Question #156530

ABCD is a square and P, Q are the midpoints 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 and (iv) AC.


1
Expert's answer
2021-01-25T09:53:30-0500


To find AB and AD we have to make some equations. First let us sign BP = x, AD = 2x, QD = y, AB = 2y. From the triangle ABP and from triangle AQD we can write two equations using the Pythagoras Theorem:  

4y2+x2=a24y^2+x^2=a^2

4x2+y2=b24x^2+y^2=b^2


x2=a24y2x^2=a^2−4y^2

4(a24y2)+y2=b24(a^2−4y^2)+y^2=b^2


4a216y2+y2=b24a^2−16y^2+y^2=b^2


4a215y2=b24a^2−15y^2=b^2


y2=(4a2b2)/15y^2=(4a^2−b^2)/15


y=(4a2b2)/15;2y=2(4a2b2)/15y=\sqrt{(4a^2−b^2)/15} ; 2y=2\sqrt{(4a^2−b^2)/15}


x2=a24(4a2b2)/15x^2=a^2−4(4a^2−b^2)/15


x2=(15a216a2+4b2)/15x^2=(15a^2−16a^2+4b^2)/15


x2=(4b2a2)/15x^2=(4b^2−a^2)/15


x=(4b2a2)/15;2x=2(4b2a2)/15x=\sqrt{(4b^2−a^2)/15} ; 2x=2\sqrt{(4b^2−a^2)/15}


So AB=2y=2(4a2b2)/15AB=2y=2\sqrt{(4a^2−b^2)/15}


AD=2x=2(4b2a2)/15AD=2x=2\sqrt{(4b^2−a^2)/15}


From the triangle ABD we can find BD using the Pythagoras Theorem: 


BD=AC=2(4a2b2)/15+(4b2a2)/15=BD=AC=2\sqrt{(4a^2−b^2)/15+(4b^2−a^2)/15}=


=2(4a2b2+4b2a2)/15=2(3a2+3b2)/15=2(a2+b2)/5=2\sqrt{(4a^2−b^2+4b^2−a^2)/15}=2\sqrt{(3a^2+3b^2)/15}=2\sqrt{(a^2+b^2)/5}


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