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

 

 May 1, 2022
 #1
avatar+220 
+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=\(24\sqrt {73}\over 73\)

 May 1, 2022
 #2
avatar+1365 
+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 \(\color{brown}\boxed{24 \sqrt{73} \over 73}\)

 May 1, 2022

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