# Another puzzle with area

Squares $$ABCD, DCGH, BEFG$$ and $$ELKM$$ are positioned as shown on the picture. Find the area of triangle $$DGK$$ if you know that the area of square $$ABCD$$ is $$20$$. Here's another solution using the fact that $$\triangle ABC$$ and $$\triangle ABD$$ have the same area if $$AB \parallel CD$$: The reason is that $$CE = DF$$, so $$\frac12 AB \cdot CE = \frac12 AB \cdot DF$$.

Now, since $$DG \parallel EK$$ and $$BD \parallel EG$$, we have $$\triangle DGK = \triangle DEG = \triangle BEG$$ in area: The result is hence

$$2 \cdot 20 = 40$$.

• Yes, well this is basicly the same solution as hexomino's. Apr 17 at 6:15
• @Greedoid It may use the same principle as Hex's proof, but it applies that principle in a different manner resulting in different triangles. Apr 17 at 7:20

Here is a quick way to do it

The locus of the point $$K$$ as the side length of the square $$EMKL$$ varies is a line which is parallel to $$DG$$. Hence, if we consider $$DG$$ as the base of the triangle $$DGK$$ then its perpendicular height, and thus its area, is invariant with respect to the size of the square $$EMKL$$.

This means that we could perform the calculation with the side length $$EM$$ being anything we like and the answer would be the same. If we make it so that the point $$M$$ coincides with the point $$F$$, it is easy to compute that the area of triangle $$DGK$$ is twice the area of the square $$ABCD$$ and thus the answer is $$40$$.

• I like this puzzle - at first sight there is a constraint missing but the answer comes out cleanly :) I think this puzzle would be suitable for some math channel such as 3Blue1Brown Apr 17 at 0:01
• Not worth a separate answer but I'd say the most opportunistic placement of M would be as the midpoint of E and F. Apr 17 at 7:37

Well, this answer is a bit late, but still quite simple I believe:

Let's introduce a coordinate system with origin at $$A$$, having $$AL$$ and $$AH$$ as $$x$$ and $$y$$ axes respectively. Also, let $$AB=1$$ (well, the $$ABCD$$ square has now an area of $$1$$, not $$20$$, but we can just scale the final result 20 times, i.e. assuming that we scaled down the entire picture $$\sqrt{20}$$ times on each axis - otherwise we had to deal with a bunch of square roots). Now, $$D$$ has coordinates $$(0, 1)$$, $$G$$ is $$(1,2)$$ (since the squares $$ABCD$$ and $$DCGH$$ are equal with side $$1$$), and $$K$$ is $$(s + 3, s)$$ where $$s$$ is the side of $$ELKM$$ square, which we do not know. (The $$BEFG$$ square must have the side of $$2$$, because $$BG=BC+CG=1+1=2$$, so $$E$$ is at $$(3, 0)$$).
Now calculate the area of $$DGK$$ triangle from the coordinates of its vertices using the well-known formula: $$A=\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|,$$ where $$x_i$$ and $$y_i$$ are the coordinates of $$i$$-th vertex ($$i=1,2,3$$).
Plugging $$x_1=0, y_1=1, x_2=1, y_2=2, x_3=s+3, y_3=s$$ gives $$A=\frac12|0(2-s)+1(s-1)+(s+3)(1-2)|=\frac12|s-1-s-3|=\frac12\times4=2.$$ Thus, the final result (which does not depends on $$s$$) is $$2\times20=40$$.