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Rhombus problem

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by karthikpandian19 » Wed Mar 21, 2012 12:36 am
Image

ABCD is a rhombus - see figure. ABE is a right triangle. AB is 10m, The ratio of the length of CE to the length of EB is 2 to 3.
What is the area of the trapezoid AECD?

(wanting for solved answer w/o choices)
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by Anurag@Gurome » Wed Mar 21, 2012 12:44 am
karthikpandian19 wrote:Image

ABCD is a rhombus - see figure. ABE is a right triangle. AB is 10m, The ratio of the length of CE to the length of EB is 2 to 3.
What is the area of the trapezoid AECD?

(wanting for solved answer w/o choices)
Area of the trapezoid AECD = (1/2) * (AD + CE) * AE
ABCD is a rhombus so all its 4 sides will be equal, so AB = CB = DC = AD = 10
CE : EB = 2 : 3
If CE = 2x, EB = 3x, then CE + EB = 5x = 10 implies x = 2
CE = 4, EB = 6
In triangle ABE, AB² = AE² + EB²
10² = AE² + 6²
AE² = 100 - 36 = 64
AE = 8

So, area of the trapezoid AECD = (1/2) * (AD + CE) * AE = (1/2) * (10 + 4) * 8 = 14 * 4 = 56 sq meters
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by karthikpandian19 » Wed Mar 21, 2012 12:59 am
Anurag,

I have the question here with the formula ....
Area of the trapezoid AECD = (1/2) * (AD + CE) * AE
1/2 * (Base1 + Base2) * Height

and this height implies AE ..... How is AE the height for the trapezoid AECD...Can you throw some light on it (even if i m missing somewhere?)

Anurag@Gurome wrote:
karthikpandian19 wrote:Image

ABCD is a rhombus - see figure. ABE is a right triangle. AB is 10m, The ratio of the length of CE to the length of EB is 2 to 3.
What is the area of the trapezoid AECD?

(wanting for solved answer w/o choices)
Area of the trapezoid AECD = (1/2) * (AD + CE) * AE
ABCD is a rhombus so all its 4 sides will be equal, so AB = CB = DC = AD = 10
CE : EB = 2 : 3
If CE = 2x, EB = 3x, then CE + EB = 5x = 10 implies x = 2
CE = 4, EB = 6
In triangle ABE, AB² = AE² + EB²
10² = AE² + 6²
AE² = 100 - 36 = 64
AE = 8

So, area of the trapezoid AECD = (1/2) * (AD + CE) * AE = (1/2) * (10 + 4) * 8 = 14 * 4 = 56 sq meters
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by Anurag@Gurome » Wed Mar 21, 2012 1:13 am
karthikpandian19 wrote:Anurag,

I have the question here with the formula ....
Area of the trapezoid AECD = (1/2) * (AD + CE) * AE
1/2 * (Base1 + Base2) * Height

and this height implies AE ..... How is AE the height for the trapezoid AECD...Can you throw some light on it (even if i m missing somewhere?)
Image

Hope this figure answers your query.
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by karthikpandian19 » Wed Mar 21, 2012 1:17 am
Got it...now thank you
Anurag@Gurome wrote:
karthikpandian19 wrote:Anurag,

I have the question here with the formula ....
Area of the trapezoid AECD = (1/2) * (AD + CE) * AE
1/2 * (Base1 + Base2) * Height

and this height implies AE ..... How is AE the height for the trapezoid AECD...Can you throw some light on it (even if i m missing somewhere?)
Image

Hope this figure answers your query.
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by Anurag@Gurome » Wed Mar 21, 2012 1:20 am
karthikpandian19 wrote:Got it...now thank you
Welcome!
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by GMATGuruNY » Wed Mar 21, 2012 5:04 am
Redraw the figure so that it's easier to see the relationships:
Image

Since CE:EB = 2:3 = 4:6, EB=6.
∆ABE is a 6-8-10 triangle in which AE=8.
Thus, the height of both rhombus ABCD and ∆ABE is 8.
Area of rhombus ABCD = bh = 10*8 = 80.
Area of ∆ABE = (1/2)bh = (1/2)(6)(8) = 24.
Thus, trapezoid AECD = rhombus ABCD - ∆ABE = 80-24 = 56.

Alternate method:
Area of a trapezoid = (b1 + b2)/2 * h.
Thus, the area of trapezoid AECD = (4+10)/2 * 8 = 56.
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by ronnie1985 » Wed Mar 21, 2012 9:16 am
BE = (3/5)*10 = 6 = > AE^2 = 100-36 =64 = > AE = 8
CE = 4 AD = 10 AE = 8 and DC = 10 Area Trapezoid = (1/2) * (AD+CE) * (AE) = (1/2)*(10+4)*8 = 56 m^2
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