# Thread: Urgent Math Help, due tomorrow!!!

1. ## Urgent Math Help, due tomorrow!!!

Hi everyone,

I'm new here and I'm soooo happy I found this forum where people can help me!
I'm in 12th grade and I'm a pre-calc student.

There are two problems that are due tomorrow before 8:30 AM.
I have wasted maybe 5 pieces of paper trying to figure out the first problem.
Here it goes, and I pray someone is here to guide me!!

The lengths of the sides of a triangle are 13, 13, 10. The sides of a second triangle are 13, 13 and x, where x is not equal to 10. If the two triangles have equal areas, what is the value of x?

I started off by finding out the area of triangle one, by the pythagorean theorem. The area is 60.
Can anyone guide me on how to find the value of x?

The second question is:

Find the missing coordinate for the given info
P (x, -3) ; Q (9,6) and the slope between P and Q is 3.
I did it with the slope formula y1-y0 divided by x1-x0 equals slope which has to be 3. It gave me x=6.

So, please, can you help my poor mathematics soul? haha thanks in advance!!

2. Originally Posted by heiress28
Hi everyone,

I'm new here and I'm soooo happy I found this forum where people can help me!
I'm in 12th grade and I'm a pre-calc student.

There are two problems that are due tomorrow before 8:30 AM.
I have wasted maybe 5 pieces of paper trying to figure out the first problem.
Here it goes, and I pray someone is here to guide me!!

The lengths of the sides of a triangle are 13, 13, 10. The sides of a second triangle are 13, 13 and x, where x is not equal to 10. If the two triangles have equal areas, what is the value of x?

I started off by finding out the area of triangle one, by the pythagorean theorem. The area is 60.
Can anyone guide me on how to find the value of x?

The second question is:

Find the missing coordinate for the given info
P (x, -3) ; Q (9,6) and the slope between P and Q is 3.
I did it with the slope formula y1-y0 divided by x1-x0 equals slope which has to be 3. It gave me x=6.

So, please, can you help my poor mathematics soul? haha thanks in advance!!
For the first question:

Area = sqrt(s(s-a)(s-b)(s-c)), (Heron's formula).

s = (a + b + c)/2

(13 + 13 + 10)/2 = 18

sqrt(18(18-13)(18-13)(18-10))

A = 60.

s = (13 + 13 + x)/2 = (x + 26)/2

sqrt((x + 26)/2*((x + 26)/2 - 13)*((x + 26)/2 - 13)*((x + 26)/2 - x)

A = [sqrt((676 - x^2) - x^2) * |x|]/4

60 = [sqrt((676 - x^2) - x^2) * |x|]/4

Solve for x:

You'll get the following values:

x = -10, -24, 10, 24...

Obviously area can't be negative. You're left with 10, 24.

x = 24.

3. Originally Posted by heiress28

Find the missing coordinate for the given info
P (x, -3) ; Q (9,6) and the slope between P and Q is 3.
I did it with the slope formula y1-y0 divided by x1-x0 equals slope which has to be 3. It gave me x=6.
The slope formula is,
$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$
Let $\displaystyle P(x,-3)$ be $\displaystyle (x_1,y_1)$
Let $\displaystyle Q(9,6)$ be $\displaystyle (x_2,y_2)$
Thus,
$\displaystyle 3=\frac{6-(-3)}{9-x}$
Thus,
$\displaystyle \frac{3}{1}=\frac{9}{9-x}$
Do cross multiply,
$\displaystyle 3(9-x)=9$
Divide by 3,
$\displaystyle 9-x=3$
Thus,
$\displaystyle -x=-6$
Thus,
$\displaystyle x=6$

4. ## This is AfterShock's Post, I'm just makeing it easier to read

For the first question:

$\displaystyle \text{Area} = \sqrt{s(s-a)(s-b)(s-c)}$ (Heron's formula).

$\displaystyle s = \frac{a + b + c}{2}$

$\displaystyle \frac{13 + 13 + 10}{2} = 18$

$\displaystyle A = \sqrt{18(18-13)(18-13)(18-10)}$

$\displaystyle A = 60$

$\displaystyle s = \frac{13 + 13 + x}{2} = \frac{x + 26}{2}$

$\displaystyle A=\sqrt{\frac{x + 26}{2}\cdot\left(\frac{x + 26}{2} - 13\right)\cdot\left(\frac{x + 26}{2} - 13\right)\cdot\left(\frac{x + 26}{2} - x\right)}$

$\displaystyle A = \frac{\sqrt{\left(676 - x^2\right) - x^2} \cdot |x|}{4}$

$\displaystyle 60 = \frac{\sqrt{\left(676 - x^2 \right) - x^2} \cdot |x|}{4}$

Solve for x:

You'll get the following values:

x = -10, -24, 10, 24...

Obviously area can't be negative. You're left with 10, 24.

x = 24.

5. Hello, heiress28!

Here's an "eyeball" solution to #1 . . .

The lengths of the sides of a triangle are $\displaystyle 13,\,13,\,10.$
The sides of a second triangle are $\displaystyle 13,\,13,\,x$, where $\displaystyle x \neq 10$
If the two triangles have equal areas, what is the value of $\displaystyle x$?

The triangle looks like this:
Code:
            *
*|*
* | *
*  |  *
13 *   |   * 13
*    |12  *
*     |     *
.*      |      *
*-------+-------*
5       5

Using Pythagorus, we find that the altitude is 12.

Cut the triangle along its altitude
. . and reassemble the two pieces like this:
Code:
                      *
*   |   *    13
*       |5      *
*           |           *
*---------------+---------------*
12             12