Since each side of ΔACD has length 3 then ACD is an equilateral triangle and its each angle is 60°.

Now, the are of equilateral triangle is \(area_{equilateral}=side^2*\frac{\sqrt{3}}{4}=9\frac{\sqrt{3}}{4}\) (for more check math-triangles-87197.html);

Next, since angle A is 60° then right triangle ABE is a 30°-60°-90° right triangle. Now, in a right triangle where the angles are 30°, 60°, and 90° the sides are always in the ratio \(1 : \sqrt{3}: 2\), the leg opposite 30° (AB) corresponds with \(1\) and the leg opposite 60° (BE) corresponds with \(\sqrt{3}\) (the hypotenuse AE corresponds with 2). So, since \(AB=1\) then \(BE=\sqrt{3}\), and the area of ABE is \(\frac{1}{2}*AB*BE=\frac{\sqrt{3}}{2}\);

The area of of region BCDE is the area of ACD minus the area of ABE: \(9\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{2}=9\frac{\sqrt{3}}{4}-2\frac{\sqrt{3}}{4}=7\frac{\sqrt{3}}{4}\).

Answer: B.

Hope it's clear.
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equilateral triangle hence Angle A = 60 deg
thus, in Triangle ABE, AB = 1 BE = 3^(1/2) and AE = 2

Area BCDE = Area ACD - Area ABE

Area ACD = 3^(1/2)/4 * (side)^2

Area ABE = 1/2 * 3^(1/2) * 1

we get 3.5/2 * 3^(1/2) multiplying numerator and denominator by 2 we get B

In Triangle ABE one angle is 90, angle BAE is 60 (as ACD is equilateral) hence other angle is 30.

Sin 30 = 1/2 , Sin 60 = Root 3 / 2

Sin 30 = side opposite to the 30 degrees / Hypotenuse
Sin 60 = side opposite to the 60 degree / Hypotenuse

Solving we get Ab = 1 , AE = 2 , BE = 1

Area of ABE = 1/2 * 1 * Root 3

BCDE = ACD - ABE
solving we get option B
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Solution

Given:

• In triangle ACD, each side length = 3
• AB = 1

To find:

• The area of quadrilateral BCDE

Approach and Working:
Area of BCDE = area of ACD – area of ABE

• Area of ACD = \(\frac{√3 * 3^2}{4} = \frac{9√3}{4}\)
• Area of ABE = \(\frac{1}{2} * 1 * BE = \frac{BE}{2}\)

And, we know angle A = 60 degrees. So, we can say that ABE is a 30 – 60 – 90 degrees triangle.

• Thus, AB : BE : AE = 1 : √3 : 2

o Implies, BE = √3

• So, area of ABE = \(\frac{√3}{2}\)

Therefore, area of BCDE = \(\frac{9√3}{4} - \frac{√3}{2} = \frac{7√3}{4}\)

Hence, the correct answer is Option B

Answer: B


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what is the formula of area of an irregular quadrilateral ?

Hi dave13, How have you been?

No - there is no such formula ( at least not that I know of, or within the scope of GMAT)

So, this question is testing three things -
1. Area of an equilateral triangle. (Also the fact that all angles are equal to 60 degrees in one)
2. 30-60-90 triangle side ratios
3. Area of a right-angled triangle.

My thought process -
Okay. We need to find the area of an unual looking quadrilateral. Let's try to find the area of the two regular triangles that we know of and subtract the smaller one from the larger triangle to find the unusual area.
The side of the equilateral triangle is 3 so its area is \(\frac{s^2}{4}*\sqrt{3}\) remember this - comes handy
\(\frac{3^2}{4}*\sqrt{3}\)
\(\frac{9}{4}*\sqrt{3}\)

The smaller triangle is 30-60-90 (one angle is 90 degrees and one angle is common between itself and the equilateral triangle)

Okay, so the side ratios of a 30-60-90 triangle are 1, \(\sqrt{3}\) and 2 with the smallest side being opposite the smallest angle.
This implies area of right angle triangle is -
\(\frac{1}{2}b*h\)
\(\frac{1}{2}*\sqrt{3}\)

So the area we are interested in will be the difference -
\(\frac{9}{4}*\sqrt{3} - \frac{1}{2}*\sqrt{3}\)
\(\frac{7}{4}*\sqrt{3}\)


Hope this helps. Have a nice day.

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Now since its an equilateral triangle area = \(\frac{9}{4} \sqrt{3}\)

area of smaller triangle = 1/2 * 1 * \(\sqrt{3}\), since it will be a 30-60-90 triangle

Area of region will be

\(\frac{9}{4} \sqrt{3}\) - 1/2 * 1 * \(\sqrt{3}\)

\(\frac{7}{4} \sqrt{3}\)
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Use the information given to determine that two triangles exist:
1. An Equilateral triangle with lengths of 3
2. A 30-60-90 triangle with base of 1 and height of \(\sqrt{3}\)
Find the area of Equilateral triangle using\(\frac{S^2\sqrt{3}}{4}\) and deduct the area of the 30-60-90 triangle.

Done.

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[quote="boomtangboy"]
If each side of ΔACD above has length 3 and if AB has length 1, what is the area of region BCDE?

(A) \(\frac{9}{4}\)

(B) \(\frac{7}{4} \sqrt{3}\)

(C) \(\frac{9}{4} \sqrt{3}\)

(D) \(\frac{7}{2} \sqrt{3}\)

(E) \(6 + \sqrt{3}\)

PS18302.01

Great question;

Please find attached the explanation for the question.

Answer B

PS: Angle A is 60 degree because triangle ACD is an equilateral triangle. Therefore In triangle BAE angle E has to be 30 degree.

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