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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 $BE$ if $BE=EC$.

[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(

 

2)

In $\triangle ABC,$ we know the side lengths $AB=9\sqrt 2$$BC=10\sqrt 2$$CA=11\sqrt 2$. Find the height of $\triangle ABC$ from $A$ to $BC.$

 Nov 2, 2017
 #1
avatar+7352 
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

 Nov 2, 2017
 #2
avatar+98173 
+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

 

 

cool cool cool

 Nov 2, 2017
edited by CPhill  Nov 3, 2017

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