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In the rectangular coordinate system above, the area of triangular

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

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New post 27 Mar 2018, 01:02
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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New post 19 Apr 2018, 14:30
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nss123 wrote:
Image

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


[Reveal] Spoiler:
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:
Image

So, the area of PQR = 28 - (3.5 + 6 + 6) = 12.5

Answer: A
Cheers,
Brent
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Brent Hanneson – Founder of gmatprepnow.com

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Re: In the rectangular coordinate system above, the area of triangular   [#permalink] 19 Apr 2018, 14:30

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In the rectangular coordinate system above, the area of triangular

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