# Math Help - are of a trapezoid below an equilateral triangle

1. ## are of a trapezoid below an equilateral triangle

question: an equilateral triangle abc sits atop an isosceles trapezoid abde which has a height of 6 cm. if the area of the trapezoid abde is twice the area of the triangle abc, what is the area of the trapezoid?

this for the most part seems simple to me. the area of the trapezoid is twice the area of the equilateral triangle, thus the area of the trapezoid= ((3^2)/2)(L^2) with L being a length of one of the sides of the equilateral triangle. Now for the trapezoid, we can also construct another equation for its area. we know that in the area lies a box 6cm x L. the rest is the two triangles on the side. Where I am stuck is what the area of these two triangles are??? if we set ((3^2)/2)(L^2) = 6cm x L + (area of two triangles) then couldn't we find the area of the trapezoid? I just don't know how to find the base of these triangles.

2. Originally Posted by dudeosu
question: an equilateral triangle abc sits atop an isosceles trapezoid abde which has a height of 6 cm. if the area of the trapezoid abde is twice the area of the triangle abc, what is the area of the trapezoid?

this for the most part seems simple to me. the area of the trapezoid is twice the area of the equilateral triangle, thus the area of the trapezoid= ((3^2)/2)(L^2) with L being a length of one of the sides of the equilateral triangle. Now for the trapezoid, we can also construct another equation for its area. we know that in the area lies a box 6cm x L. the rest is the two triangles on the side. Where I am stuck is what the area of these two triangles are??? if we set ((3^2)/2)(L^2) = 6cm x L + (area of two triangles) then couldn't we find the area of the trapezoid? I just don't know how to find the base of these triangles.
1. Draw a sketch!

2. The area of an equilateral triangle with side length L is calculated by:

$a_{\Delta} = \frac14\cdot \sqrt{3} \cdot L^2$

3. The area of a trapezium is calculated by:

$a_{trap}=\dfrac{L+B}2 \cdot h$

In your case since h = 6:

$a_{trap}=3(L+B)$

4. $a_{trap}=2\cdot a_{\Delta}$

Are you sure that you posted the complete question?

3. yes that is all

4. As earboth is saying, we need more information. There is no set relationship between L and B so we can't finish the problem. Is it possible that the equilateral triangle plus the trapezoid together form a triangle as well?

-Dan

5. I'm 99.9% sure this'll turn out to be 2 equilateral triangles; only way solving is possible.

Larger triangle has sides = 6(sqrt(3) + 1), smaller = 6(sqrt(3) + 1) / sqrt(3).
Makes areas = ~116.3538 and ~38.7846, a 3:1 ratio.

6. yes, the small triangle and the trap form a larger triangle. obviously an isosceles one at least. we can construct several equations, with the trapezoid area being equal to both twice the area of the small triangle and (b1+b2)(h)/2. you can also find that the area of the large triangle is the sum of the trap and the small triangle, and is also (b2 + h)/2. you can isolate b2 as a variable but from here i am stuck.

7. Originally Posted by dudeosu
yes, the small triangle and the trap form a larger triangle. obviously an isosceles one at least. we can construct several equations, with the trapezoid area being equal to both twice the area of the small triangle and (b1+b2)(h)/2. you can also find that the area of the large triangle is the sum of the trap and the small triangle, and is also (b2 + h)/2. you can isolate b2 as a variable but from here i am stuck.
1. You have to determine the values of 3 distances:
L the base of the equilateral triangle
h the height of the equilateral triangle
B the base of the isosceles trapezium

2. Compare the areas of the large and the small equilateral triangles:

$\frac12 \cdot B (h+6)=3 \cdot \frac12 \cdot L \cdot h$ <--- [1]

3. Compare the areas of small equilateral triangle and the trapezium:

$\dfrac{B+L}2 \cdot 6 = \frac12 \cdot \sqrt{3} \cdot L^2$ <--- [2]

4. Since the bases of the samll triangle and the trapezium are parallel you can set up the following ratio:

$\dfrac h{h+6}=\dfrac LB$ <--- [3]

5. Solve the system of these 3 equations.

8. Agree; will result in L = 6[sqrt(3) + 1] / sqrt(3) and B = 6[sqrt(3) + 1]

9. how did you get that the area of the large triangle is three times the area of the small triangle? otherwise, i understand all else.

10. By calculating the areas...

11. Originally Posted by dudeosu
question: an equilateral triangle abc sits atop an isosceles trapezoid abde which has a height of 6 cm. if the area of the trapezoid abde is twice the area of the triangle abc, what is the area of the trapezoid?
Originally Posted by earboth

Compare the areas of small equilateral triangle and the trapezium:

$\dfrac{B+L}2 \cdot 6 = \frac12 \cdot \sqrt{3} \cdot L^2$ <--- [2]
Shouldn't the equation be like this

$\dfrac{B+L}2 \cdot 6 = 2(\frac12 \cdot \sqrt{3} \cdot L^2)$

since the area of the trapezoid should be double the area of the small equilateral triangle, or am I missing something really obvious?

12. Originally Posted by dudeosu
... if the area of the trapezoid abde is twice the area of the triangle abc, what is the area of the trapezoid?

...
Originally Posted by dudeosu
how did you get that the area of the large triangle is three times the area of the small triangle? otherwise, i understand all else.
According to the text of the question (and some additional information we got during the discussion) the small triangle has 1 area unit, the trapezium has 2 area units. If the small triangle and trapezium form the large triangle this must contain 3 area units.

Originally Posted by scounged
Shouldn't the equation be like this

$\dfrac{B+L}2 \cdot 6 = 2(\frac12 \cdot \sqrt{3} \cdot L^2)$

since the area of the trapezoid should be double the area of the small equilateral triangle, or am I missing something really obvious?
No.

The area of an equilateral triangle with base length L is calculated by:

$a_{\Delta}=\frac14 \cdot \sqrt{3} \cdot L^2$ (see post #2)

So

$2 a_{\Delta}=2 \cdot \frac14 \cdot \sqrt{3} \cdot L^2 = \frac12 \cdot \sqrt{3} \cdot L^2$

13. $\mbox{Oh! I forgot that}~\sin{60}=\displaystyle{\frac{\sqrt{3}}{2}}~\ mbox{and not just}~\sqrt{3}$

14. how do you simplify these three equations? i have tried in every possible way i can think of and have come to a dead end.

15. Originally Posted by earboth
1 $\frac12 \cdot B (h+6)=3 \cdot \frac12 \cdot L \cdot h$ <--- [1]

$\dfrac{B+L}2 \cdot 6 = \frac12 \cdot \sqrt{3} \cdot L^2$ <--- [2]

$\dfrac h{h+6}=\dfrac LB$ <--- [3]
Start by simplifying a bit:
B(h + 6) = 3Lh [1]
6(B + L) = sqrt(3)L^2 [2]
L(h + 6) = Bh [3]

I suggest this as next step:
[1]: h + 6 = 3Lh / B
[3]: h + 6 = Bh / L
So:
3Lh / B = Bh / L
B^2 = 3L^2
Now use that and [2] : OK?