Solve $ac+bd-ad-bc$ Given $\dfrac{3}{4}$, $\sqrt{a^2+c^2}=15$

[anemone]

Assume that $a, b, c, d$ are positive integers and $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3}{4}$, and \(\displaystyle \sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15\), find $ac+bd-ad-bc$.

 

[kaliprasad]

Re: Find ac+bd-ad-bc

Assume that $a, b, c, d$ are positive integers and $\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{3}{4}$, and \(\displaystyle \sqrt{a^2+c^2}-\sqrt{b^2+d^2}=15\), find $ac+bd-ad-bc$.

Spoiler

Sqrt(a^2+c^2) = 5/4 c
Sqrt(b^2+d^2) = 5/4 d
so 5/4(c-d) = 15or c-d = 12
ac + bd – ad –bc
= ¾ c^2 + ¾ d^2 – ¾ cd – ¾ cd
= ¾(c-d)^2 = ¾ * 12^2 = 108