# Area of Parallelograms and Rhombi

Parallelograms are confusing shapes, mostly because the word itself has too many syllables. Why does a shape with *four* sides need *five* syllables? Come on, geometry.

Luckily, **parallelograms** are defined only by the fact that they're quadrilaterals (don't even get us started on the number of syllables in *that* word) made of two sets of parallel lines. Because of this simple fact, opposite sides in a parallelogram are congruent. But you knew that.

As wonderful as that is, it doesn't really help us find the area. What we need when we find the area is the length of the **base** and the **height**, which must be perpendicular to the base. All we have to do is multiply them together, and we end up with the area formula for a parallelogram. So our formula is:

*A* = *bh*

Getting déjà vu? Looks like the area of a rectangle or like twice the area of a triangle, doesn't it? Well, there's a reason for that.

So really, a parallelogram is just two triangles wearing one big trenchcoat pretending to be something they're not. Since each one's area is given as ½*bh*, the area of both of them together is 2(½*bh*) = *bh*. We can see right through their trickery.

If we also sever the parallelogram from the top vertex down to the base at a right angle, we can take the triangle we've cut off and fill in the gap on the other side. That'll give us a parallelogram with four right angles. Or, you know, a rectangle. Like this:

Since a rectangle's area is given as *A* = *lw*, we can use this formula. Only now, we know that *l* = *b* and *w* = *h*. Swapping out those values gives us the area of a parallelogram: *A* = *bh*. Do you really need any more proof?

### Sample Problem

If the base of the Centaurus A galaxy is 16,500 light-years and the height of the galaxy is 10,000 light-years, what's the area of this parallelogram-shaped galaxy?

A parallelogram is a parallelogram, no matter how big or small. Even a galaxy-sized parallelogram has an area.

*A* = *bh A* = 16,500 light-years × 10,000 light-years

*A*= 165,000,000 light-years

^{2}

### Sample Problem

What's the area of this parallelogram?

Well, we know that the base is 10 cm, but the height is not immediately clear. If we draw in the height, it forms a 30-60-90 triangle. How convenient.

We can use the ratios of the triangle's sides to find the height of the parallelogram (which, in our case, is the long side of the 30-60-90 triangle). Since we know the hypotenuse and the long leg have a ratio of , we can solve for the height by creating a proportion.

Now that we have the height, we can solve for the area of the parallelogram.

*A* = *bh*

While the square and rectangle paid enough to get their own sections, there's another type of parallelogram whose area we haven't yet talked about. It also happens to be a quadrilateral's favorite automobile: a **rhombus**.

We're only kidding. They prefer Ferraris.

We can think of a rhombus as a special parallelogram, just as a square is a special type of rectangle. A rhombus has four equal sides and diagonals that are perpendicular to each other. Basically, they look like diamonds. So if you think about it, *rhombi* are a girl's best friend.

Since rhombi are special types of parallelograms, we can use the formula *A* = *bh* to find their areas, just like we used *A* = *lw* to find the area of a square. However, the angles of the shape aren't all the same. That means the base and height of a rhombus will **not** be the same, so we can't go around squaring sides willy-nilly. We'll still have to find the base and the height of the rhombus.

What's so special about rhombi if we have to use the exact same formula? Well, their diagonals, of course. Rhombi have a special characteristic in that half of the product of their diagonals is their area.

*A* = ½*d*_{1}*d*_{2}

### Sample Problem

A rhombus has a side length of 4 m and a height of 3 m. What's the area of the rhombus?

Sometimes, problems try and trick you by using "side" and "base" interchangeably. In the case of rhombi, all the side lengths are the same, so they're interchangeable. Just remember that for a rhombus, base and side are the same, but the height is neither of those.

So in this case, *b* = 4 m and *h* = 3 m, and that's all we need. Area, here we come.

*A* = *bhA* = 4 m × 3 m

*A*= 12 m

^{2}

### Sample Problem

The same rhombus (*b* = 4 m and *h* = 3 m) has a diagonal of 8 m. What is the length of the other diagonal?

We've already solved for the area of the rhombus. If we haven't, we could easily do so using *A* = *bh* = 4 m × 3 m = 12 m^{2}. Now we can use the formula with the diagonals and find the length of the diagonal whose measurement we don't have. It doesn't matter which diagonal is *d*_{1} and which is *d*_{2}.

*A* = ½*d*_{1}*d*_{2}

12 m^{2} = ½ × 8 m × *d*_{2}

3 m = *d*_{2}

The other diagonal is 3 m long.