GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In the rectangular coordinate system above, the area of tria
[#permalink]
06 Jul 2018, 08:39
06 Jul 2018, 08:39
Expert Reply
00:00
Question Stats:
68% (00:57) correct
31% (00:56) wrong based on 19 sessions
Hide Show timer Statistics
Attachment:
xy.jpg [ 11.62 KiB | Viewed 3710 times ]
In the rectangular coordinate system above, the area of triangular region PQR is
A. \(\frac{ac}{2}\)
B. \(\frac{(c-a)(b-d)}{2}\)
C. \(\frac{(c+a)(b+d)}{2}\)
D. \(\frac{c(b-d)}{2}\)
E. \(\frac{c(b+d)}{2}\)
Re: In the rectangular coordinate system above, the area of tria
[#permalink]
07 Jul 2018, 08:00
07 Jul 2018, 08:00
Carcass wrote:
Attachment:
The attachment xy.jpg is no longer available
In the rectangular coordinate system above, the area of triangular region PQR is
A. \(\frac{ac}{2}\)
B. \(\frac{(c-a)(b-d)}{2}\)
C. \(\frac{(c+a)(b+d)}{2}\)
D. \(\frac{c(b-d)}{2}\)
E. \(\frac{c(b+d)}{2}\)
Here,
The length of PR = \(\sqrt{(a- a)^2 + (b - d)^2} = \sqrt{(b - d)^2} = b - d\)
Let us take point M, whose co -ordinates are = (a , 0) (Since it lies in the same x co-ordinate as of point P and Point R . It also lies in the same y coordinate of point Q)
Now the length of MC = \(\sqrt{(c- a)^2 + (0 - 0)^2} = \sqrt{(c - a)^2} = c - a\)
Now the area of △PQR = \(\frac{1}{2} * base * altitude = \frac{1}{2} * PR * MC = \frac{(b - d)(c - a)}{2}\)
Attachments
Untitled.png [ 5.34 KiB | Viewed 3657 times ]
gmatclubot
Re: In the rectangular coordinate system above, the area of tria [#permalink]
07 Jul 2018, 08:00
Moderators:
