Answer to Question #157068 in Geometry for Kelvin Sowah

Question

ABCD is a square and P, Q are the mid points of BC, CD respectively. If AP = a and AQ = b, find in terms of a and b, the directed line segments 

I. AB

II. AD

III. BD 

IV. AC

Answer 1

$Let\:ABCD\:be\:a\:square.$

$Let\:equal\:sides\:of\:square=x$

$Given\:that\:P\:and\:Q\:are\:middle\:points\:of\:BC\:and\:CD\:respectively$

$BC=CD=x$

$BP=\frac{x}{2}\:and\:QD\:=\frac{x}{2}$

$Given\:that\:AP=a\:and\:AQ=b\:$

$Now\:in\:right\:angled\:triangle\:ABP$

$By\:pythagorous\:theorem$

$\left(AP\right)^2=\left(AB\right)^2+\left(BP\right)^2$

$a^2=x^2+\left(\frac{x}{2}\right)^2$

$a^2=\frac{5x^2}{4}$

$Now\:in\:right\:angled\:triangle\:ADQ$

$By\:pythagorous\:theorem$

$\left(AQ\right)^2=\left(AD\right)^2+\left(DQ\right)^2$

$b^2=x^2+\frac{x^2}{4}$

$b^2=\frac{5x^2}{4}$

$a^2=b^2=\frac{5x^2}{4}\:or\:a=b=\frac{\sqrt{5x}}{2}$

$\:x=\frac{2a}{\sqrt{5}}\:or\:x=\frac{2b}{\sqrt{5}}$

$I.\:AB=x=\frac{2a}{\sqrt{5}}\:or\:\frac{2b}{\sqrt{5}}$

$II.\:AD=x=\frac{2a}{\sqrt{5}}\:or\:\frac{2b}{\sqrt{5}}$

$III.\:For\:BD$

$\left(BD\right)^2=\left(BC\right)^2+\left(CD\right)^2$

$=x^2+x^2$

$=2x^2$

$=2\times\:\frac{4a^2}{5}$

$BD=\sqrt{2}\times\frac{2a}{\sqrt{5}}\:or\:\sqrt{2}\times\frac{2b}{\sqrt{5}}$

$IV.\:AC$

$For\:AC$

$\left(AC\right)^2=x^2+x^2=2x^2$

$\:AC=\frac{2\sqrt{2a}}{\sqrt{5}}\:or\:\frac{2\sqrt{2b}}{\sqrt{5}}$