24. As shown in Figure 7, the area of $\square ABCD$ is 15. $P$ and $Q$ are the projections of points $A$ and $C$ on line $BD$, and $S$ and $R$ are the projections of points $B$ and $D$ on line $AC$. Given that $PQ=6$ and $RS=8$, let $BD=d$. If $d^2=m+n\sqrt{p}$ where $m, n \in \mathbf{Z}_{+}$ and $\sqrt{p}$ is in its simplest radical form, then the value of $m+n+p$ is ( ).
(A) 81
(B) 89
(C) 97
(D) 105
(E) 113
24. A.

Let $A C$ and $B D$ intersect at point $X, \angle A X D=\theta$. Then $X A=X C=\frac{3}{\cos \theta}, X B=X D=\frac{4}{\cos \theta}$.
Thus, $S_{\square A B C D}=2 X C \cdot X D \sin \theta$
$$
\begin{array}{l}
=\frac{24}{\cos ^{2} \theta} \cdot \sin \theta=15 \\
\Rightarrow 5 \sin ^{2} \theta+8 \sin \theta-5=0 \\
\Rightarrow \sin \theta=\frac{-4+\sqrt{41}}{5} .
\end{array}
$$
$$
\begin{array}{c}
\text { By } B D=B X+X D=\frac{X S}{\cos \theta}+\frac{X R}{\cos \theta}=\frac{R S}{\cos \theta} \\
\Rightarrow d^{2}=\frac{64}{\cos ^{2} \theta}=\frac{64}{\frac{8 \sin \theta}{5}}=\frac{40}{\sin \theta} \\
=\frac{200}{-4+\sqrt{41}}=32+8 \sqrt{41} .
\end{array}
$$

Therefore, $m+n+p=32+8+41=81$.