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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
@ bunuel - pls help. Is it always that area of any quadrilateral will be its bxh?
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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meenakshi1 wrote:
@ bunuel - pls help. Is it always that area of any quadrilateral will be its bxh?


No, that's not true. Please check here: math-polygons-87336.html

Hope it helps.
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


[spoiler=]Source: GMATPrep Exam Pack 1

OA: A


Theory: The area of any # of trianlge(s) between
A.2 fixed parallel lines, and
B.With the same base is ALWAYS the same.

The first point makes sure that the height of the given triangles is always the same;the second point mandates the base lenght to be the same for all of them.

From the given image, try sliding the point R towards Q till the line segment RP co-incides with the side QP of the rectangle and thus, RT becomes the diagonal.
Now, if we were told to find the area of the newly formed triangle PRT, it would be nothing but \(\frac{1}{2}*24\)

Hope this helps.
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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I'm sorry but I have a bit of a question on this:

(1) The are of PQST is 24 (.5*B*H) so that's the area of the triangular region: sufficient

(2) The length of the line segment RT is 5.

RT is 5
If you drop a perpendicular line down from R to PT that forms a 90 degree triangle. Let's call that point X.
Can we then assume a 3-4-5 triangle? I got 90 degrees and a 5 right?
Then triangle XRT is similar to XRP because they share the same side RX so you can set up proportions to find length PX and XT to get length PT and since you know RX is 4. b*h*.5 is the answer? thus sufficient???
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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SweetiePi wrote:
I'm sorry but I have a bit of a question on this:

(1) The are of PQST is 24 (.5*B*H) so that's the area of the triangular region: sufficient

(2) The length of the line segment RT is 5.

RT is 5
If you drop a perpendicular line down from R to PT that forms a 90 degree triangle. Let's call that point X.
Can we then assume a 3-4-5 triangle? I got 90 degrees and a 5 right?
Then triangle XRT is similar to XRP because they share the same side RX so you can set up proportions to find length PX and XT to get length PT and since you know RX is 4. b*h*.5 is the answer? thus sufficient???


No, the red part is not correct. Hypotenuse of 5 does not necessarily means that the legs must be 3 and 4.
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


You may also prove this algebraically. Area of the rectangle is the sum of the all three embedded triangles:

(1/2*RQ*ST) + (1/2*RS*ST) + (1/2*PT*ST). But, you will notice that RQ +RS = QS= PT; hence the sum of area of triangle PQR and RST is equal to the area of PRT. Thus, the area of triangle PRT is half of the rectangle.
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


Area of triangle = 1/2bh.
Area of rectangle = bh

1) Gives bh and is therefore suff.

2) Cannot get b or h. Insuff.

A

Kudos if you agree!
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


I picked E because I thought we couldn't assume R was on QS since it is a Data Sufficiency Question and therefore, we don't know the height of the triangle. Just as we couldn't assume PQST was a quadrilateral until statement 1. I'm confused on what we can and cannot assume in these questions. Does anybody have any good advice on this?
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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jwang1191 wrote:
yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


I picked E because I thought we couldn't assume R was on QS since it is a Data Sufficiency Question and therefore, we don't know the height of the triangle. Just as we couldn't assume PQST was a quadrilateral until statement 1. I'm confused on what we can and cannot assume in these questions. Does anybody have any good advice on this?


R is shown to be on QS so it is actually on QS. This will never be a trick on the GMAT.
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


Tricky question ahhh

What the trap in this question is getting you to think that you need a particular hypotenuse length in order to calculate the respective areas- HOWEVER if you think about the question from a formulaic perspective then you will see that the area of a right triangle

1/2 BH essentially equals the area of the square

It doesn't actually matter what the height and length are- 6 and 4 or 8 and 3 because they will multiply to 24 which can just be substituted for BH in

1/2 BH

And will result in the same answer

A
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


Target question: What is the area of triangular region PRT?

Statement 1: The area of rectangular region PQST is 24
The area of ∆PRT = (base)(height)/2
Statement 1 tells us that (base)(height) = 24
Since the base and the height of the rectangle are the SAME as the base and the height of the triangle PRT, we can use this information to find the area of ∆PRT.
The area of ∆PRT = (base)(height)/2 = 24/2 = 12
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The length of line segment RT is 5
The area of ∆PRT = (base)(height)/2
Statement 2 tells us NOTHING about the base and height, so we can't use this to find the area of ∆PRT.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
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Hi Bunuel,

If it is not always type that the area of a quadrilateral is base*height then how is that property being used on this question? I answered E to this question because i was using the formula for rectangle's area as length * width. I am not able to follow the solution of this problem as I'm unable to understand the concept that is applied in regards to the area of a quadrilateral being base*height. Isn't the formula for the area of a rectangle length*width? I apologize that i'm asking such a basic question, but I'm unclear on this problem.

Thank you!

Bunuel wrote:
meenakshi1 wrote:
@ bunuel - pls help. Is it always that area of any quadrilateral will be its bxh?


No, that's not true. Please check here:

Hope it helps.
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
Gmatstudent2018 wrote:
Hi Bunuel,

If it is not always type that the area of a quadrilateral is base*height then how is that property being used on this question? I answered E to this question because i was using the formula for rectangle's area as length * width. I am not able to follow the solution of this problem as I'm unable to understand the concept that is applied in regards to the area of a quadrilateral being base*height. Isn't the formula for the area of a rectangle length*width? I apologize that i'm asking such a basic question, but I'm unclear on this problem.

Thank you!

Bunuel wrote:
meenakshi1 wrote:
@ bunuel - pls help. Is it always that area of any quadrilateral will be its bxh?


No, that's not true. Please check here:

Hope it helps.


Gmatstudent2018, We know that L*W = 24 for the rectangle. Let's say that PQ=ST=Length and QS=PT=Width

For the triangle, let's assume our Base is line PT. So our Base=Width. Cool?

Now we need to find out what our Height would be for the triangle.

Our height would start at vertex R and be a line that is perpendicular with our Base... maybe draw a dotted line on your scratch piece of paper.

Now, how does your dotted line compare with the PQ and ST?
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
When I saw this question, the figure that came to my mind was that of two different rectangles each of which were divided equally into triangles.

See attached image and solution. Not the most elegant one, but worked for me.
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IMG_20200216_001153.jpg
IMG_20200216_001153.jpg [ 2.95 MiB | Viewed 42781 times ]

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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
VeritasPrepBrandon wrote:
It statement 1, is it supposed to say "The area of the rectangular region PQST is 24? If so, statement 1 gives you enough information to prove that the area is 12. This is because the area of a quadrilateral is equal to base * height. Because the triangle is inscribed within the rectangle, we know that the base * height of the triangle must also equal 24. But the area of a triangle is equal to 1/2*base*height, so its area must equal 1/2*24 = 12.

Try some numbers to prove this to yourself. Say that the base of the rectangle is 24, and its height 1. That makes the base of the triangle 24 and its height 1, and 1/2*24*1 = 12. How about if the base is 6 and the height is 4? 1/2*6*4 = 12. The area of the triangle will always be 12.

Statement 2, on the other hand, provides neither the base nor the height, and is this insufficient.

This makes the correct answer A.

I hope this helps!!!



I know the GMAT can be literal, so how do we know that the triangle is inscribed on the rectangle? Bunuel
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In the figure shown, what is the area of triangular region PRT? [#permalink]
BrentGMATPrepNow wrote:
yuvrajsub wrote:
Attachment:
image.png
In the figure shown, what is the are of the triangular region PRT?

(1) The area of rectangular region PQST is 24.
(2) The length of line segment RT is 5.


Target question: What is the area of triangular region PRT?

Statement 1: The area of rectangular region PQST is 24
The area of ∆PRT = (base)(height)/2
Statement 1 tells us that (base)(height) = 24
Since the base and the height of the rectangle are the SAME as the base and the height of the triangle PRT, we can use this information to find the area of ∆PRT.
The area of ∆PRT = (base)(height)/2 = 24/2 = 12
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: The length of line segment RT is 5
The area of ∆PRT = (base)(height)/2
Statement 2 tells us NOTHING about the base and height, so we can't use this to find the area of ∆PRT.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer = A

Cheers,
Brent


Hi Brent, for S2, why cant we draw a perpendicular line from VERTEX R downward to PT, thereby splitting the triangle into two 45-45-90 triangles. Once we have two 45-45-90 triangles, we can solve the question by using information from Statement 2? (EG: RT the "hypotenuse" will be 5, so height will be 5/ \sqrt{2}, etc)
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Re: In the figure shown, what is the area of triangular region PRT? [#permalink]
Bunuel wrote:
SweetiePi wrote:
I'm sorry but I have a bit of a question on this:

(1) The are of PQST is 24 (.5*B*H) so that's the area of the triangular region: sufficient

(2) The length of the line segment RT is 5.

RT is 5
If you drop a perpendicular line down from R to PT that forms a 90 degree triangle. Let's call that point X.
Can we then assume a 3-4-5 triangle? I got 90 degrees and a 5 right?
Then triangle XRT is similar to XRP because they share the same side RX so you can set up proportions to find length PX and XT to get length PT and since you know RX is 4. b*h*.5 is the answer? thus sufficient???


No, the red part is not correct. Hypotenuse of 5 does not necessarily means that the legs must be 3 and 4.


Because it can be a 45-45-90 triangle or anything else..
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