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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
nss123 wrote:

In the rectangular coordinate system above, the area of triangular region PQR is

(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25


So area of Trapezium = 1/2 * sum of parallel sides * height
= 1/2 * (OQ + O'R) * 7
= 49/2
=24.5

Now from this we need to subtract the two tingles to give area of triangle QPR
= 24.5 - ( 6*2)
= 12.5

Answer A.
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
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Multiple methods exist to solving this problem.

The easiest, at least for me, was pythag triples. See my diagram

We can actually deduce the distance of each point using the x-y axis, giving us 5,5 (legs of the triangle).
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In the rectangular coordinate system above, the area of triangular [#permalink]
Pathfinder wrote:
I came to the same result but my way of solving was a little bit different.

Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.

Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\).

We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5.



How do we know that PGR is the right triangle? It is also can be isosceles?
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In the rectangular coordinate system above, the area of triangular [#permalink]
nss123 wrote:


In the rectangular coordinate system above, the area of triangular region PQR is

(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25


Attachment:
IMAGE PT1.jpg


I FOUND THE EASIEST AND QUICKEST METHOD EVER .It took me 1 minute to solve .

step 1 write down all the coordinates in vertical format inside MODE (because area has to be positive always ) and multiply it with 1/2

1/2 * MODE/ X1 - X2 X1-X3 /MODE here 1/2* MODE / 0 - 7 0-4 / MODE
................/ Y1 - Y2 Y1-Y3 /....................................../ 3 - 4 3-0 /

Step2 (upon solving you get ) 1/2* / A B / SO cross multiply you get 1/2 * /AD - BC / . Here, 1/2 * MODE / -7 -4/
.................................................../ C D /........................................................................................... /-1 -3/

Your answer is 25/2 = 12.5 . This method is quickest , just be careful with signs .

TO UNDERSTAND IT IN DETAILS WATCH IT ON BELOW LINK
https://www.youtube.com/watch?v=It9Vd3UFYVg
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
No need to calculate anything here. The area of the largest triangle inscribed in a rectangle is half the area of the rectangle. Here we have a rectangle of area 7*4 = 28. The largest inscribed triangle happens to be if Q = (0.4) ,i.e base of triangle = length of rectangle. Since the area of the largest triangle possible is 14, the answer has to be less than 14. I.e A.
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
Bunuel wrote:
gottabwise wrote:
Pathfinder wrote:
I came to the same result but my way of solving was a little bit different.

Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.

Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\).

We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5.


Nookway...thanks for the colored graph. It's a great visual and helpful reminder of how to be smarter/more efficient.

Pathfinder...is it necessary to determine whether PQR is a right triangle? I'm unsure about the relevance.

Method - distance formula for PQ & PR (because they're perpendicular), area formula (1/2B*H) = 1/2(PQ)(PR)=1/2(5)(5)


PQR just happened to be right triangle, so if we noticed this fact we could use properties of a right triangle to solve the problem (for example the way Pathfinder did). On the other hand solution provided by nookway works no matter whether PQR is right or not, also it requires much less calculations.



How do we know that the triangle is right? Could it not have been an isosceles triangle?
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
Expert Reply
Franchise wrote:
How do we know that the triangle is right? Could it not have been an isosceles triangle?


I believe it's answered on the previous pages. Please re-read the discussion. Hope it helps.
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
How can we state that the triangle in this problem is a right angle triangle?
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In the rectangular coordinate system above, the area of triangular [#permalink]
use the determinant method 1/2{(x1-x2)(y2-y3)-(x2-x3)(y1-y2)}.this is the easiest method and the modular sign has to be used becs area cant be negative
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In the rectangular coordinate system above, the area of triangular [#permalink]
I have found that these questions are easier to solve using the Discriminant property.

Set all X's and Y's as separate columns in the matrix, with a column of 1s, solve for the 3x3 discriminant and appropriately multiply by +/- 1/2.
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
This is the easiest way to do this problem... do not look any further Team.

You need to memorize your pythagorean triples (3,4,5) (5,12,13) and if you don't have these memorized to so now or else you will be in very deep trouble. Memorize (3,4,5) & (5,12,13)

It is very easy to determine the triangle centerQP which will give a hypotenuse of 5 (height of Point Q is (0,3), length of Point P is (0,4), your hypotenuse is easily 5.
This same concept goes line QR being they hypotenuse of 5 for another triangle that you can visualize.

Now, this is the part that messed me up initially-- do not try to find the hypotenuse of triangle QPR... its asking for the Area. finding the hyp will result in 5squareroot2 which is a similar answer but A=1/2bh and is sufficient to find with the 1/2 5*5

Hope this helps.
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
JeffTargetTestPrep wrote:
In the rectangular coordinate system above, the area of triangular region PQR is

(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25

We begin by drawing a rectangle that circumscribes the given triangle, creating 3 right triangles, which we label as A, B, and C. Notice that each of these 3 triangles is a right triangle. To determine the area of triangular region PQR, we can subtract the combined areas of triangles A, B, and C from the area of the rectangle.



Let’s determine the area of each right triangle.

Triangle A:

Area = base x height x 1/2

A = 7 x 1 x ½ = 3.5

Triangle B:

A = 4 x 3 x ½ = 6

Triangle C:

A = 3 x 4 x ½ = 6

The sum of the areas of triangles A, B, and C is 3.5 + 6 + 6 = 15.5

Finally we need the area of the rectangle:

Area = length x width

Area = 7 x 4 = 28.

So the area of triangle PQR is 28 – 15.5 = 12.5.

Answer: A



Great explanation JeffTargetTestPrep. In calculation triangle area, how to decide using which value to minus which value between P (4,0) , R(7,4) for e.g? Thanks
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In the rectangular coordinate system above, the area of triangular [#permalink]
Bunuel wrote:
gottabwise wrote:
Pathfinder wrote:
I came to the same result but my way of solving was a little bit different.

Using Pythagorean theorem we can easily find that PQ and PR are 5 each. Now we have to determine is the angle PQR right angle.

Yes it is. How we know that. Using Pythagorean theorem again we can determine that QR is\(5\sqrt{2}\) simply by solving \(\sqrt{1^2 + 7^2}\) = \(\sqrt{50}\) = \(5\sqrt{2}\).

We know that PQ and QR are both 5 and their base is \(5\sqrt{2}\) and that diagonale of the square is \(a\sqrt{2}\). So, triangle PQR must be half of the square with the base of 5 or 12,5.


Nookway...thanks for the colored graph. It's a great visual and helpful reminder of how to be smarter/more efficient.

Pathfinder...is it necessary to determine whether PQR is a right triangle? I'm unsure about the relevance.

Method - distance formula for PQ & PR (because they're perpendicular), area formula (1/2B*H) = 1/2(PQ)(PR)=1/2(5)(5)


PQR just happened to be right triangle, so if we noticed this fact we could use properties of a right triangle to solve the problem (for example the way Pathfinder did). On the other hand solution provided by nookway works no matter whether PQR is right or not, also it requires much less calculations.



Can we find out the coordinates of the point on the line QR where a Perpendicular from P, if drawn would meet? Then that would be our Height for the triangle, no matter if it's right-angled or not!!
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
BrentGMATPrepNow wrote:
nss123 wrote:


In the rectangular coordinate system above, the area of triangular region PQR is

(A) 12.5
(B) 14
(C) 10√2
(D) 16
(E) 25


Attachment:
IMAGE PT1.jpg


Let's draw a rectangle around the triangle (as shown below) and then subtract from the rectangle's area (28) the areas of the 3 right triangles that surround the triangle in question.

We get the following:


So, the area of PQR = Area of rectangle - (area of 3 right triangles)
= 28 - (3.5 + 6 + 6)
= 12.5

Answer: A
Cheers,
Brent


Hi BrentGMATPrepNow, to clarify how do we work out base * height values from the given cocordinate points and do we ignore minus sign if getting a negative value ? Thanks Brent
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
Expert Reply
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Kimberly77 wrote:
Hi BrentGMATPrepNow, to clarify how do we work out base * height values from the given cocordinate points and do we ignore minus sign if getting a negative value ? Thanks Brent


If you get a negative value, then turn it into a positive value
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In the rectangular coordinate system above, the area of triangular [#permalink]
BrentGMATPrepNow wrote:
Kimberly77 wrote:
Hi BrentGMATPrepNow, to clarify how do we work out base * height values from the given cocordinate points and do we ignore minus sign if getting a negative value ? Thanks Brent


If you get a negative value, then turn it into a positive value


Get it thanks BrentGMATPrepNow
Also how do we work out base * height values from the given cocordinate points?
Lastly for rectangular area of l * W, is it just using R (7,4) here? Thanks Brent
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
Expert Reply
Top Contributor
Kimberly77 wrote:
BrentGMATPrepNow wrote:
Kimberly77 wrote:
Hi BrentGMATPrepNow, to clarify how do we work out base * height values from the given cocordinate points and do we ignore minus sign if getting a negative value ? Thanks Brent


If you get a negative value, then turn it into a positive value


Get it thanks BrentGMATPrepNow
Also how do we work out base * height values from the given cocordinate points?
Lastly for rectangular area of l * W, is it just using R (7,4) here? Thanks Brent


For this question you have to figure out the base and the height of each triangle.
The top left triangle has a horizontal base with length 7 and a height of 1, which means the area is 3.5
The bottom left triangle has a horizontal base with length 4 and a height of 3, which means the area is 6
The bottom right triangle has a horizontal base with length 4 and a height of 4, which means the area is 6
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Re: In the rectangular coordinate system above, the area of triangular [#permalink]
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