If triangle ABD is an equilateral triangle and AB = 6 and CE = 18, wha

Let Height of Triangle = Height of Trapezoid = h

Area of triangle \(= \frac{1}{2} * 6 * h\)

Area of trapezoid \(= \frac{1}{2} (6+18)*h\)

Ratio \(= \frac{6}{24} = \frac{1}{4}\)

Answer = B

The answer is B.
Triangle ABD area equal: 6* square root 3/2 * 6/2 = 9*square root of 3
ABCE area equal: Height * ( AB+ CE)/2 = Height* ( 6+18)/2 = Height * 12.
Height = 6* square root of 3/2
ABCE Area = 36*squareroot of 3
=> ABD area/ABCE area =1/4

There maybe a short cut approach?

This is a isosceles trapezium, however I'm facing difficulty to prove it via calculation

Area of a Trapezium = 1/2 * ( Sum of lengths of Two parallel Lines) * Distance between two Lines.

Distance between two lines = height of equilateral Triangle.

So Answer = area of Equilateral Triangle / Area of Trapezium = 1/4.

Hope This Helps!!!!

Regards
Narayana Raju G

We can let the height of the triangle (which is also the height of the trapezoid) = h.

Let’s first determine the area of triangle ABD in terms of h:

area = (1/2)(base)(height) = (1/2)(6)(h) = 3h

Next we can determine the area of trapezoid BACE in terms of h:

area = (1/2)(base 1 + base 2)(height) = (1/2)(18 + 6)(h) = 12h

Thus, the fraction of the trapezoid that is shaded is 3h/12h = 1/4.

Answer: B
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Although Area of Trapezoid formula will make this easier, it is not required. I thought I'd post the way to do this without that formula:

Method Using Only Triangle Formulas

SHADED TRIANGLE

AB=6, and that triangle is equilateral, so AB=BD=DA=6.

Our height will be 3\(\sqrt{3}\). We can make Right Triangles and determine this. (See picture)

Height = 3\(\sqrt{3}\) for all three triangles

Area of ABD = \(\frac{1}{2}\) * b * h ----> \(\frac{1}{2}\) * 6 * 3\(\sqrt{3}\) ----> 9\(\sqrt{3}\)


OTHER TWO TRIANGLES

We know that CAD=DBE. They are similar triangles. It is given that CE=18, so CD=DE=9.

Base = 9

Area of CAD = \(\frac{1}{2}\) * 9 * h

Area of CAD = \(\frac{1}{2}\) * 9 * 3\(\sqrt{3}\) = \((27\sqrt{3})/2\)

We have two of these triangles, so we multiply \((27\sqrt{3})/2\) by 2 and get 27\(\sqrt{3}\)


TOTAL AREA OF ALL THREE TRIANGLES

27\(\sqrt{3}\) + 9\(\sqrt{3}\) = 36\(\sqrt{3}\)

FRACTION SHADED

9\(\sqrt{3}\) / 36\(\sqrt{3}\) = \(\frac{1}{4}\)

Answer: B

Answer is 1/4

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I struggled on this question massively. And I don't know why my approach didn't yield the same answer.

First draw two altitudes for each triangle to get a rectangle in the middle. The base of the two right triangles that flank the sides are necessarily 6 each.

Then, find the height.

Area(trapezoid) = (18 + 6) (3√3) / 2

Take the ratio: 9√3 / (18 + 6) (3√3) / 2 <--- 9√3) is the area of the equilateral triangle.

What am I doing wrong???
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hi, I think you may have made a calculation error.
It is right.