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

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In the rectangular coordinate system above, the area of tria [#permalink]  06 Jul 2018, 08:39
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90% (01:02) correct 9% (01:09) wrong based on 11 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. $$\frac{c(b+d)}{2}$$
[Reveal] Spoiler: OA

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Re: In the rectangular coordinate system above, the area of tria [#permalink]  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 = $$\frac{1}{2} * base * altitude = \frac{1}{2} * PR * MC = \frac{(b - d)(c - a)}{2}$$
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Re: In the rectangular coordinate system above, the area of tria   [#permalink] 07 Jul 2018, 08:00
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