In rectangle ABCD, shown here, CE is perpendicular to BD . If BC = sqrt(3) and DC = 2, what is CE?
+2666 By the Pythagorean Theorem, \(BD = \sqrt 7\)
Let \(CE = x\) and \(DE = y \)
We have the following system:
\(x^2+y^2=4\)
\(x^2+(\sqrt7 - y)^2=3\)
Solve for x by solving this system.
+2666 By the Pythagorean Theorem, \(BD = \sqrt 7\)
Let \(CE = x\) and \(DE = y \)
We have the following system:
\(x^2+y^2=4\)
\(x^2+(\sqrt7 - y)^2=3\)
Solve for x by solving this system.
