+618 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 $DEC$.
+128460
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
+128460
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
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
