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# In the rectangular coordinate system above, the area of triangular

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

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27 Mar 2018, 00:02
nss123 wrote:
Attachment:
IMAGE PT1.jpg
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

For me here estimation works as following:-
we know base is 7 and by looking at figure we can say maximum height is 4 and minimum is 3 and our triangle height will like somewhere in between so
D and E out as maximum is $$\frac{1}{2}* 7*4= 14$$ and minimum is$$\frac{1}{2}*7*3= 10.5$$
now we can also eliminate C AND B as from the figure itself we know its some where around between so height can be 3.4,3.5 or 3.6 and we can also say$$\sqrt{2}$$ is also not possible then only option A left.
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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19 Apr 2018, 13:30
Top Contributor
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 = 28 - (3.5 + 6 + 6) = 12.5

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

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03 May 2018, 06:48
I was struggling with this question- but I've found another method to solving this question, as I was unsure if PQR is a right triangle or not. It is to do with lengths and midpoint calculation.

first lets solve for the length of the base which is P (0,3) and R(7,4) = (√(0-7)²+(3-4)²=> √49+1=> √50=> 5√2

For the height- fist establish the mid point on the base PR= (0+7)/2, (3+4)/2=> 3.5,3.5
height is therefore length from Q(4,0) to midpoint (3.5,3.5)= √(4-3.5)²+(0-3.5)² => √(.5)²+(-.35)² => √.25+12.25=> √12.50
This can be further calculated as √5x5x5x5x2 x 10⁻²=> hence Height= 10⁻¹x 25√2

Put all together in triangle formula= 1/2 x 5√2 x 25√2 x 10⁻¹=> 125 x 10⁻¹= 12.50
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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13 Aug 2018, 07:06
Top Contributor
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

We're trying to find the area of triangle PQR.

Let's draw a rectangle around the triangle (as shown above) 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 = 28 - (3.5 + 6 + 6) = 12.5

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

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20 Jan 2019, 23:02
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

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

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19 Jul 2019, 18:20
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).
Attachments

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In the rectangular coordinate system above, the area of triangular  [#permalink]

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16 Oct 2019, 07:12
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]

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03 Jan 2020, 23:05
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
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Re: In the rectangular coordinate system above, the area of triangular  [#permalink]

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04 Jan 2020, 21:02
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]

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21 Jan 2020, 06:28
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]

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21 Jan 2020, 06:35
Franchise wrote:
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]

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25 Jan 2020, 12:00
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]

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25 Feb 2020, 22:40
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
In the rectangular coordinate system above, the area of triangular   [#permalink] 25 Feb 2020, 22:40

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