Question 1111119
THE PICTURE:
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
rectangle(-1,-1,1,1),locate(-0.2,0.2,shaded),
triangle(-1,0,0,1,1,-1),locate(-0.2,0,area),
locate(-1.07,-1,A),locate(1.01,-1,B),locate(1.01,1.09,C),
locate(-1.07,1.09,D),locate(-1.07,0.04,M),locate(-0.02,1.1,N)
)}}}
 
THE FIFTH_GRADER'S SOLUTION:
If we connect M to the midpoint of BC,
and N to the midpoint of AB, we split the square into 4 smaller squares, each one being {{{1/4}}} of square ABCD.
Joining 2 of those small squares we would get a rectangle
that is {{{1/2}}} of square ABCD.
One of those 4 smaller squares
is divided into two equal triangular parts by diagonal MN,
so each of those two triangles is {{{1/8}}} of square ABCD.
Triangle MND is {{{1/8}}} of square ABCD.
{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
rectangle(-1,-1,1,0),rectangle(0,0,1,1),
triangle(-1,0,0,1,-1,1),locate(-0.05,-0.4,1/2),
locate(0.4,0.6,1/4),locate(-0.7,0.8,1/8),locate(-0.4,0.4,1/8),
locate(-1.07,-1,A),locate(1.01,-1,B),locate(1.01,1.09,C),
locate(-1.07,1.09,D),locate(-1.07,0.04,M),locate(-0.02,1.1,N)
)}}}
The diagonal of a rectangle that is {{{1/4+1/4=1/2}}} of square ABCD
(such as diagonal MB),
divides the rectangle into two equal triangles.
Each of those triangles is {{{1/4}}} of square ABCD.
Triangle ABM is {{{1/4}}} of square ABCD, and so is triangle NCB.

{{{drawing(300,300,-1.1,1.1,-1.1,1.1,
triangle(-1,-1,-1,0,1,-1),triangle(1,-1,-1,0,1,0),
rectangle(0,0,1,1),locate(-0.5,-0.5,1/4),locate(0.5,-0.3,1/4),
triangle(-1,0,0,1,-1,1),locate(0.4,0.6,1/4),
locate(-0.7,0.8,1/8),locate(-0.4,0.4,1/8),
locate(-1.07,-1,A),locate(1.01,-1,B),locate(1.01,1.09,C),
locate(-1.07,1.09,D),locate(-1.07,0.04,M),locate(-0.02,1.1,N)
)}}}
The unshaded area is triangles ABM + NCB + MND, and amounts to
{{{1/4+1/4+1/8=5/8}}} of the square.
So, the shaded area is
{{{1-5/8=3/8}}} of the square.
Obviously the ratio of shaded area to unshaded area is
{{{3/8}}}{{{":"}}}{{{5/8}}} or {{{3:5}}}
 
THE EXPECTED SOLUTION is probably as follows
As ABCD is a square with side length {{{x}}} cm",
{{{AB=BC=CD=AD=x}}}{{{cm}}} .
As "M and N are the midpoints of AD and CD respectively",
{{{AM=MD=DN=NC=x/2}}}{{{cm}}} .
The unshaded area is made up of triangles {{{ABM}}} , {{{NCB}}} , and {{{MND}}} ,
so its total area is easy to calculate.
{{{area(ABM)}}}{{{"="}}}{{{(1/2)*AB*AM}}}{{{"="}}}{{{(1/2)*"( x"}}}{{{"cm )"}}}{{{" "*" "}}}{{{"("}}}{{{x/2}}}{{{"cm )"}}}{{{"="}}}{{{x^2/4}}}{{{cm^2}}} .
{{{area(NCB)}}}{{{"="}}}{{{(1/2)*CB*NC}}}{{{"="}}}{{{(1/2)*"( x"}}}{{{"cm )"}}}{{{" "*" "}}}{{{"("}}}{{{x/2}}}{{{"cm )"}}}{{{"="}}}{{{x^2/4}}}{{{cm^2}}} .
{{{area(MND)}}}{{{"="}}}{{{(1/2)*MD*ND}}}{{{"="}}}{{{(1/2)*"("}}}{{{x/2}}}{{{"cm )"}}}{{{" "*" "}}}{{{"("}}}{{{x/2}}}{{{"cm )"}}}{{{"="}}}{{{x^2/8}}}{{{cm^2}}} .
{{{unshaded}}}{{{area}}}{{{"="}}}{{{(x^2/4+x^2/4+x^2/8)}}}{{{cm^2}}}{{{"="}}}{{{(5/8)x^2}}}{{{cm^2}}}
 
Calculating the shaded area as the area of triangle MNB may seem difficult,
However,
{{{area(MNB)}}}{{{"="}}}{{{shaded}}}{{{area}}}{{{"="}}}{{{area(ABCD)}}}{{{"-"}}}{{{unshaded}}}{{{area}}}
{{{area(ABCD)}}}{{{"="}}}{{{AB*BC}}}{{{"="}}}{{{"( x"}}}{{{"cm )"^2}}}{{{"="}}}{{{x^2}}}{{{cm^2}}}
So, {{{area(MNB)}}}{{{"="}}}{{{shaded}}}{{{area}}}{{{"="}}}{{{x^2}}}{{{cm^2}}}{{{"-"}}}{{{(5/8)x^2}}}{{{cm^2}}}{{{"="}}}{{{(3/8)x^2}}}{{{cm^2}}} .
The ratio of those two area values, both in {{{cm^2}}} is
{{{((3/8))/((5/8))}}}{{{"="}}}{{{(3/8)*(8/5)}}}{{{"="}}}{{{highlight(3/5)}}} or {{{highlight(3:5)}}}