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In rhombus EFGH, if HE = 13 and HP=12,

What is PE?

In rhombus EFGH, if HE = 13 and HP=12, What is PE?-example-1
User Felix Weis
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PE=5. You need to know an important property of a rhombus: the two diagonals are perpendicular to each other. Angle EPH, same as the other four angles around the center P, is equal to 90 degrees. Since we know HE=13 and HP=12, we use Pythagorean's Theorem to solve for PE=\sqrt(13^2-12^2)=5.
User SMTF
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