# Thread: Finding X usin an algebraic method

1. ## Finding X usin an algebraic method

Hi can somebody help me find X in this diagram using an algebraic method?
Please also explain how its done!
Firstly the shape is a sqaure, and a line has been attached to 3 corners of the square from a random dot inside the square, the smallest line is 3 units, the mid one is 4 units and the largest is 5 units large.

2. I have a method, but it might be messy.

The sum of those areas of those three peices equals to the area of the square. That gives you just an equation in terms of x. By "Heron's Formula"

3. Hello, hossein!

I have an approach . . . but is it acceptable?

Firstly the shape is a sqaure.
Lines have been drawn to 3 corners of the square from a point inside the square.
The smallest line is 3 units, the mid one is 4 units, and the largest is 5 units.

Place the square on a coordinate system.
Let the side of the square be $a$.
Code:
    (0,a)           (a,a)
A * - - - - - - - * B
|               |
|     * P       |
|   (x,y)       |
|               |
|               |
|               |
- C * - - - - - - - * D -
(0,0)           (a,0)
Using the Distance Formula, we have:

. . $\begin{array}{ccccc}PA = 3: & x^2 + (y-a)^2 & = & 9 & {\bf (1)} \\
PB = 4: & (x-a)^2 + \,(y-a)^2 & = & 16 & {\bf(2)} \\
PC = 5: & x^2 + \,y^2 & = & 25 & {\bf(3)}\end{array}$

Subtract (1) from (3): . $y^2 - (y-a)^2 \:=\:16$
. . which simplifies to: . $y \:=\:\frac{a^2+16}{2a}$ . (4)

Subtract (1) from (2): . $(x-a)^2-x^2 \:=\:7$
. . which simplifies to: . $x \:=\:\frac{a^2-7}{2a}$ .(5)

Substitute (4) and (5) into (3): . $\left(\frac{a^2-7}{2a}\right)^2 + \left(\frac{a^2+16}{2a}\right)^2\:=\:25$

. . which simplifies to: . $2a^4 - 82a^2 + 305 \:=\:0$ . . . a quadratic in $a^2$.

Quadratic Formula: . $a^2 \;=\;\frac{-(\text{-}82) \pm\sqrt{(\text{-}82)^2 - 4(2)(305)}}{2(2)} \;=\;\frac{82 \pm \sqrt{4284}}{4}\;=\;\frac{41 \pm\sqrt{1071}}{2}$

Then: . $a \;=\;\sqrt{\frac{41 \pm\sqrt{1071}}{2}}\;=\;\begin{Bmatrix}6.07149637 \\ 2.033944893 \end{Bmatrix}$

. . The second root is too small be the side of the square.

Therefore, the side of the square is about $6.07$ units.

Edit: corrected errors in last four lines.
.

4. hey Soroban!
I kindda undrestand wut u did, but I dont think its a right answer.because if u draw it out using a protractor the side turns out to be 5.8 .so in a way I know the answer is 5.8 but I still have to show where I got it from. and the HINT that we got was that we have to use the method that we learned in grade 10 math.

5. Originally Posted by hossein
hey Soroban!
I kindda undrestand wut u did, but I dont think its a right answer.because if u draw it out using a protractor the side turns out to be 5.8 .so in a way I know the answer is 5.8 but I still have to show where I got it from. and the HINT that we got was that we have to use the method that we learned in grade 10 math.
you mind being more specific about what method you used. depending on what country you are from--and what part of the country--there are many different things you could learn in this respect

6. A similar solution as the Soroban's, but without using coordinates.
Let $ABCD$ be the square and $P$ the point inside such as $AP=3,BP=4,DP=5$.
Draw a line through $P$ parallel with $AB$ which intersects $AD, \ BC$ in $M$ and $N$.
Draw a line through $P$ parallel with $AD$ which intersects $AB, \ DC$ in $P$ and $Q$.
Let $AB=x,MP=y,AM=z$. Then $PB=x-y,MD=x-z$.
Applying Pitagora in triangles where $AP,BP,DP$ are hypotenuses, we have the following system:
$\displaystyle \left\{\begin{array}{lll}y^2+z^2=9\\z^2+(x-y)^2=16\\y^2+(x-z)^2=25\end{array}\right.$
From here you can follow the solution of Soroban.

7. Originally Posted by Soroban
Hello, hossein!

I have an approach . . . but is it acceptable?

Place the square on a coordinate system.
Let the side of the square be $a$.
Code:
    (0,a)           (a,a)
A * - - - - - - - * B
|               |
|     * P       |
|   (x,y)       |
|               |
|               |
|               |
- C * - - - - - - - * D -
(0,0)           (a,0)
Using the Distance Formula, we have:

. . $\begin{array}{ccccc}PA = 3: & x^2 + (y-a)^2 & = & 9 & {\bf (1)} \\
PB = 4: & (x-a)^2 + \,(y-a)^2 & = & 16 & {\bf(2)} \\
PC = 5: & x^2 + \,y^2 & = & 25 & {\bf(3)}\end{array}$

Subtract (1) from (3): . $y^2 - (y-a)^2 \:=\:16$
. . which simplifies to: . $y \:=\:\frac{a^2+16}{2a}$ . (4)

Subtract (1) from (2): . $(x-a)^2-x^2 \:=\:7$
. . which simplifies to: . $x \:=\:\frac{a^2-7}{2a}$ .(5)

Substitute (4) and (5) into (3): . $\left(\frac{a^2-7}{2a}\right)^2 + \left(\frac{a^2+16}{2a}\right)^2\:=\:25$

. . which simplifies to: . $2a^4 - 82a^2 + 305 \:=\:0$ . . . a quadratic in $a^2$.

Quadratic Formula: . $a^2 \;=\;\frac{-(\text{-}82) \pm\sqrt{(\text{-}82)^2 - 4(2)(305)}}{2(2)} \;=\;\frac{84 \pm \sqrt{4284}}{4}\;=\;\frac{42 \pm\sqrt{1071}}{2}$

Then: . $a \;=\;\sqrt{\frac{42 \pm\sqrt{1071}}{2}}\;=\;\begin{Bmatrix}6.112533695 \\ 2.153353624 \end{Bmatrix}$

. . The second root is too small be the side of the square.

Therefore, the side of the square is about $6.11$ units.

i thought about doing this, but like you, i wasn't sure it would be acceptable for some reason

8. the canadian skool system, ONTARIO..does that help??

9. Okay, I'll try again . . .

Are you allowed to use the Pythagorean Theorem?

Let the side of the square be $a$.

From $P$ draw perpendicular $PE$ to $AB\!:\;\;h = PE$
. . Let $x \,= \,AE$, then $a-x \:=\:EB$
Code:
          x   E     a-x
A * - - - + - - - - - - * B
| *     :           * |
|  3*   :h      *4    |
|     * :   *         |
|       *             |
|       P             |

In right triangle $AEP$, we have: . $x^2 + h^2 \:=\:3^2$ . [1]

In right triangle $BEP$, we have: . $(a-x)^2 + h^2 \:=\:4^2$ .[2]

Subtract [1] from [2]: . $(a-x)^2 - x^2 \:=\:7\quad\Rightarrow\quad x \:=\:\frac{a^2-7}{2a}$ . [3]

From $P$ draw perpendicular $PF$ to $AD$; $x = PF$
. . Let $y \,= \,AF$, then: $a-y \:= \:FD$
Code:
    A * - - - - - - -
| *
y |   *3
|     *
F + - - - * P
|   x  *
|     *
|    *
a-y |   *5
|  *
| *
|*
D * - - - - - - -

In right triangle $PFA\!:\;x^2 + y^2 \:=\:3^2$ . [4]

In right triangle $PFD\!:\;x^2 + (a-y)^2 \:=\:5^2$ . [5]

Subtract [4] from [5]: . $(a-y)^2 - y^2 \:=\:16\quad\Rightarrow\quad y \:=\:\frac{a^2-16}{2a}$ . [6]

Substitute [3] and [6] into [4]: . $\left(\frac{a^2-7}{2a}\right)^2 + \left(\frac{a^2-16}{2a}\right)^2 \:=\:9$

. . which simplifies to: . $2a^4-82a^2 + 305\:=\:0$

. . . the same equation as before.

I think I just repeated red_dog's solution . . . *sigh*

Also, I see a typo and the end of my original post.
. . I'll correct it . . .

10. thanx guys I GOT THE right answer!! thanx so much