Question 969785
EQUILATERAL TRIANGLE:
The area of a triangle with sides of lengths {{{a}}} , {{{b}}} , and {{{c}}} ,
opposite angles of measures {{{A}}} , {{{B}}} , and {{{C}}} ,
can be calculated as {{{area=b*c*sin(A)}}} :
{{{drawing(300,300,-1,9,-1,9,
green(triangle(0,0,3,8,3,0)),
green(rectangle(3.5,0.5,3,0)),
triangle(0,0,3,8,8,0),locate(3.1,3.5,green(h=c*sin(A))),
locate(3.85,0,b),arrow(3.8,-0.25,0,-0.25),arrow(4.2,-0.25,8,-0.25),
locate(5.5,4.5,c),locate(1.25,4.5,a),locate(8.2,0.2,A),
locate(-0.4,0.2,C),locate(2.9,8.5,B)
)}}}
In the case of an equilateral triangle,
all sides have the same length, {{{b}}} ,
and all angles measure {{{60^o}}} ,
so {{{area=b^2*sin(60^o)=b^2sqrt(3)/2=approximately}}}{{{0.866b^2}}} .
 
If your equilateral triangle inside a square looks like this {{{drawing(300,300,-1,11,-1,11,
green(rectangle(0,0,10,10)),
red(triangle(0,0,10,0,5,8.66)))}}} ,
then {{{b=s=7}}} and {{{area=s^2*sin(60^o)=(sqrt(3)/2)*s^2=approximately}}}{{{0.866*s^2}}} .
{{{area=7^2*sqrt(3)/2=49sqrt(3)/2=approximately}}}{{{0.866*49=42.434}}} .
However, it may not be the expected answer,
but you may be able to fit a slightly larger triangle if you "tilt" it,
like this {{{drawing(300,300,-1,11,-1,11,
green(rectangle(0,0,10,10)),
red(triangle(0,0,10,2.68,2.68,10)))}}} .
 
CIRCLE:
The larges circle that you would be able to fit inside a square has a diameter as long as the side of the square:
{{{drawing(300,300,-1,11,-1,11,
green(rectangle(0,0,10,10)),
red(circle(5,5,5)))}}} If the square side length is {{{s}}}, the circle radius is {{{r=s?2}}}
The area of a circle of radius {{{r}} is {{{pi*r^2}}} ,
so a circle of radius {{{s/2}} would have an area of
{{{area=pi*(s/2)^2=pi*s^2/4}}} .
In particular, for {{{s=7}}}
the area would be {{{area=pi*7^2/4=approximately}}}{{{38.485}}} .
 
For the triangle and circle described above,
the ratio of their areas would be
{{{(sqrt(3)/2)*s^2/(pi*s^2/4)=(sqrt(3)/2)*s^2*(4/pi*s^2)=sqrt(3)*4*s^2/(2pi*s^2)=2sqrt(3)/pi=approximately}}}{{{1.103}}} .
(The side of the square {{{s=7}}} did not matter, as long as we use the same square size to fit the triangle and the circle).
 
LARGER TRIANGLE:
{{{drawing(300,300,-1,11,-1,11,
green(rectangle(0,0,10,10)),
red(triangle(0,0,10,2.68,2.68,10)),
locate(0.1,5,s),locate(5,0,s),
locate(6.3,10,x),locate(9.5,6.3,x),
locate(0.5,10,s-x),locate(8.5,1.2,s-x),
locate(5,1.9,b),locate(6,6.34,b)
)}}} In the right triangle corners, according to the Pythagorean theorem,
{{{b^2=x^2+x^2}}} and {{{b^2=(s-x)^2+s^2}}} , so
{{{x^2+x^2=(s-x)^2+s^2}}}
{{{x^2+x^2=s^2-2sx+x^2+s^2}}}
{{{x^2=-2sx+2s^2}}}
{{{x^2+2sx=2s^2}}}
{{{x^2+2sx+s^2=2s^2+s^2}}}
{{{(x+s)^2=3s^2}}} , and since {{{x+s>0}}} and {{{s>0}}}
{{{x+s=sqrt(3s^2)}}}
{{{x+s=s*sqrt(3)}}}
{{{x=s(1-sqrt(3))}}} , and {{{x^2=s^2(1-sqrt(3))^2}}} .
Now, we knew that for a triangle, {{{area=(sqrt(3)/2)b^2}}} ,
and that this tilted triangle has {{{b^2=x^2+x^2=2x^2}}} ,
so {{{b^2=2(1-sqrt(3))^2*s^2}}} and {{{area=(sqrt(3)/2)2(1-sqrt(3))^2*s^2=sqrt(3)(1-sqrt(3))^2*s^2=approximately}}}{{{0.928s^2}}} .
That makes the area of thte tilted triangle
approximately {{{0.928*7^2=45.472}}} ,
and the ratio of areas for the tilted triangle and circle would be
{{{sqrt(3)(1-sqrt(3))^2*s^2/(pi*s^2/4)=sqrt(3)(1-sqrt(3))^2*s^2*(4/pi*s^2)=4sqrt(3)(1-sqrt(3))^2/pi}}} .
That ratio is approximately {{{1.182}}} .