The bases of trapezoid ABCD are \overline{AB} and \overline{CD}. Let P be the intersection of diagonals \overline{AC} and \overline{BD}. If the areas of triangles ABP and CDP are 8 and 18, respectively, then find the area of trapezoid ABCD.

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To find the area of trapezoid ABCD, we need to use the given information about the areas of triangles ABP and CDP.

Let's denote the height of trapezoid ABCD as 'h', and let 'x' represent the distance from point P to the base \overline{AB}.

Since triangles ABP and CDP share the same height 'h', we know that the ratio of their areas is equal to the ratio of their bases.

The area of triangle ABP is given as 8, so the ratio of its area to the base \overline{AB} is 8:1, or simply 8:1. This means that the length of \overline{BP} is 1/8 times the length of \overline{AB}.

Similarly, the area of triangle CDP is given as 18, so the ratio of its area to the base \overline{CD} is 18:1, or simply 18:1. This means that the length of \overline{DP} is 1/18 times the length of \overline{CD}.

Now, notice that \overline{AP} and \overline{PC} are diagonals of the trapezoid, and they intersect at point P. Since the diagonals of a trapezoid divide each other proportionally, we can conclude that the ratio of \overline{AP} to \overline{PC} is the same as the ratio of \overline{AB} to \overline{CD}.

Thus, we have the following ratio:

\overline{AP} : \overline{PC} = \overline{AB} : \overline{CD} = 8:18 = 4:9 (dividing both sides by the greatest common factor, 2)

Let's denote the length of \overline{AB} as 'a' and the length of \overline{CD} as 'b'. Using the ratio 4:9, we can write the following equation:

a : b = 4 : 9

Since the bases of a trapezoid are parallel, we can conclude that the ratio of the lengths of the bases is the same as the ratio of the lengths of the parallel sides. Therefore, we can say:

\overline{AB} : \overline{CD} = a : b = 4 : 9

The area of a trapezoid can be calculated using the formula:

Area = (1/2) * (sum of the bases) * (height)

In this case, the sum of the bases is equal to \overline{AB} + \overline{CD} = a + b.

We know that the lengths of the bases are in a 4:9 ratio, so we can express \overline{AB} and \overline{CD} as:

\overline{AB} = (4 / (4 + 9)) * (a + b) = (4 / 13) * (a + b)

\overline{CD} = (9 / (4 + 9)) * (a + b) = (9 / 13) * (a + b)

Substituting these values into the area formula, we get:

Area = (1/2) * (4 / 13) * (a + b) * h + (1/2) * (9 / 13) * (a + b) * h

Simplifying, we have:

Area = (13 / 26) * (a + b) * h

We still need to find the value of (a + b) * h to calculate the area of trapezoid ABCD.

We can look at triangles ABP and CDP to find the value of (a + b) * h.

In triangle ABP, the base is \overline{AB} = a, and the height is 'h'. So the area is given as 8:

Area_ABP = (1/2) * a * h = 8

Multiplying both sides of this equation by 2/h, we get:

a = 16/h

In triangle CDP, the base is \overline{CD} = b, and the height is 'h'. So the area is given as 18:

Area_CDP = (1/2) * b * h = 18

Multiplying both sides of this equation by 2/h, we get:

b = 36/h

Now we can substitute these values of 'a' and 'b' into the expression (a + b) * h:

(a + b) * h = (16/h + 36/h) * h = (52/h) * h = 52

Finally, substituting this value back into the trapezoid area formula, we get:

Area = (13/26) * (52) = 26

Therefore, the area of trapezoid ABCD is 26 square units.

Let x=AB and y=CD

Triangle PAB is similar to PCD so the area of PAB over the area of PCD is x^2/y^2

Plugging in 8 and 18 gives us x/y=2/3

So AP/PC=AB/CD=2/3. Triangle ABP and BCP have the same height to AC so...

The area of ADP/ABP=3/2

So the area of ADP=3/2*8=12

That means the area of the trapezoid is 8+18+12+12=50

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