark(D,EE,A,10));label(](/api/ssl-img-proxy?src=https%3A%2F%2Flatex.artofproblemsolving.com%2F3%2F9%2F8%2F398d187fbcf22d8df8eadf9cd55bcd1e6b47e7e9.png)
2)
In 
we know the side lengths 
, 
, 
. Find the height of 
from 
to 
+9481 
Since they form a straight line,
∠AEB + 90° + ∠DEC = 180° → ∠AEB = 90° - ∠DEC
Since there are 180° in a triangle,
∠EDC + 90° + ∠DEC = 180° → ∠EDC = 90° - ∠DEC
Therefore, ∠AEB = ∠EDC and triangle AEB is similar to triangle EDC by AA similarity.
So...
AB = 120/35 * EC And EC = BE
AB = 120/35 * BE
And from the Pythagorean theorem,
AB2 + BE2 = 1202 Substitute 120/35 * BE in for AB .
( 120/35 * BE )2 + BE2 = 1202
576/49 * BE2 + BE2 = 14400
BE2( 576/49 + 1) = 14400
BE = √[ 14400 / (576/49 + 1) ] = 33.6
+130071
Here's one method....but....maybe not the best one....!!!
Let's find the area of a triangle with side lengrhs of AB = 9, BC =10 and CA = 11
Using Heron's Formula the semi-perimeter is [ 30] / 2 = 15
And the area of this triangle will be
sqrt [ 15 * 6 * 5 * 4 ] = sqrt [ 1800] = 30sqrt (2)
So......the height of this triangle will be
30sqrt (2) = 10 * height / 2
30sqrt (2) = 5 * height
height = [30sqrt (2) / 5] = 6sqrt (2)
But....the given triangle has a scale factor of sqrt (2) of this triangle
So....the height of ABC drawn from A to BC = 6 sqrt (2) * sqrt(2) =
6 * 2 =
12 units
