Algebra 3 HELP!!!?

Question

The illustration found in the link( the pic should be below but if it isn't the link will work) is the smallest possible Quadrilateral composed of squares. Can you find it's perimeter given only the two lenghths shown. LINK BELOW ON NEX T LINE CLICK TO ENLARGE PIC http://i128.photobucket.com/albums/p199/av611/11-03-2006060546PM.jpg




[IMG]http://i128.photobucket.com/albums/p199/av611/11-03-2006060546PM.jpg[/IMG]

Answer 1

I wish I had Adobe Acrobat so I could e-mail you the lengths of the squares because explaining this without the diagram will be confusing, but I will try.

The important thing is to remember that the problem stated that this entire figure is a QUADRILATERAL THAT IS NOT DRAWN TO SCALE. So you can't assume that this is a square.

Let's start with the side marked 9cm and 8cm (the bottom side). The small square in between the two is 1cm around (the difference between the 9 and 8 cm squares). Because of that 1 cm square, the larger square on top of the 9cm square is 10cm around (the side of the 9cm square + the side of the 1cm square). So on the left hand side of the quadrilateral, you now have a side of 9cm and 10cm.

We established the 1 cm square. The square on top of the 8 cm square is 7cm in length (8-1). So the last square on the bottom of the quadrilateral is 15cm (8cm+7cm). So you now have the total length of the bottom length of the quadrilateral (9+8+15)== 32cm. You also now have a partial length of the right hand side of the quadrilateral.

The square on top of the 7cm square is 4cm around. Subtract the 1cm side from the 7cm side (6cm), then subtract it from the 10cm side...(4cm). Now that you know that the square is 4cm around, then the remaining length on the 7cm side of the square beneath the 4cm square is 3cm. (7-4). We already established the 15cm square. Add the 3cm length to 15cm to get 18cm and the remaining side of the right part of the quadrilateral, 33cm (18 +15).

Now that you know 18cm square, subtract the 4cm square from it to leave you a length of 14cm. That leaves you with the remaining side of the left hand of the quadrilateral (10+9+14) ==33cm. It also leaves you with the remaining upper side of the quadrilateral, 14cm to give you (18+14)cm == 32cm

So you have 32+32+33+33==130cm, the perimeter of the quadrilateral, which is a rectangle.

This is just one way to solve this problem. There are others. For example, once you deduced the 4cm square, you could have added that side to the 10cm side to give you 14cm.

Answer 2

a million) every person be responsive to that's a rectangle, and that's perimeter is 24, so: s1 + s1 + s2 + s2 = 24 We additionally be responsive to that s2 = s1 + 3 so: s1 + s1 + s1+3 + s1+3 = 24 4s1 + 6 = 24 4s1 = 18 s1 = 18/4 = 9/2 = 4 a million/2 s2 = 7 a million/2 2) every person be responsive to the three factors are: x, x-a million, x+a million And every person be responsive to the perimter is 17 in. so: x + x-a million + x+a million = 17 3x = 17 x = 17/3 = 5 2/3 So the three factors are 5 2/3, 4 2/3, 6 2/3 3) Supplementary angles upload to one hundred eighty ranges, so: x = the unique perspective 3x + 20 = the supplement to the unique perspective the two angles greater = one hundred eighty, so: x + 3x+20 = one hundred eighty 4x + 20 = one hundred eighty 4x = sixteen- x = 40 it somewhat is supplement is one hundred 40 desirable of fulfillment! :)

Answer 3

The whole thing is square (so it looks to be), so 9cm and 8cm =17 which looks to be half of the bottom line. 17 x 2=34
perimeter = 4s s = 34
P = 4 x 34
P= 136 cm

Answer 4

one side is equal to 32 cm....
9 + 8 + 15 = 32 cm for one side
multiply it by 4 to get the perimeter is eaqual to 128 cm

Answer 5

it's possible, certainly. the thing itself isn't a square, the dimensions are about 32x33, i believe, but check me on it. i was just doing that in my head.

Answer 6

It's 130 trust me.

Answer 7

No.