# Question: Diagonal of Rectangle

## Comment on Diagonal of Rectangle

### I Solved in different way. I

I Solved in different way. I considered ED as x and hence BE is 20-x
Now CE^2 = 16^2 - (20-x)^2
Again one more equation is CE^2 = 12^2 - x^2
Solving these two equations, I am getting x = 7.2, with that I found CE as 9.some thing, so answer is E.

### Can you pls explain the

Can you pls explain the solution using Similar Triangles?

### You bet.

You bet.

I'll compare ∆ABD with ∆EDC (they are similar)

Here's WHY they're similar.

First notice that ∠CED is 90º, since it's on the line with another 90º angle. So, ∆ABD and ∆EDC both SHARE a 90-degree angle.

Next, notice that, since AB || CD, we know that ∠ABD = ∠CDE So, ∆ABD and ∆EDC both SHARE another angle.

Since ∆ABD and ∆EDC already SHARE two angles, they must also share the 3rd angle. So, we know that ∆ABD and ∆EDC are similar triangles.

Let's focus on ∆ABD for a while.
We know that AB = 12
Then we can apply the Pythagorean Theorem to see that BD = 20

Now let's COMPARE corresponding sides.
We'll let x = the length of side EC

First, sides BD and CD are CORRESPONDING sides in the two similar triangles.
So, we'll compare their two lengths: 20 and 12

Second, sides AD and CE are CORRESPONDING sides in the two similar triangles.
So, we'll compare their two lengths: 16 and x

We can now write: 20/12 = 16/x

When we solve this equation for x, we get x = 9.6 (answer choice E)

### Hi Brent,

Hi Brent,

I am not able to figure out this explanation,

"First notice that ∠CED is 90º, since it's on the line with another 90º angle. So, ∆ABD and ∆EDC both SHARE a 90-degree angle.
Next, notice that, since AB || CD, we know that ∠ABD = ∠CDE So, ∆ABD and ∆EDC both SHARE another angle."

Angle CED = 90 degree is directly given. I didn't understand the property you cited for the same. Moreover, i am oblivious to the second lowdown when two angles are parallel, then the cited angles are equal. Please help.

### We're not told that ∠CED is

We're not told that ∠CED is 90º, but we can quickly deduce this, since ∠CEB is 90º

Since AB || CD, there are some equal angles that get created when the line BD intersects those parallel lines.
For more on that, go to 6:33 of the following video: https://www.gmatprepnow.com/module/gmat-geometry/video/858

Does that help?

Cheers,
Brent

### Hi Brent

Hi Brent
Does this Qs fall into the league of 700-800?

### Yes definitely. In fact, it's

Yes definitely. In fact, it's probably closer to 800 than it is to 700.

Cheers,
Brent

### Instead of noticing BA || to

Instead of noticing BA || to CD is it safe to say that

Angle BAD = Angle BDC since side AD = side BC

Equal lengths means equal angles. Same formula

20/12 = 16/X

### No, we definitely can't make

No, we definitely can't make that conclusion.
Since ABCD is a rectangle, ∠BAD is 90°. However, ∠BDC is definitely less than 90°

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

I solved this in a different manner:

1) Find BD via PT or common PT triplets (i.e.: multiple of 3-4-5)
16^2+12^2=BD^2
BD^2=400
BD=20

2) Recognize that BC and AD are two parallel lines that are bisected by the line BD. This means that angle ABD and angle BDA are equal. If these two angles are equal, we can infer that angles ABD and CDE are also equal, given that the corner angles of rectangles must each be 90 degrees.

3) We now have two triangles (BAD and CED) that share the same angles. We can now apply the relative side length ratio such that:

EC/16=12/20
EC=192/20
EC=96/10=9.6

Thanks!

### Sweeeeeeeet solution!!!

Sweeeeeeeet solution!!!

### Dear sir,

Dear sir,

Can we compare traianhle BCD and traiagle DEC for similarities and then find ratios and then find solution.

Is that going to work.

### Great idea!

Great idea!
Those two triangles are, indeed, similar. So, your approach should work.

Cheers
Brent

### https://gmatclub.com/forum

https://gmatclub.com/forum/pat-will-walk-from-intersection-x-to-intersection-y-along-a-route-that-68374.html

Hi Brent, can you please share your solution to this question?

Thanks
Kashaf

### https://gmatclub.com/forum/in

https://gmatclub.com/forum/in-the-figure-above-what-is-the-area-of-region-pqrst-1-pq-rs-294329.html

Hi Brent, can you please solve this question? According to me, the answer should be D.

Thanks
Kashaf

### Hi Brent,

Hi Brent,
In the following question, why does the formula for diagonal of cube not apply?
https://gmatclub.com/forum/the-surface-distance-between-2-points-on-the-surface-of-a-cube-is-the-305873.html

### Question link: https:/

The formula for the diagonal of a box/cube (covered here https://www.gmatprepnow.com/module/gmat-geometry/video/869) helps us determine the length of a straight line (passing THROUGH THE INTERIOR of the box/cube) that connects two opposite vertices.

In the linked question, we're asked to find the shortest path when traveling in ON THE SURFACE of the cube.

Does that help?

### It's so easy to miss these

It's so easy to miss these little things. Thank you for the help though! :)