I have this problem:

enter image description here

So my development was:

Denote side of rectangle with: $2a, 2b$.

So, $4ab= 64, ab = 16$

Denote shaded region with $S$

Denote area of triangle $DGH = A_1$ and triangle $FBE = A_2$.

So, $A_1 + A_2 + S = 64$

$S = 64 - A_1 - A_2$

The triangles $A_1, A_2$ are congruent because $LAL$ congruence criterion.

The area of $A_1$ and $A_2$, is the same and i got it with this way:

Since, the $\angle{GDH} = 90$ and the median from this angle to the base $HG$, that is the altitude of the triangle $DGH$, will measure the half of the $HG$ side.

And the $HG$ side by Pythagorean theorem, will be $\sqrt{a^2 + b^2}$, that will be the base of the triangle.

And the altitude will be: $\frac{\sqrt{a^2 + b^2}}{2} $,

So the Area of $A_1 = \frac{a^2 + b^2}{4}$

So, $A_1 + A_2 = \frac{a^2 + b^2}{2}$

Then, $64 - (\frac{a^2 + b^2}{2}) = S$

And, $-(a^2 - 8ab + b^2) = 2S$

And I have not been able to continue from here, what should I do? Thanks in advance.

  • 1
    $\begingroup$ $\triangle DGH$ is similar to $\triangle DCA$. Since the ratio of their sides is $1:2$, the ratio of the areas is $1:4$, so area of $\triangle DGH$ is 1/8 of the whole rectangle. $\endgroup$ – achille hui Jul 22 '18 at 22:59
  • 1
    $\begingroup$ Note that there is an error in your development: The median to the base HG is not the altitude of the triangle DGH, except when the triangle is isosceles. In general, the altitude is $ab/\sqrt{a^2+b^2}$, hence S is $ab/2$. $\endgroup$ – TonioElGringo Jul 23 '18 at 9:22
  • $\begingroup$ My error was assume that is a isosceles triangle and all its secundary elements coincide $\endgroup$ – Eduardo S. Jul 23 '18 at 13:47

The area of hatched region of the rectangle is the area of the rectangle minus the sum of the areas of the right triangles $BEF$ and $DGH$. Now if you glue these triangles along their hypotenuses, you obtain a rectangle with dimensions half the dimensions of the big rectangle, hence with area equal to ¼ area of the big rectangle.

Hence the required area is ¾ big area, i.e. $\color{red}{48}$.

  • $\begingroup$ Nice solution, thanks $\endgroup$ – Eduardo S. Jul 22 '18 at 23:23


enter image description here

  • there are eight small red triangles all with the same area

  • six of them are shaded

  • $\begingroup$ How you know that? $\endgroup$ – Eduardo S. Jul 22 '18 at 23:17
  • 7
    $\begingroup$ @Mattiu Each has perpendicular sides half the lengths of the sides of the original rectangle $\endgroup$ – Henry Jul 22 '18 at 23:19

enter image description here

If you notice that if you combine two right triangles then they occupy the area of $\dfrac14$ of the total area.

So, the area of the shaded region is $=64-\dfrac14(64)=64-16=48$

  • $\begingroup$ What do you mean with $G = E $ ? $\endgroup$ – Eduardo S. Jul 22 '18 at 23:16
  • $\begingroup$ Length of the midpoint from the vertices $\endgroup$ – Key Flex Jul 22 '18 at 23:17
  • $\begingroup$ Well, what is the lenght of the rigth triangle ? is the Hypotenuse? $\endgroup$ – Eduardo S. Jul 22 '18 at 23:21
  • $\begingroup$ The length of the right triangle is nothing but the length of the midpoints $\endgroup$ – Key Flex Jul 22 '18 at 23:22

Hint. Because the triangles connecting the midpoints with the vertices are all congruent, we have:

$$\color{grey}{A}=64-2A_{DHG}$$ $$\color{grey}{A}=2A_{DHG}+A_{GHEF}$$

Addinng the two relations:


But, $A_{GHEF}=\frac{A_{ABCD}}{2}$ (I'll let you figure out why yourself as an (easy) exercise), and therefore:



By Midpoint theorem, area of triangle DHG is $\frac{1}{4}$ of triangle ACD. So, $$ar(DHG)=0.25*ar(ACD)=64/(2*4)=8$$

By symmetry, ar(unshaded region)=$8*2=16$ So, ar(Shaded region)=$64-16=48$


If your sidelengths are $2a$ and $2b$, how do you think you could represent $|\triangle GHD|$ and $|\triangle EFB|$ in terms of $a$ and $b$? (Keeping in mind that $E,G,H,F$ are midpoints.)


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