Re: finding the area of abd

Quote:

Originally Posted by

**slapmaxwell1** Attachment 23064
the trouble i am finding with this particular problem is that i do not know how to solve without using plug/chug.

What do you mean by that? Do you already know how to solve it?

Re: finding the area of abd

Quote:

Originally Posted by

**scounged** What do you mean by that? Do you already know how to solve it?

i thought i did, but its not working out...

Re: finding the area of abd

Ok. In the problem it says that |AB| = 2|BC|. I'd solve this one by letting |BC| equal a, and calculate the area of the triangles ACE and ABD, in terms of a. Can you do that?

Re: finding the area of abd

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?

Re: finding the area of 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.

Re: finding the area of abd

Hello, slapmaxwell1!

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` A E`

o * * * o

* |\ *

* | \ 20 *

D o 2| \ *

* | \ *

* | 1 \ *

B o-----o C

* *

* *

o

F

$\displaystyle ABC$ is a right triangle with $\displaystyle BC\!:\!AB \,=\,1\!:\!2$

$\displaystyle AC\!E, AB\!D, BC\!F$ 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: .$\displaystyle \frac{\Delta AB\!D}{20} \:=\:\frac{4}{5} \quad\Rightarrow\quad \Delta AB\!D \:=\:16\;\hdots\;\text{answer (B)}$