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

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Manager
Joined: 23 Sep 2016
Posts: 231
Re: In the rectangular coordinate system above, the area of triangular [#permalink]

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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.
SVP
Joined: 12 Sep 2015
Posts: 2308
Re: In the rectangular coordinate system above, the area of triangular [#permalink]

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19 Apr 2018, 14:30
Expert's post
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

[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:

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

Cheers,
Brent
_________________

Brent Hanneson – Founder of gmatprepnow.com

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

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