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

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

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New post 08 Jul 2018, 20:40
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A
B
C
D
E

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59% (01:15) correct 41% (01:25) wrong based on 83 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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New post 08 Jul 2018, 20:59
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\(Area = \frac{1}{2} * base * height\)

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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New post 09 Jul 2018, 03:15
\(Base = c + a\)

\(Height = b + d\)

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


Answer C.


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

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New post 09 Jul 2018, 03:21
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, 03:21
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