Answer to Question #305607 in Geometry for Mike

Question #305607

A square ABCD and PQ are the midpoints of BC and CD. If AP=a and AQ=b .find in terms of a and b the directed line segment AB


1
Expert's answer
2022-03-04T05:41:10-0500

The square is presented on the picture:



Denote by xx the length of the side of the square. a=AP=AB2+BP2=x2+14x2=x52a=|AP|=\sqrt{|AB|^2+|BP|^2}=\sqrt{x^2+\frac{1}{4}x^2}=x\frac{\sqrt{5}}{2}. b=AQ=x52b=|AQ|=x\frac{\sqrt{5}}{2} . Thus, we received that a=b=x52a=b=x\frac{\sqrt{5}}{2}. Thus, we get: x=2a55=2b55x=\frac{2a\sqrt{5}}{{5}}=\frac{2b\sqrt{5}}{{5}}. The length of ABAB is 2a55=2b55\frac{2a\sqrt{5}}{{5}}=\frac{2b\sqrt{5}}{{5}}

Answer: 2a55=2b55\frac{2a\sqrt{5}}{{5}}=\frac{2b\sqrt{5}}{{5}}


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