Do you know your ABCD's?

$\begin{aligned} (a^2 + b^2)(c^2 + d^2) = (ac - bd)^2 + (ad + bc)^2 & = 85^2 \\ |ad + bc|^2 & = 85^2 - 40^2 \implies |ad + bc| = \boxed{75} \end{aligned}$

Consider two complex numbers $z_1=(a+ib)$ and $z_2=(c+id)$ . Therefore, $z_1z_2=(ac-bd)+i(ad+bc)$ .

In polar form $z_1z_2=re^{i \theta}$ , where $r=\sqrt{a^2+b^2}\sqrt{c^2+d^2}$ .

From the given data, $r=85$ . Also, by definition, $r^2=(ac-bd)^2+(ad+bc)^2$ , i.e.,

$(ad+bc)^2=85^2-40^2\\ =(85+40)(85-40)\\ =125 \times 45\\ =125 \times 5 \times 9\\ =(25 \times 3)^2\\ =75^2$ .

Therefore $|ad+bc|=75$ .

A little observation and you'll find (a,b,c,d) as (6,7,9,2). So |ad+bc|= |6x2+7x9|= 75.

Here's yet another way of solving this: since I've spent perhaps too much time doing linear algebra, these expressions looked like dot products and determinants. Consider the following matrix product: $\begin{bmatrix} a & b \\ d & c \end{bmatrix} \begin{bmatrix} a & d \\ b & c \end{bmatrix} = \begin{bmatrix} a^2 + b^2 & ad + bc \\ ad + bc & c^2 + d^2 \end{bmatrix}.$ Using the fact that $a^2 + b^2 = c^2 + d^2 = 85$ , letting $x = ad + bc$ , and taking the determinant of both sides, we have: $\begin{aligned} \begin{vmatrix} a & b \\ d & c \end{vmatrix} \begin{vmatrix} a & d \\ b & c \end{vmatrix} &= \begin{vmatrix} 85 & x \\ x & 85 \end{vmatrix} \\ (ac - bd)^2 &= 85^2 - x^2 \\ x^2 &= 5625 \end{aligned}$ Thus $x = \pm 75$ , and therefore $|ad + bc| = |x| = \boxed{75}$ .

$a^2+b^2=c^2+d^2=85$

$c^2=85-d^2$

$d^2=85-c^2$

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$|ac-bd|=40$

$(|ac-bd|)^2=40^2$

$a^2c^2+b^2d^2-2abcd=40^2$

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$a^2(85-d^2)+b^2(85-c^2)-2abcd=40^2$

$85a^2-a^2d^2+85b^2-b^2c^2-2abcd=40^2$

$85a^2+85b^2-(a^2d^2+b^2c^2+2abcd)=40^2$

$85(a^2+b^2)-(a^2d^2+b^2c^2+2abcd)=40^2$

$85(85)-(a^2d^2+b^2c^2+2abcd)=40^2$

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$(a^2d^2+b^2c^2+2abcd)=85^2-40^2$

$(ad+bc)^2=75^2$

$ad+bc=\pm75$

$|ad+bc|=\boxed{75}$

a=6. b=7, c=9 and d=2, There fore, /ad +bc/ = 6x2 + 7x9 = 12 + 63 =75

$a=\sqrt{85}\cos {t}$ , $b=\sqrt{85}\sin {t}$ , $d=\sqrt{85}\cos {s}$ , $c=\sqrt{85}\sin {s}$ , $ac-bd=85\cos {(t+s)}$ , $ad+bc=85\sin {(t+s)}$ , $cos {(t+s)}=\frac{40}{85}$ , $ad+bc=85\sqrt{1-{cos {(t+s)}}^2}=75$