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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
1
Kudos
Seeing figure we can determine thaat AD= 3k and AB is K
given area ABCD = 12
3k^2 = 12
k=2
#1 is sufficient
as by using distance formula we can find sides of ∆AED
#2
AD=6 ;
3k=6
k=2
sufficient
IMO D

In the rectangular coordinate system above, what is the area of triangular region AED?

(1) The area of region ABCD is 12.
(2) AD = 6
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
1
Kudos
(1) The area of region ABCD is 12.
we know that the area of the triangle is the same if the base and height are the same.
we have an area of ABCD which is 12 so the area above portion will also be 12.
The area of a rectangular will be 24 and the area of the triangle will be half of rectangle i.e. 12.
Sufficient.

(2) AD = 6
AD is 6 i.e. 4k - k = 6
k=2
We can identify height
h=4
area of triangle = 1/2*4*6 = 12
Sufficient

Answer: D
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
Quote:

In the rectangular coordinate system above, what is the area of triangular region AED?

(1) The area of region ABCD is 12.
(2) AD = 6


(1) The area of region ABCD is 12: \(3k(k)=12…k^2=4…k=2\), sufic.
(2) AD = 6; \(AD=3k=6…k=2\), sufic.

Answer (D)
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
Let point(k,6k) in the coordinate system is called point F.

Area of Δ AED = Area of ΔAEF - Area of ΔDEF
 \(= \frac{1}{2} * AF * EF – 1/2 * AF * DF\)
 \(= \frac{1}{2} * EF *(AF – DF)\)
 \(= \frac{1}{2} * (3k - k) * (6k – k – (4k – k))\)
 \(= \frac{1}{2} * 2k * 2k\)
 \(= \frac{1}{2} * 4k^2\)
So, k = ?

(1) The area of region ABCD is 12.

Area of region ABCD = AB * AD
 = (2k - k) * (4k - k)
 = 3k*k
 3k*k = 12
 K = 2

SUFFICIENT.

(2) AD = 6
AD = 4k – k = 3k
 3k = 6
 K = 2

SUFFICIENT.

Answer (D).
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
to find area of the triangle = height X base AD/2= (2k X 3K )/2= 3k^2

1)sufficient : we are given AD x k = 3k^2 =12
2)sufficient AD=3k given so we can find 3k^2
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
In the coordinate system,
The area of triangle of AED = \(\frac{5k*2k}{2}-\frac{2k*2k}{2}\)=\(5k^{2}-2k^{2}\)=\(3k^{2}\) ???

Statement1: ABCD is a rectangle --> k*3k=12 --> k=2
--> \(3k^{2}\)=12
Sufficient

Statement2: AD=3k=6
k=2
The same result as Statement1
Sufficient

The answer is D.
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
Hi guys, a rather silly question that I had..

From (1) we find k=2 and with this we can find all the co-ordinates of the triangle. With distance formula we can find the lengths of each side of triangle. But since area of triangle is bh/2, how do we know which is side is the height and which is the base?

Could someone please explain.
Thanks.
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
mokiburnsred wrote:
Hi guys, a rather silly question that I had..

From (1) we find k=2 and with this we can find all the co-ordinates of the triangle. With distance formula we can find the lengths of each side of triangle. But since area of triangle is bh/2, how do we know which is side is the height and which is the base?

Could someone please explain.
Thanks.

every side of the triangle can be treated as base we are treating the bottom side as base as it coincides with the rectangle and we can find its value easily
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
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Re: In the rectangular coordinate system above, what is the area of triang [#permalink]
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