Today, we’ll quickly look at how a rectangle and parallelogram can have the same area.

In the diagram to the above, the parallelogram and rectangle share a vertex (D), one vertex of the rectangle (E) is on a side of the parallelogram, and one vertex of the parallelogram (C) is on a side of the rectangle. That is enough information to guarantee that the rectangle and parallelogram have equal area.

Here’s an argument why. In the diagram below, notice I have constructed segment EQ, which is perpendicular to CD. This segment is the height of the parallelogram, so that times the length of CD would be the area of the parallelogram.

Look at ΔDGC and ΔEQD. Those two triangles are similar. Why?

Well, first of all, ∠QDG and ∠EDQ are complementary: they both add up to the 90° angle of ∠EDG. Also, ∠QDG and ∠QCG are complementary, because they are the acute angles of a right triangle. Since ∠EDQ and ∠QCG are both complementary to the same angle (∠QDG), they are congruent: ∠EDQ≅ ∠QCG.

Since we know ∠EDQ≅ ∠QCG and we know ∠EQD≅ ∠G (both right angles), we know two angles in ΔDGC are congruent to two angles in ΔEQD. By the AA Similarity Theorem, they must be similar triangles.

ΔDGC ~ ΔEQD

Similar triangles have proportional sides. In particular, we can set up a proportion:

After cross-multiplying, we get two equal products. (ED)*(DG) = the area of the rectangle. (EQ)*(DC) = the area of the parallelogram. Therefore, those two areas are equal.

That’s a more equation-based way of proving the areas equal. Here’s another HUGE idea, which is much more appealing for visual thinkers. Imagine extending one pair of sides in a parallelogram like railroad rails:

If we slide either side along its “rail”, the shape of the parallelogram will change, but the area will stay the same, because the base (the length on the rail) doesn’t change, the height (the distance between the parallel lines) doesn’t change.

The purple and the orange parallelograms must have exactly the same area. That’s a HUGE geometry idea.

Now, think about our diagram, with some “rails” added.

Now, when we slide AB upward, so that A coincides with E, that will make both AD and BC perfectly vertical:

Now, if we slide BC down, so it exactly coincides with FG, then the parallelogram will exactly coincide with the rectangle, which means they must have identical areas.

This “sliding method” can be a very handy shortcut with parallelogram areas.

### Most Popular Resources

Hi Mike 🙂

Could you please provide some questions related to this theory of (rectangle & parallelogram same area ) to practice ?

Hi Vipul,

I don’t think that we have any practice problems on our blog specifically related to this concept, but I will forward your request on to our blog writers! In the meantime, you can check out these other geometry practice questions 🙂

Dear Mike,

Thank you for posting this. We we can be sure that when ever there are to intersection points as shown in the diagram rectangle and parallelogram will be equal?

Furthermore, I am unable to understand the explanation of equality of angles particularly this:

“Also, ∠QDG and ∠QCG are complementary, because they are the acute angles of a right triangle. Since ∠EDQ and ∠QCG are both complementary to the same angle (∠QDG), they are congruent: ∠EDQ≅ ∠QCG. ”

Thanks again for posting!

Dear Sunny,

I’m happy to respond. 🙂 As long as one vertex is common to both shapes, and each has another vertex on a side of the other, then the areas must be equal.

When we say that “∠QDG and ∠QCG are complementary” we are saying that:

(∠QDG) + (∠QCG) = 90°

Think about right triangle CDG. As with any triangle, the three angles have to add up to 180°. Well, ∠G = 90°, so the other two angles must have a sum of 90°. That’s always true in a right triangle: the two acute angles always have a sum of 90°, which is to say, they are complementary.

From the angle at D, we have (∠QDG) + (∠EDQ) = 90°

From right triangle CDG, we have (∠QDG) + (∠QCG) = 90°

Look at those two equations — the first term and the 90° are equal. If you solved both equations for [90° – (∠QDG)], then you could set them equal, and you would have

∠EDQ = ∠QCG

Does all this make sense?

Mike 🙂

Yes, thank you! 🙂

Dear Sunny,

You are quite welcome. Best of luck to you!

Mike 🙂

Really a mind boggling question..!! \m/

But such questions could really show up on the real test in form of Quant. comparison ??

And what abt an idea of putting frequently and recently asked questions in real GRE given by your students ???

Dear AArendy,

This is Mike, Magoosh’s GMAT expert. Yes, this sort of thing *could* show up on a very challenging GRE Quant section.

As far as your other suggestion, it’s a bit problematic. First of all, after the stress of taking a full GRE, most people are simply not able to recall questions with any accuracy. Of course, taking pictures or any kind of recording is explicitly forbidden. Even if someone had a photographic memory, and could walk out of the test and memory-dump several questions, it would be 100% illegal for us to put live GRE questions on this blog. ETS would sue us and win. So, instead of that, here’s what happens. The GRE expert, Chris, is deeply familiar with GRE math questions, and periodically he re-takes the GRE to check on it. Then, when he writes his own questions, he has those standards in mind, the standard of GRE Quant questions; thus, even though someone at Magoosh creates those questions, they are an excellent representation of what you will see on the test. Those are the math questions throughout this entire blog.

Does that make sense?

Mike 🙂