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Rectangle

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In rectangle ABCD, shown here, CE is perpendicular to BD . If BC = sqrt(3) and DC = 2, what is CE?

May 1, 2022

#1
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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.

May 2, 2022

#1
+2540
0

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.

BuilderBoi May 2, 2022