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In the rectangular coordinate system above, the area of tria
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06 Jul 2018, 08:39
06 Jul 2018, 08:39
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65% (00:57) correct
35% (00:54) wrong based on 20 sessions
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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
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07 Jul 2018, 08:00
07 Jul 2018, 08:00
Carcass wrote:
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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 = \(\frac{1}{2} * base * altitude = \frac{1}{2} * PR * MC = \frac{(b - d)(c - a)}{2}\)
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gmatclubot
Re: In the rectangular coordinate system above, the area of tria [#permalink]
07 Jul 2018, 08:00
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