how do u do these word problems?

**polakio92** · March 29th 2008, 08:35 AM

a) a parabola that has its vertex on the line y=3x+1 and whose equation is of the form y=3(x-h)^2 + k passes through the point (1,10) . what are the equations of all such parabolas?

b) square ABCD is inscribed in square EFGH forming four congruent right triangles ( the inner square is kinda of crooked inside the larger square).
if the perimeter of the square EFGH is 64 and the area of square ABCD is 130, determine the values of x and y.

c) a rectangle is cut to form a square. the perimeter of the rectangle is 54 cm, and the area of the cut-off piece is 36 cm sq. what are the dimensions of the original rectangle?

**Moo** · March 29th 2008, 10:17 AM

Hello,

You have to found to values : h and k.

First of all, you know that the summit is the point where the derivate of the parabola is 0.

So calculate dy/dx and get the value for which it's 0. It'll give you the value of h.

Then the parabola goes through the point (1,10). So replace x and y in y=3(x-h)^2 + k and you will get a relation between h and k.

**polakio92** · March 29th 2008, 11:18 AM

Originally Posted by Moo

Hello,

You have to found to values : h and k.

First of all, you know that the summit is the point where the derivate of the parabola is 0.

So calculate dy/dx and get the value for which it's 0. It'll give you the value of h.

Then the parabola goes through the point (1,10). So replace x and y in y=3(x-h)^2 + k and you will get a relation between h and k.

well how do i find h and k

**Soroban** · March 29th 2008, 12:34 PM

Hello, polakio92!

b) Square $ABCD$ is inscribed in square $EFGH$ forming four congruent right triangles.
If the perimeter of the square $EFGH$ is 64 and the area of square $ABCD$ is 130,
determine the values of $x\text{ and }y.$

Code:

    E   x   A       y       F
    * - - - * - - - - - - - *
    |      *    *           |
    |     *         *       | x
    |    *              *   |
  y |   *                   * B
    |  *                   *|
    | *                   * |
    |*                   *  |
  D *                   *   | y
    |   *              *    |
  x |       *         *     |
    |           *    *      |
    * - - - - - - - * - - - *
    H     16-x      C   x   G
    : - - - -  16 - - - - - :

The sides of EFGH are divided into the segments $x\text{ and }y.$

The perimeter of $EFGH$ is 64.
. . Hence: . $x + y \:=\:16\quad\Rightarrow\quad y \:=\:16-x\;\;{\color{blue}[1]}$

The area of $ABCD$ is 130.
. . Hence: . $AB = BC = CD = AD = \sqrt{130}$
In right triangle $AFB\!:\;\;x^2 + y^2 \:=\:AB^2\quad\Rightarrow\quad x^2 + y^2 \:=\:130\;\;{\color{blue}[2]}$

Substitute [1] into [2]: . $x^2 + (16-x)^2 \:=\:130 \quad\Rightarrow\quad x^2 - 16x + 63 \:=\:0$

. . which factors: . $(x-7)(x-9) \:=\:0$

. . and has roots: . $x \;=\;7,\,9$

Therefore: . $\{x,\,y\} \;=\;\{7,\,9\}$

**polakio92** · March 29th 2008, 01:08 PM

thank you so much! hopefully i can solve the rest of the word problems i have..

**Soroban** · March 29th 2008, 02:29 PM

Hello again, polakio92!

c) A rectangle is cut to form a square.
The perimeter of the rectangle is 54 cm, and the area of the cut-off piece is 36 cm².
What are the dimensions of the original rectangle?

Code:

            x         y
      * - - - - - * - - - *
      |           |       |
      |           |       |
    x |           |       | x
      |           |       |
      |           |       |
      * - - - - - * - - - *
            x         y

The perimeter of the rectangle is 54: . $4x + 2y \:=\:54\quad\Rightarrow\quad y \:=\:27 - 2x\;\;{\color{blue}[1]}$

The area of the cut-off piece is 36 cm²: . $xy \:=\:36\;\;{\color{blue}[2]}$

Substitute [1] into [2]: . $x(27-2x) \:=\:36\quad\Rightarrow\quad 2x^2 - 27x + 36 \:=\:0$

. . which factors: . $(x-12)(2x-3)\:=\:0$

. . and has roots: . $x \;=\;12,\:\frac{3}{2}$

and has y-values: . $y \;=\;3,\:24$

There are two solutions: . $(x,y) \;=\;(12,\,3)\text{ and }(1.5,\:24)$

The original rectangle was: . $\boxed{12 \times 15}\;\text{ or }\;\boxed{1.5 \times 25.5}$

Code:

           12         3
      * - - - - - * - - - *
      |           |       |
      |           |       |
   12 |           |       | 12
      |           |       |
      |           |       |
      * - - - - - * - - - *
           12         3


        1½             24
      * - * - - - - - - - - - - - - *
   1½ |   |                         | 1½
      * - * - - - - - - - - - - - - *
        1½             24

Both of them satisfy the problem!

**polakio92** · March 29th 2008, 02:54 PM

Originally Posted by Soroban

Hello again, polakio92!

Code:

            x         y
      * - - - - - * - - - *
      |           |       |
      |           |       |
    x |           |       | x
      |           |       |
      |           |       |
      * - - - - - * - - - *
            x         y

The perimeter of the rectangle is 54: . $4x + 2y \:=\:54\quad\Rightarrow\quad y \:=\:27 - 2x\;\;{\color{blue}[1]}$

The area of the cut-off piece is 36 cm²: . $xy \:=\:36\;\;{\color{blue}[2]}$

Substitute [1] into [2]: . $x(27-2x) \:=\:36\quad\Rightarrow\quad 2x^2 - 27x + 36 \:=\:0$

. . which factors: . $(x-12)(2x-3)\:=\:0$

. . and has roots: . $x \;=\;12,\:\frac{3}{2}$

and has y-values: . $y \;=\;3,\:24$

There are two solutions: . $(x,y) \;=\;(12,\,3)\text{ and }(1.5,\:24)$

The original rectangle was: . $\boxed{12 \times 15}\;\text{ or }\;\boxed{1.5 \times 25.5}$

Code:

           12         3
      * - - - - - * - - - *
      |           |       |
      |           |       |
   12 |           |       | 12
      |           |       |
      |           |       |
      * - - - - - * - - - *
           12         3


        1½             24
      * - * - - - - - - - - - - - - *
   1½ |   |                         | 1½
      * - * - - - - - - - - - - - - *
        1½             24

Both of them satisfy the problem!

you've saved me once again. thanks!

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