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

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Joined: 02 Sep 2009
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In the rectangular coordinate system above, the area of triangular reg  [#permalink]

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08 Jul 2018, 21:40
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Difficulty:

45% (medium)

Question Stats:

58% (01:17) correct 42% (01:26) wrong based on 77 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. $$\frac{c(b+d)}{2}$$

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

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08 Jul 2018, 21:59
3
$$Area = \frac{1}{2} * base * height$$

Let the point where PR intersect x-axis is S. The co-ordinates of S will be (a,0)

the base of the triangle is PR
the height of the triangle is QS

Co-ordinates of P and Q are (a,b) and (a,d) respectively

base = |b| + |d| = b-d (since |b| = b as it is y co-ordinate in 2nd quadrant and |d|=-d as it is the y co-ordinate in 3rd quadrant)
height = |c| - |a| = c-a (since |c| = c as it is x co-ordinate in 1st quadrant and |a|=-a as it x co-ordinate is in 2nd quadrant)

Hence $$Area = \frac{(c-a)(b-d)}{2}$$

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

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09 Jul 2018, 04:15
$$Base = c + a$$

$$Height = b + d$$

$$Area =$$ $$\frac{(c+a)(b+d)}{2}$$

Answer C.

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

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09 Jul 2018, 04:21
GyMrAT wrote:
$$Base = c + a$$

$$Height = b + d$$

Thanks,
GyM

The value of a will be -1, -2 ,-3 etc since it is x-coordinate and it is in left half of x-axis. So we need to add |a| to c. If we simply add 'a' to value of c, we don't get the correct height. Same is the case for b+d
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Re: In the rectangular coordinate system above, the area of triangular reg &nbs [#permalink] 09 Jul 2018, 04:21
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# In the rectangular coordinate system above, the area of triangular reg

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