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# Medians help

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Triangle $ABC$ has $AB=BC=5$ and $AC=6$. Let $E$ be the foot of the altitude from $B$ to $\overline{AC}$ and let $D$ be the foot of the altitude from $A$ to $BC$. Compute the area of triangle $DEC$.

michaelcai  Aug 7, 2017
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#1
+76288
+1

Look at the following pic :

Note that  angle ADC  = angle BEC  and angle BCE  = angle DCA

Therefore.....by AA congruency, triangle BEC  is similar to triangle ADC

And triangle BEC  is a Pythagorean Triple 3-4-5 right triangle  with EC  = 3  and BC  = 5

So

And  DC / CA  =  EC / CB

DC / 6  =  3 / 5

DC  =  18 / 5  =   3.6

And the sine of  angle  BCE  = 4/5  = sine  of angle DCE

Therefore....the area of triangle  DEC  = (1/2) EC * DC sinDCE  =

(1/2) (3) (3.6) (4/5)  = 3 * 1.8 * .8  = 2.4 * 1.8  =  4.32  units^2

CPhill  Aug 7, 2017
edited by CPhill  Aug 7, 2017
#2
+76288
0

Here's a way to solve this using no trig at all....

Refer to the following pic :

Note that  angle ADC  = angle BEC  and angle BCE  = angle DCA

Therefore.....by AA congruency, triangle BEC  is similar to triangle ADC

And triangle BEC  is a Pythagorean Triple 3-4-5 right triangle  with EC  = 3  and BC  = 5

So

And  DC / CA  =  EC / CB

DC / 6  =  3 / 5

DC  =  18 / 5  =   3.6

Now......draw EF parallel to BC.....angle EFC = angle BEC and angle FCE = angle BCE....so by AA congruency, triangle FCE  is similar to triangle ECB......and the hypotenuse of FCE = CE = 3.....and the hypotenuse of ECB = CB = 5.....so the scale factor of triangle FCE to triangle ECB  = 3/5

And in triangle ECB the side opposite angle EBC = 3  and the side opposite the similar angle FEC in triangle FCE - FC -  is 3/5 ths of this  = (3/5) (3)  = 9/5  = 1.8.......but  DC was shown to be 3.6....so DC - FC = DF ...and..... 3.6 - 1.8  = 1.8  = DF

So...by SAS..triangle DFE is comgruent to triangle CFE...so....angle EDF = angle ECF.....but amgle ECF  = angle CAB...so angle EDF = angle CAB

So....in triangles ABC and DEC angles BCA, BAC  are equal to angles EDC, ECD.....thus, by AA congruency, triangle ABC  is similar to triangle DEC

But the ratio of the base of triangle DEC  to triangle ABC =  3.6 / 6  =  3 / 5.....so this is the scale factor of triangle DEC to triangle ABC

And the area of ABC  = (1/2) * 6 * 4  = 12 units^2

So....the area of DEC will be equal to

(scale factor of triangle DEC to triangle ABC)^2 * area of ABC  =

(3/5)^2 * 12  =

(9 / 25) * 12  =

108 / 25  =

4 + 8/25  =

4 + 32/100 =

4.32  units^2

CPhill  Aug 8, 2017
#3
+1

Here's another way.

Pythagoras says

BD^2 + AD^2 = 5^2 = 25

CD^2 + AD^2 = 6^2 = 36

Subtract, CD^2 - BD^2 = 36 - 25 = 11

(CD - BD)(CD + BD) = 11,

but CD + BD = 5,

so CD - BD = 11/5.

Solving CD + BD = 5 and CD - BD = 11/5 simultaneously, CD = 18/5 and BD = 7/5.

The triangles AED and DEC have the same area, (same base length and height), so

area BAD + 2*area DEC = area ABC = 6*4/2 = 12. (BE = 4).

area BAD = BD*AD/2 = (7/5)*sqrt(25 - (7/5)^2)/2 = (7/5)*(24/5)/2 = 84/25

2*area DEC = 12 - 84/25 = 216/25,

so area DEC = 108/25 = 4.32 .

Tiggsy

Guest Aug 8, 2017

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