# Finding perimeter of a square.

• Feb 17th 2009, 03:41 PM
NYCKid09
Finding perimeter of a square.
A square on a coordinate grid has horizontal and vertical sides. The upper left vertex of the square is located at (-3,4). The perimeter of the square is 40 units.

What are the coordinates of the other three vertices?

b.)What percent of the area is in each quadrant?

c.) Suppose you translate the square one unit to the right. Does this translation change the percent of the now existing in the third quadrant?

I got this for A.

Upper Right Vertex (7,4)
Lower Right Vertex (7,-6)
Lower Left Vertex (-3,-6)
• Feb 17th 2009, 03:46 PM
masters
Quote:

Originally Posted by NYCKid09
A square on a coordinate grid has horizontal and vertical sides. The upper left vertex of the square is located at (-3,4). The perimeter of the square is 40 units.

What are the coordinates of the other three vertices?

b.)What percent of the area is in each quadrant?

c.) Suppose you translate the square one unit to the right. Does this translation change the percent of the now existing in the third quadrant?

No idea. Never had to solve perimeter when given coordinates.

(Worried)

Hi NYCkid09,

If the perimeter of the square is 40, each side is 10. The area is $\displaystyle 10^2=100$ square units.

After sketching a graph, it is easy to locate vertices at (-3, -6), (7, -6), and (7, 4).

The area in QI is 7 X 4 = 28 which is 28%
The area in QII is 3 X 4 = 12 which is 12%
The area in QIII is 3 X 6 = 18 which is 18%
The area in QIV is 6 X 7 = 42 which is 42%

Translating 1 unit to the right changes the QIII area to 2 X 6 = 12 which changes to 12%
• Feb 17th 2009, 04:15 PM
Soroban
Hello, NYCKid09!

Quote:

A square on a coordinate grid has horizontal and vertical sides.
The upper left vertex of the square is located at (-3,4).
The perimeter of the square is 40 units.

(a) What are the coordinates of the other three vertices?

(b) What percent of the area is in each quadrant?

(c) Suppose you translate the square one unit to the right.
Does this translation change the percent of area in the third quadrant? yes

I got this for (a):. . $\displaystyle \begin{array}{cc} \text{upper left: }(\text{-}3,4) & \text{upper right: }(7,4) \\ \text{lower left: }(\text{-}3,\text{-}6) & \text{lower right: }(7,\text{-}6) \end{array}$ . . . . Good!

(b) The total area is: .$\displaystyle 10 \times 10 \,=\,100\text{ units}^2$

In quadrant 1, the area is: .$\displaystyle 7\times 4 \,=\,28 \quad\Rightarrow\quad 28\%$

In quadrant 2, the area is: .$\displaystyle 3\times 4 \,=\,12\quad\Rightarrow\quad 12\%$

In quadrant 3, the area is: .$\displaystyle 3 \times 6 \,=\,18 \quad\Rightarrow\quad 18\%$

In quadrant 4, the area is: .$\displaystyle 7\times 6 \,=\,42 \quad\Rightarrow\quad 42\%$

(c) If the square is translated one unit to the right,

. . the vertices would be: .$\displaystyle \begin{array}{ccc}(\text{-}2,4)&& (8,4) \\ (\text{-}2,\text{-}6) & & (8,\text{-}6) \end{array}$

The area is quandrant 3 is now: .$\displaystyle 2\times 6 \,=\,12 \quad\Rightarrow\quad 12\%$

• Feb 17th 2009, 04:27 PM
NYCKid09
Quote:

Originally Posted by masters
Hi NYCkid09,

If the perimeter of the square is 40, each side is 10. The area is $\displaystyle 10^2=100$ square units.

After sketching a graph, it is easy to locate vertices at (-3, -6), (7, -6), and (7, 4).

The area in QI is 7 X 4 = 28 which is 28%

The area in QII is 3 X 4 = 12 which is 12%

The area in QIII is 3 X 6 = 18 which is 18%

The area in QIV is 6 X 7 = 42 which is 42%

Translating 1 unit to the right changes the QIII area to 2 X 6 = 12 which changes to 12%

You interpreted it as they are only moving Quadrant 3 to the right one unit, and not the whole square?

I dont really understand the wording of that question.
• Feb 18th 2009, 04:30 PM
HallsofIvy
A square has 4 side, all of the same length. If the length is x, then the perimeter is 4x. If the perimeter is 40, then the length of each side is 40/4= 10.

If the upper left vertex of the square is (-3, 4) then the other three vertices are (-3+10, 4)= (7, 4), (-3, 4-10)= (-3, -6), and (7, 4-10)= (-3+10, -6)= (-7, -6).