In rectangle ABCD, shown here, CE is perpendicular to BD. If BC = 3 and DC = 8, what is CE?
+128473 DBC is a right triangle.....so....
BD = sqrt [ DC^2 + BC^2 ] = sqrt [ 8^2 + 3^2 ] = sqrt (73)
Let EB = x and DE = sqrt (73) - x
And EBC and EDC are also right triangles
So
CE^2 = BC^2 - EB^2 = 3^2 - x^2 = 9 - x^2 (1)
and
CE^2 = DC^2 - DE^2 = 8^2 - (sqrt (73) - x )^2 = 64 - x^2 + 2xsqrt (73) - 73 =
-x^2 + 2xsqrt (73) - 9 (2)
Set (1) = (2)
9 - x^2 = -x^2 + 2 sqrt (73) x - 9 simplify
18 = 2 sqrt (73) x
9 = sqrt (73) x
x = 9 / sqrt (73) = EB
So
CE = sqrt [ BC^2 - EB^2 ] = sqrt [ 3^2 - (9/sqrt (73))^2 ] = sqrt [ 9 - (81 / 73 ) ] ≈ 2.8
