ok im still having trouble with this problem...i figured out that one side is sqrt(5), but then thats it. so the sides of the shades triangle are 1*k, 2*k and sqrt(5)*k.
im using k because it is saying its a ratio. so what i am asking is this, how can i use this info to find the area of the triangle abd?
Ok; to start, you do not need triangle BCF; disregard it.
Let a = BC; then AB = 2a (ratio 1:2) ; still with me?
Let AC = b
Using triangle ACE:
since sides = b, area = b^2 SQRT(3) / 4 ; since that area is given as 20:
b^2 SQRT(3) / 4 = 20
b = SQRT[80 / SQRT(3)] ; that'll give you b = ~6.796
Using right triangle ABC:
b^2 = a^2 + (2a)^2
a = SQRT(b^2 / 5) ; that'll give you a = ~4.298
Using triangle ABD:
since sides = 2a, area = a^2 SQRT(3) ; that'll give you area = 16.00000000....
Hope you followed...come back with questions if not.
Hello, slapmaxwell1!
$\displaystyle ABC$ is a right triangle with $\displaystyle BC\!:\!AB \,=\,1\!:\!2$Code:A E o * * * o * |\ * * | \ 20 * D o 2| \ * * | \ * * | 1 \ * B o-----o C * * * * o F
$\displaystyle AC\!E, AB\!D, BC\!F$ are equilateral triangles.
The area of $\displaystyle \Delta AC\!E = 20.$
Find the area of $\displaystyle \Delta AB\1D.$
. . $\displaystyle (A)\;18\qquad (B)\;16 \qquad (C)\;15 \qquad (D)\;14 \qquad (E)\;12$
Pythagorus says: $\displaystyle AC = \sqrt{5}$
The ratio of the areas of two similar polygons
. . is the square of the ratios of their sides.
The ratio of the sides of $\displaystyle \Delta AB\!D$ and $\displaystyle \Delta AC\!E$ is.$\displaystyle 2\!:\!\sqrt{5}$
The ratio of their areas is.$\displaystyle 2^2\!:\!(\sqrt{5})^2 \:=\:4\!:\!5$
We have: .$\displaystyle \frac{\Delta AB\!D}{20} \:=\:\frac{4}{5} \quad\Rightarrow\quad \Delta AB\!D \:=\:16\;\hdots\;\text{answer (B)}$