10. Rectangle $ABCD$ with sides $AB=1$ and $AD=10$ serves as the base of pyramid $SABCD$, and the edge $SA=4$ is perpendicular to the base. Find a point $M$ on edge $AD$ such that triangle $SMC$ has the smallest perimeter. Find the area of this triangle.

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# Solution:

Let's lay off $A T=A S$ on the extension of edge $A B$. For any position of point $M$ on side $A D$, $T M=S M$, so the minimum value of the sum $S M+M C$ will be when $M=T C \cap A D$.

Let's introduce the notation

$A B=a, A D=b, A S=c, A M=x$. From

$\triangle T A M \sim \triangle C D M$ it follows that

$\frac{T A}{C D}=\frac{A M}{D M}, \frac{c}{a}=\frac{x}{b-x}$,

$x=\frac{b c}{a+c}=A M, \frac{a b}{a+c}=D M$.

Draw $A K \perp T C$ and connect $K$ and $S$.

$A K=\frac{T A \cdot A M}{\sqrt{T A^{2}+A M^{2}}}=\frac{c x}{\sqrt{c^{2}+x^{2}}}=$

$=\frac{b c}{\sqrt{(a+c)^{2}+b^{2}}} \cdot S K=\sqrt{A S^{2}+A K^{2}}=\sqrt{c^{2}+\frac{b^{2} c^{2}}{(a+c)^{2}+b^{2}}}=c \frac{\sqrt{(a+c)^{2}+2 b^{2}}}{\sqrt{(a+c)^{2}+b^{2}}}$.

$M C=\sqrt{D C^{2}+M D^{2}}=\sqrt{a^{2}+\frac{a^{2} b^{2}}{(a+c)^{2}}}=\frac{a \sqrt{(a+c)^{2}+b^{2}}}{a+c}$.

$S_{\triangle M C S}=\frac{1}{2} \cdot M C \cdot S K=\frac{1}{2} \cdot \frac{a c}{a+c} \cdot \sqrt{(a+c)^{2}+2 b^{2}}=\frac{1}{2} \cdot a c \sqrt{1+2\left(\frac{b}{a+c}\right)^{2}}$.

Answer: when $a=1, b=10, c=4 \quad x=8, S_{\triangle M C S}=6$.