# [SOLVED] trapezoid problem area

• Jun 4th 2008, 05:32 PM
roxmysox
[SOLVED] trapezoid problem area
this is a tough problem and i just can't seem to solve it. please help, thank you

there is a trapezoid EFZD. top left is E, top right is F, bottom right is Z, and bottom left is D. it is an isosceles trapezoid. the intersection of the diagonals EZ and FD is at point Q. the trapezoid is the lateral face of a frustum and the frustum has base areas of 20 and 32. triangle EQF has an area of 12. find the area of the trapezoid EFZD. we don't know the shape of the bases of the frustum. the bases can be square, triangular, hexagonal, or anything else.
• Jun 5th 2008, 04:16 AM
earboth
Quote:

Originally Posted by roxmysox
this is a tough problem and i just can't seem to solve it. please help, thank you

there is a trapezoid EFZD. top left is E, top right is F, bottom right is Z, and bottom left is D. it is an isosceles trapezoid. the intersection of the diagonals EZ and FD is at point Q. the trapezoid is the lateral face of a frustum and the frustum has base areas of 20 and 32. triangle EQF has an area of 12. find the area of the trapezoid EFZD. we don't know the shape of the bases of the frustum. the bases can be square, triangular, hexagonal, or anything else.

I've sketched the trapezium.

1. The base area B and the top area T are similar. Therefore the proportion of these areas is:

$\frac BT = \frac{20}{32} = \left( \frac{EF}{DZ}\right)^2$

2. The triangles EFQ and DZQ are similar and therefore their areas have the same proportion as the areas B and T. This proportion is valid for every pair of corresponding length in the two triangles. Therefore

$\left( \frac{x}{h-x}\right)^2 = \frac{20}{32}$

3. The equation to calculate the area of triangle EFQ is:

$\frac12 \cdot EF \cdot x = 12$

4. The area of the trapezium is:

$A_T=\frac12 \cdot (EF+DZ) \cdot h$

from #2 you know: $h = \sqrt{\frac85} \cdot x + x = x \left(\sqrt{\frac85} + 1\right)$

from #1 you know: $DZ = \sqrt{\frac85} \cdot EF$

5. Plugin these terms into the equation of #4:

$A_T=\frac12 \cdot \left(EF+\sqrt{\frac85} \cdot EF \right) \cdot x\left(\sqrt{\frac85} + 1\right)$

$A_T=\frac12 \cdot EF \left(1+\sqrt{\frac85} \right) \cdot x\left(\sqrt{\frac85} + 1\right)$

$A_T=\underbrace{\frac12 \cdot EF \cdot x}_{= 12} \left(\sqrt{\frac85} + 1\right)^2$ ....... according to #3

$A_T=12 \left(\sqrt{\frac85} + 1\right)^2~\approx~ 61.55$
• Jun 5th 2008, 07:44 AM
roxmysox
thanks so much.
but one thing, i don't understand how you get $
DZ = \sqrt{\frac85} \cdot EF
$
and $
h = \sqrt{\frac85} \cdot x + x = x \left(\sqrt{\frac85} + 1\right)
$
• Jun 5th 2008, 12:01 PM
earboth
Quote:

Originally Posted by roxmysox
thanks so much.
but one thing, i don't understand how you get $
DZ = \sqrt{\frac85} \cdot EF
$
and $
h = \sqrt{\frac85} \cdot x + x = x \left(\sqrt{\frac85} + 1\right)
$

to #1:
$\frac BT = \frac{20}{32} = \left( \frac{EF}{DZ}\right)^2$

$\left( \frac{EF}{DZ}\right)^2 = \frac{20}{32} = \frac58$

$\left( \frac{DZ}{EF}\right)^2 = \frac85$

$\frac{DZ}{EF} = \sqrt{\frac85}~\iff~ DZ = \sqrt{\frac85} \cdot EF$

to #2:
$\left( \frac{x}{h-x}\right)^2 = \frac{20}{32}=\frac58$

$\left( \frac{h-x}{x}\right)^2 =\frac85$

$\frac{h-x}{x} =\sqrt{\frac85}$

$h-x= \sqrt{\frac85} \cdot x$

$h= \sqrt{\frac85} \cdot x + x = x\left(\sqrt{\frac85} + 1 \right)$