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 .

2)

In we know the side lengths , , . Find the height of from to

Guest Nov 2, 2017

#1**0 **

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,

AB^{2} + BE^{2} = 120^{2} Substitute 120/35 * BE in for AB .

( 120/35 * BE )^{2} + BE^{2} = 120^{2}

576/49 * BE^{2} + BE^{2} = 14400

BE^{2}( 576/49 + 1) = 14400

BE = √[ 14400 / (576/49 + 1) ] = **33.6**

hectictar Nov 2, 2017

#2**+2 **

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

CPhill Nov 2, 2017