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.
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.
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:
3. Compare the areas of small equilateral triangle and the trapezium:
4. Since the bases of the samll triangle and the trapezium are parallel you can set up the following ratio:
5. Solve the system of these 3 equations.
The area of an equilateral triangle with base length L is calculated by:
(see post #2)