Lesson: Quadrilaterals

Comment on Quadrilaterals

When it says area of rhombus at the end: Multiply two diagonals / 2, aren't diagonals in a rhombus of the same size?
greenlight-admin's picture

The only time the diagonals of a rhombus are equal length is when that rhombus is a square (a square is a very special kind of rhombus in which the angles are all 90 degrees).

In all other cases, the diagonals of a rhombus are not equal length. To see what I mean, check out the rhombus here: http://i1168.photobucket.com/albums/r500/GMATPrepNow/rhombus_zpssslp0yuu...

Parallelogram PQRS has a perimeter of 24. PQ = 2QR and Angle PQR = 120 degrees. Parallelogram TQUV is formed by cutting out part of parallelogram PQRS so that angle TQU and PQR share the same angle. If TQUV is half the area of PRQS and TQ = 2QU, what is the perimeter of TQUV to nearest 0.1?

could you please explain this.?
greenlight-admin's picture

It's a very tricky question to explain without a diagram, but here goes....

Since the perimeter is 24 and the 2 longest sides are twice the length of the 2 smallest sides, the parallelogram has the dimensions 4-8-4-8

The 120-degree angle in one corner tells us that the adjacent corner has a 60-degree angle (which is a special angle).

So, if we make the side with length 8 the BASE, then we can see that the height of the parallelogram = 2√3

So, the area of the big parallelogram (PQRS) = (8)(2√3) = 16√3

The area of the small parallelogram (TQUV) is HALF the area of the big parallelogram (PQRS)

So, the area of the small parallelogram (TQUV) = 16√3/2 = 8√3

One side of the small parallelogram is twice the other side. So, let's let 2x be the length of its base, and let x = length of the slanted side.

Since the angle between them is 60 degrees (as with the large parallelogram) we can use special triangle properties to determine that (x√3)/2 is the height of the small parallelogram.

At this point, we can say...

Area of TQUV = (2x)(x√3)/2 = 8√3 [from earlier calculations]

Simplify: x²√3 = 8√3

Divide both sides by √3 to get: x² = 8

So x = √8 = 2√2

In other words, the shorter side of the small parallelogram (TQUV) has length 2√2, which means the longer side has length 4√2

So, the TOTAL perimeter = 2√2 + 4√2 + 2√2 + 4√2 = 12√2 ≈ 17

Thank you very much


Can you please explain the above?
i am getting answer as D, in forum its A!.
greenlight-admin's picture

Link to question: http://gre.myprepclub.com/forum/qotd-9-the-length-of-a-leg-of-an-isoscel...

The answer is A.
If you show me your calculations, perhaps we can see what the issue is.

Let "a" be the leg of isosceles right angle triangle
then its area R=a^2/2

Let "b" be the side of square then its area R=b^2

we are given that a^2/2 = b^2

a^2 = 2 b^2

which means a^2 is greater than b^2, answer is A.

what i thought is, what if a = 0 and b = 0, on that
case area is equal :) so D was what i opted. But late
found that no geometry figure is possible with 0 length!:)
greenlight-admin's picture

Yes, the sides have to be longer than zero :-)

Is there another way to explain this problem?
greenlight-admin's picture

You bet!
Here's my step-by-step solution: https://gre.myprepclub.com/forum/qotd-9-the-length-of-a-leg-of-an-isosce...



Hi Brent - can you explain why we do this to find the width of the entire rectangle: 18+(2)(3)=24. I don't get why we're multiplying it by 2 instead of just doing 18+3.
greenlight-admin's picture

Hi okookiez,

The quick answer is that, since the walkway is 3 feet wide, we must add 3 to one side of our rectangular region and add 3 to the other side of our region.

I created a step-by-step solution that (I believe) answers your question: https://gre.myprepclub.com/forum/the-gure-above-represents-a-rectangular...

Does that help?


Do the diagonals of a rectangle create four 45-45-90 triangles?
greenlight-admin's picture

The diagonals of a rectangle can create four 45-45-90 triangles ONLY WHEN the rectangle is a SQUARE.
If the rectangle is NOT a square, then the diagonals will NOT create four 45-45-90 triangles.

Here's a graphic that shows this: https://imgur.com/1yZFqov


Thanks Brent! Is there anything we should know about the triangles created by the diagonals of a rectangle?
greenlight-admin's picture

The main thing you need to know is that the diagonal of a rectangle will divide the shape into two RIGHT triangles.

A rectangular playground with the area 2400m^2 is surrounded by a running track on the three sides and a wall on the fourth side. The length of the running track is 140m. Which of the following could be the longest possible length of the wall?
greenlight-admin's picture

This question requires some answer choices, but we can answer it nonetheless.

Let x = the width of the playground
Let y = the length of the playground

Since the area of the playground is 2400, we can write: xy = 2400

The running track has length 140.
Since the running track is only on 3 sides, we can write: x + x + y = 140

ASIDE: We can also write x + y + y = 140, but that route will give us the same answers.

We now have two equations: xy = 2400 and 2x + y = 140

Solve the second equation for y to get: y = 140 - 2x

Now take the first equation, xy = 2400 and replace y with (140 - 2x).
We get: x(140 - 2x) = 2400
Expand to get: 140x - 2x² = 2400
Rearrange to get: 2x² - 140x + 2400 = 0
Divide both sides by 2 to get: x² - 70x + 1200 = 0
Factor to get: (x - 30)(x - 40) = 0
So, EITHER x = 30 OR x = 40

Let's find the corresponding y-values by using our equation y = 140 - 2x
If x = 30, then y = 140 - 2(30) = 80
So, one possibility is that the playground has dimensions 30 by 80

If x = 40, then y = 140 - 2(40) = 60
So, another possibility is that the playground has dimensions 40 by 60

So, the longest possible length of the wall is 80 meters.


I tried some word probelm but it didn't work the. One side of the perimeter was named x and since we're looking for both sides then it's 2x
I set the perimeter equal to area
Solved for x which was incorrect. What was wrong with this approach
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/what-is-the-perimeter-in-meters-of-a-re...
Why did you set the perimeter equal to the area?

The question tells us that one rectangle has the same AREA as some other rectangle, and we're suppose to find the perimeter of that first rectangle.

Here's my full solution: https://gre.myprepclub.com/forum/what-is-the-perimeter-in-meters-of-a-re...


How was the (10,10) vertex gotten
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/in-the-xy-plane-one-of-the-vertices-of-...

Notice that, to travel from the vertex at (2, 2), to the CENTER of the square at (6, 6), we must travel 4 units UP and 4 units to the RIGHT.

If we repeat the same steps, we'll arrive at the other vertex.

Does that help?


Hi Brent,
Could you explain your solution in ( https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-of-cardboard-satisfyi-12275.html ) Please? I don't understand why the proportions are the way they are. In my assumption, I would put the ratio like ( x:y = x/2 : y ) in order to align ratio of the lengths and widths of the new rectangles. However, I understand that this is not the case.
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

We need the ratio of the sides of the BIG rectangle to equal the ratio of the sides of a SMALL rectangle.

In other words, we want:
(longest length of BIG rectangle)/(shortest length of BIG rectangle) = (longest length of SMALL rectangle)/(shortest length of SMALL rectangle)

See my diagram at https://gre.myprepclub.com/forum/a-project-requires-a-rectangular-sheet-...

longest length = x
shortest length = y
So, ratio = x/y

longest length = y
shortest length = x/2
So, ratio = y/(x/2)

The resulting EQUATION is: x/y = y/(x/2)

Does that help?



Please explain
greenlight-admin's picture


Area of a triangle= b*h/2. How do we assume the height and base both is also equal to x?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-9-the-length-of-a-leg-of-an-isosce...

We're told that the triangle is an ISOSCELES right triangle.
This means the two legs have the same length (and the hypotenuse, which is the longest side, has a different length)

Does that help?



The question states that length of the side is an integer, but it doesn't say anything about the breadth. So for each integer value of length, we get a corresponding non-integer breadth, giving us more perimeters than 5. eg: for length = 7, we get breadth = 36/7
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-10-the-length-of-each-side-of-rect...

The question says "The length of EACH side of rectangle R is an integer..."
So, all 4 sides have integer lengths.



In the solution to this question, we are making some assumptions about 2 regions being rectangles. Can you kindly explain to me why those assumptions are valid? In the video above, it has been mentioned that the opposite sides in a rectangle must be equal and parallel. So, how are we making those assumptions from the image in the question?

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/what-is-the-area-of-the-region-shown-ab...

Since all of the given angles are 90 degrees, we can conclude that opposite sides are parallel and equal length.


Brent could you please solve this.
greenlight-admin's picture

If you get a chance, could you write out an explanation for this problem: https://gre.myprepclub.com/forum/gre-math-challenge-794.html

I got the right answer, but it seemed too easy. So, I think I might have done something wrong, or just got lucky.
greenlight-admin's picture


I have a question related to this. So if we know the side lengths of any shape (ex. rectangle, square, rhombus, parallelogram, etc) we cannot make any conclusions about the diagonal lengths right?

The only thing we can say about diagonals is that for rectangles/squares the diagonals are equal. We cannot make any conclusions regarding the diagonal length and the length of the sides of the rectangles/squares right?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/rs-st-tu-ur-14558.html

Great question!

If we know the lengths of all sides in a square, then we CAN determine the length of the diagonal.
If we know the lengths of all sides in a rectangle, then we CAN determine the length of the diagonal.

If we know the lengths of all sides in a rhombus or parallelogram, then we CANNOT determine the length of the diagonal.


Hi Brent,

I wanted to know if we need to know the areas of the pentagon or hexagon for the GRE. Or should we be just able to calculate them by diving the figure into smaller parts? Thanks
greenlight-admin's picture

Hi Carla,

You don't need to know how to calculate area of a pentagon or hexagon.
As you suggest, if you're given such a question, you'll be able to divide the shape into easier shapes and then find those areas.


Hey Brent
Can I calculate the area of rectangle and then subtract the area of rhombus to find the area of shaded region and so the area of rhombus would be area of unshaded region.
greenlight-admin's picture

Hi Suryansh,

Sorry but I'm not sure what question you are referring to.
If you can provide a link to the question, I'd be happy to answer your question


sorry I just forgot to send the link.
greenlight-admin's picture

Yes, you're absolutely right.
You can calculate the area of rectangle and then subtract the area of rhombus to find the area of shaded region.



This is your answer to the above question that I believe it's correct :

Let's TEST some possible scenarios for Quantity A
Case 1:
If the perimeter is 60, then one possible scenario is that the base = 15 and the height = 15
In this case, the area = (15)(15) = 225
Here, Quantity A is greater
Case 2:
Another possible scenario is that the base = 25 and the height = 5 (perimeter still equals 60)
In this case, the area = (25)(5) = 125
Here, Quantity B is greater
Answer: D

But, the revealed answer is option D and I am totally confused!
greenlight-admin's picture

In case 1, Quantity A is great than Quantity B (since 225 > 200).
Aside: at this point, the correct answer must be either A or D
In case 2, Quantity B is great than Quantity A (since 125 < 200).

D) The relationship cannot be determined from the information given.
Since we cannot definitively say which Quantity is great, the correct answer must be D.


HI Brent,

in regards to the linked question.
Looking at the explanation of JeffTarget.
I started of in the same way yet was not able to determine how many times bigger x(x+2) is than x(x-2).
He said it that the entire rectangle is 2 times the area of the picture.
How does he come to that conclusion?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/a-at-rectangular-picture-represented-by...

The question asks us to find a value of x so that the area of the frame equals the area of the picture.
So: (frame area) = (picture area)

We also know that: (frame area) + (picture area) = area of entire rectangle
Since (frame area) = (picture area), we can also write: (picture area) + (picture area) = area of entire rectangle
In other words, 2(picture area) = area of entire rectangle

Does that help?


Yes it helps, I seem to have missed the crucial word "represented"


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