  Let $X$ be an $n$-dimensional Riemannian manifold with "large positive"
scalar curvature. In this paper, we prove in a variety of cases that if $X$
"spreads" in $(n-2)$ directions {\it "distance-wise"}, then it {\it can't} much
"spread" in the remaining 2-directions {\it "area-wise".} Here is a
geometrically transparent example of what we plan prove in this regard that
illustrates the idea. Let $g$ be a Riemannin metric on $X= S^2\times \mathbb
R^{n-2}$, for which the submanifolds $$\mbox {$\mathbb R_s^{n-2}=s\times
\mathbb R^{n-2}\subset X$ and $S^2_y= S^2\times y \subset X$}$$ are {\it
mutually orthogonal} at all intersection points $$x=(s,y)\in X=\mathbb
R_s^{n-2}\cap S^2_y.$$ (An instance of this is
$g=g(s,y)=\phi(s,y)^2ds^2+\psi(s,y)^2dy^2$.) Let the Riemannian metric on
$\mathbb R_s^{n-2}$ induced from $(X,g)$, that is $g|_{\mathbb R_s^{n-2}}$, be
{\it greater than the Euclidean} metric on $\mathbb R_s^{n-2} =\mathbb R^{n-2}$
for all $s\in S^2$. (This is interpreted as "large spread" of $g$ in the
$(n-2)$ Euclidean directions.) {\sf If the {\it scalar curvature of $g$ is
strictly greater than that of the unit 2-sphere}, $$Sc(g) \geq
Sc(S^2)+\varepsilon=2+\varepsilon, \mbox { }\varepsilon>0,$$ then, provided
$n\leq 7$, } (this, most likely, is unnecessary) {\sf there exists a smooth
{\it non-contractible} spherical surface $S\subset X$, such that
$$area(S)<area(S^2)=4\pi.$$} (This says, in a way, that $(X,g)$ "doesn't spread
much area-wise" in the 2 directions complementary to the Euclidean ones.)
