In rectangle ABCD, shown here, CE is perpendicular to BD. If BC = 3 and DC = 8, what is CE?

Guest May 1, 2022

#1**+1 **

Answer: \(24\sqrt {73}\over 73\)

Solution:

The area of the rectangle is 24. The area of BCD is half of that, which is 12.

Apply the Pythagorean theorem to find that BD²=CD²+BC²=9+64=73. BD is therefore \(\sqrt {73}\).

Multiplying BD by CE and then dividing by two should give the area of BCD (12) because BD is a side and CE is its matching altitude.

\(\sqrt {73} \cdot CE \over 2\)=12

CE=\(24\sqrt {73}\over 73\)

WhyamIdoingthis May 1, 2022

#2**+1 **

Let \(CE = x\).

By similar triangles, \(BE = {3 \over 8} x\)

Because \(\triangle BEC \) is a right triangle, we can apply the Pythagorean Theorem: \(x^2+{9 \over 64}x^2=9\)

Solving, we find \(x = \sqrt{576 \over 73}\), which can be simplified to \(\color{brown}\boxed{24 \sqrt{73} \over 73}\)

BuilderBoi May 1, 2022