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# geometry

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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
+302
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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
+2541
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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