+0

0
49
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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
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#1
+5258
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,

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

hectictar  Nov 2, 2017
#2
+78756
+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
edited by CPhill  Nov 3, 2017

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