In triangle $ABC$, $\angle ABC = 90^\circ$ and $AD$ is an angle bisector. If $AB = 90,$ $BC = x$, and $AC = 2x - 6,$ then find the area of $\triangle ADC$. Round your answer to the nearest integer.
+128408
By the Pythagorean Theorem
AC^2 = AB^2 + BC^2
(2x - 6)^2 = 90^2 + x^2
4x^2 - 24x + 36 = 8100 + x^2 rearrange as
3x^2 - 24x - 8064 = 0 divide through by 3
x^2 - 8x - 2688 = 0 complete the square on x
x^2 - 8x + 16 = 2688 + 16
(x - 4)^2 = 2704 take the positive root
x - 4 = 52
x = 56
So BC = 56 and AC = 2(56) - 6 = 106
So...since AD is an angle bisector we have that
AB / BD = AC / DC
90 / BD = 106 / DC
So
BD/ DC = 90 / 106
So...there are 90 + 106 = 196 equal parts of BC....and DC is 106 of them
So DC = 56 ( 106 /196) = 212/ 7
So the area of triangle ADC = (1/2) (DC) ( AB) = (1/2) (212/7) * 90 ≈ 1363 units^2
Here's a pic :
