How can I determine the G 1:3 using given midpoints?

[squexy]

Can someone show me how to resolve this question?

View attachment 2902
The answer is G 1:3

Thanks ^.^

 

[Deveno]

This is how I would do it: image drawing a line between E and F, this splits the larger rectangle into 2 equal smaller rectangles, since these are midpoints.

Each smaller rectangle is then split into 4 triangles. Let's say the top and bottom ones have base length $a$ and the side ones have base length $b$.

The entire larger rectangle has area:

$(a + a)(b) = 2ab$.

It is easy to see that the top and bottom triangles have height $\dfrac{b}{2}$, so their area (each) is:

$\dfrac{1}{2}\cdot a\cdot \dfrac{b}{2} = \dfrac{ab}{4}$.

We have 2 of then, so their total area is $2\cdot\dfrac{ab}{4} = \dfrac{ab}{2}$.

The two side triangles must then together have total area:

$ab - \dfrac{ab}{2} = \dfrac{ab}{2}$ (Each smaller rectangle has area $ab$).

Since the white area is just two side triangles put together (one from each smaller rectangle made from the large rectangle cut in half), it must have area $\dfrac{ab}{2}$.

Thus the shaded area is $2ab - \dfrac{ab}{2} = \dfrac{3ab}{2}$.

So the ratio of the white area to the shaded area is:

$\dfrac{\dfrac{ab}{2}}{\dfrac{3ab}{2}} = \dfrac{ab}{2}\cdot \dfrac{2}{3ab} = \dfrac{1}{3}$

 

[Prove It]

Can someone show me how to resolve this question?

View attachment 2902
The answer is G 1:3

Thanks ^.^

An easier way:

Draw the segment EF. This cuts the entire rectangle in half. Surely you can see that of these two halves, the resulting white triangles each make up 1/4 of these halves, so 1/8 of the entire rectangle's area.

Since there will be two of these triangles, the white region is 1/4 of the rectangle's area.

Compared to the remaining area of the rectangle (3/4), this gives a ratio of 1/4 : 3/4, or simply 1:3.

 

[Deveno]

An easier way:

Draw the segment EF. This cuts the entire rectangle in half. Surely you can see that of these two halves, the resulting white triangles each make up 1/4 of these halves, so 1/8 of the entire rectangle's area.

Since there will be two of these triangles, the white region is 1/4 of the rectangle's area.

Compared to the remaining area of the rectangle (3/4), this gives a ratio of 1/4 : 3/4, or simply 1:3.

It is (or at least was) not immediately apparent (to me) that the four triangles of each half of the entire rectangle all have the same area. It turns out that they do, because the formula (in terms of the sides of the "half-rectangles"), namely what I have called:

$\dfrac{ab}{4}$

is symmetric in $a$ and $b$ (on the top versus the sides, the two "swap places", $a$ is the base of one pair, and $b$ is twice the height, while on the other pair $b$ is the base, and $a$ is twice the height).

The reason I stress this, is because if $a$ and $b$ were *very* different (and not so close as they are for this diagram), it would be even less clear that all 8 triangles we can make all have the same area (although each pair of 4 clearly do).

This, in my opinion, is one of the "boons" of mathematics: it allows us to discover we knew more than we thought we did.

 

[Prove It]

It is (or at least was) not immediately apparent (to me) that the four triangles of each half of the entire rectangle all have the same area. It turns out that they do, because the formula (in terms of the sides of the "half-rectangles"), namely what I have called:

$\dfrac{ab}{4}$

is symmetric in $a$ and $b$ (on the top versus the sides, the two "swap places", $a$ is the base of one pair, and $b$ is twice the height, while on the other pair $b$ is the base, and $a$ is twice the height).

The reason I stress this, is because if $a$ and $b$ were *very* different (and not so close as they are for this diagram), it would be even less clear that all 8 triangles we can make all have the same area (although each pair of 4 clearly do).

This, in my opinion, is one of the "boons" of mathematics: it allows us to discover we knew more than we thought we did.

It IS immediately apparent when you are told that the points given are midpoints of the given sides...

 

[Deveno]

It IS immediately apparent when you are told that the points given are midpoints of the given sides...

With all due respect, I beg to differ. It may well be obvious to you, it was not obvious to me (I had to spend about a minute thinking about it).

Imagine I ask you "why it is obvious?". The two types of triangles are not congruent. How would you answer such a question?

There is nothing "harder" about what I posted, than what you suggested, in fact, they are in essence the same approach. I just filled in "more blanks" to make what I said transparent.

I suppose I could have just said: it is obvious the answer is 1:3, by similar reasoning as yours. In my mind, however, that seems a tad unhelpful.