1. ## Algebra problem

A friend recently emailed me this one (apparently it's from a game he plays)

AB=81, AD=180, AE=108, BE=135,CD=84, CE=156, BC=square root of ?

I've been poring over this for a bit, but I'm obviously out of practice.

Anyone care to have a bash?

2. Originally Posted by Davey Jones
A friend recently emailed me this one (apparently it's from a game he plays)

AB=81, AD=180, AE=108, BE=135,CD=84, CE=156, BC=square root of ?

I've been poring over this for a bit, but I'm obviously out of practice.

Anyone care to have a bash?

AB= A*B
AB=A+B

3. $\displaystyle \frac{AE}{BE} = \frac{108}{135} = \frac{A}{B}$

$\displaystyle \frac{A}{B} \times AB = \frac{108}{135} \times 81 = \frac{324}{25} = A^2$

So $\displaystyle A = \frac{18}{\sqrt{5}}$

$\displaystyle AD = 180$, so $\displaystyle D = \frac{180}{A} = 10\sqrt{5}$

$\displaystyle CD = 84$, so $\displaystyle C = \frac{84}{D} = \frac{84}{10\sqrt{5}} = \frac{42}{5\sqrt{5}}$

$\displaystyle AB = 81$, so $\displaystyle B = \frac{81}{A} = \frac{81\sqrt{5}}{18} = \frac{9\sqrt{5}}{2}$

$\displaystyle \therefore BC = \frac{9\sqrt{5}}{2} \times \frac{42}{5\sqrt{5}} = \frac{189}{5}$

4. Originally Posted by nzmathman
$\displaystyle \frac{AE}{BE} = \frac{108}{135} = \frac{A}{B}$

$\displaystyle \frac{A}{B} \times AB = \frac{108}{135} \times 81 = \frac{324}{25} = A^2$

So $\displaystyle A = \frac{18}{\sqrt{5}}$

$\displaystyle AD = 180$, so $\displaystyle D = \frac{180}{A} = 10\sqrt{5}$

$\displaystyle CD = 84$, so $\displaystyle C = \frac{84}{D} = \frac{84}{10\sqrt{5}} = \frac{42}{5\sqrt{5}}$

$\displaystyle AB = 81$, so $\displaystyle B = \frac{81}{A} = \frac{81\sqrt{5}}{18} = \frac{9\sqrt{5}}{2}$

$\displaystyle \therefore BC = \frac{9\sqrt{5}}{2} \times \frac{42}{5\sqrt{5}} = \frac{189}{5}$

why did u not squared $\displaystyle \frac{189}{5}$ because it is said that BC=is the squared of ?

i did my own calculations i think i'm lost.

ABCD=ABCD

BC= x

$\displaystyle 81*84=180*\sqrt{x}$

$\displaystyle 6804=180\sqrt{x}$

$\displaystyle \frac{6804}{180}=\sqrt{x}$

$\displaystyle (\frac{189}{5})^{2}=(\sqrt{x})^{2}$

$\displaystyle x=\frac{35721}{25}$

5. Hello, Davey Jones!

Is there a typo?
The equations are inconsistent . . .

A friend recently emailed me this one (apparently it's from a game he plays)

. . $\displaystyle \begin{array}{cccc}(1)&AB&=&81 \\ (2)&AD&=&180 \\ (3)& AE&=&108 \\ (4) & BE&=&135 \\ (5)&CD&=&84 \\ (6)& CE&=&156\\ \\ & BC&=& \sqrt{??} \end{array}$

. . $\displaystyle \begin{array}{cccc}\text{Divide (5) by (2):} & \dfrac{CD}{AD} \:=\:\dfrac{84}{180} &\Rightarrow& \dfrac{C}{A} \:=\:\dfrac{7}{15} \\ \\[-2mm] \text{Divide (6) by (3):} & \dfrac{CE}{AE} \:=\:\dfrac{156}{108} &\Rightarrow& \dfrac{C}{A}\:=\:\dfrac{13}{9} \end{array}$ . ??

6. Originally Posted by Davey Jones
A friend recently emailed me this one (apparently it's from a game he plays)

AB=81, AD=180, AE=108, BE=135,CD=84, CE=156, BC=square root of ?

I've been poring over this for a bit, but I'm obviously out of practice.

Anyone care to have a bash?
The way that I interpret this question, A, B, C, D, E are meant to be points in a plane, and AB, ..., are the distances between these points.

If that is the case, then ABE is a right-angled triangle, because $\displaystyle 81^2+108^2=135^2$. However, the remaining information is not sufficient to fix the positions of C and D. You need an extra piece of information, such as the distance DE, before you can pin down the distance BC.

7. There was a typo sorry!

CD = 60

8. Even CD=60, the equations are stil not consistent