Math Question?

Question

Ok. The foundation of a house is in the shape of 2 adjacent squares (note:these squares are stuck together at the side, but are different sizes). The perimeter of the foundation is 80m and the area is 325m^2. Find the length of a side of each square.

Answer 1

You have to set up 2 equations to answer this problem. First, you have the area of both squares adding up to 325. Then you have the total perimeter equalling 80.

if x = length of side of smaller square and
if y = length of side of larger square, then ...

x² + y² = 325. (the area part)

Now, with the two squares butted up next to each other, the 'y' square would be taller. So you have 3 sides of the y exposed fully, 3 sides of the x exposed fully, and then the little piece on the 4th side of the y square that is exposed (because the x square next to it isn't as tall). Therefore you have ...

3y + 3x + (y-x) = 80

The y-x here is the amount left on the y side that is partially covered by the x. The equation above can be reduced to ...

4y + 2x = 80 ---- and reduced again to
2y + x = 40 ----- (div. all parts by 2).

Now you have a system of equations to solve. Substitution is the most obvious approach .. so I need one of the equations to isolate a single variable... I chose the bottom one ..

2y + x = 40
2y + x - 2y = 40 - 2y
x = 40 - 2y <--- I can take this and substitue into the first equ.

That makes ...

(40 - 2y)² + y² = 325.

This equation can be expanded and solved using the quadratic formula or the reverse foil method. I will expand it a bit for you here....

(1600 - 160y + 4y²) + y² = 325
5y² - 160y + 1600 = 325 ---- divide all parts by 5
y² - 32y + 320 = 65 ---- get everything equal to 0.
y² - 32y + 255 = 0

You can solve this equation however you choose to get an answer. I will leave that for you, but warning! You will get 2 values for y that solve this equation. That's ok because there are actually two combinations of numbers that will work in this scenario!!!

Good luck with the rest of your homework. I hope what I wrote helps!!

Answer 2

Let S and s be the lengths of the sides of each of the two squares. Since the squares are adjacent, without loss of generality, if we assume that S is the length of the side of the larger square and s is the length of the side of the smaller square, then the perimeter of the foundation is: 3S + 3s + (S - s) = 4S + 2s = 80.

Also, the area of the foundation equals: S^2 + s^2 = 325.

Solving the first equation for s in terms of S, we have: s = 40 - 2S.

Substituting s = 40 - 2S into the second equation, we get:

S^2 + (40 - 2S)^2 = 325
S^2 + 1600 - 160S + 4S^2 = 325
5S^2 - 160S +1275 = 0
(5S - 85)(S - 15) = 0 ==> S = 15 or S = 85/5 = 17

S = 15 ==> s = 40 - 2S = 40 - 2(15) = 40 - 30 = 10
S = 17 ==> s = 40 - 2S = 40 - 2(17) = 40 - 34 = 6

Check:

S = 15, s = 10 ==> S^2 + s^2 = 225 + 100 = 325
==> 4S + 2s = 60 + 20 = 80, so this checks

S = 17, s = 6 ==> S^2 + s^2 = 289 + 36 = 325
==> 4S + 2s = 68 + 12 = 80, so this checks too!

The lengths of the sides of each square are either:

15 m for the larger square and 10 m for the smaller square, OR

17 m for the larger square and 6 m for the smaller square.

Answer 3

Check your answers!

4y + 2x = 80 => 2y + x = 40 => x = 40 - 2y
x^2 + y^2 = 325
(40 - 2y)^2 + y^2 = 325
4y^2 - 160y + 1600 + y^2 = 325
5y^2 - 160y + 1600 = 325, divide by 5
y^2 - 32y + 320 = 54

y^2 - 32y + 255 = 0
255 = 3*5*17
(y - 15)(y - 17) = 0 => y = 15 or y = 17

y = 15: 2y + x = 40, x = 10
15^2 + 10^2 = 225 + 100 = 325 : check

y = 17: 2y + x = 40, x = 6
17^2 + 6^2 = 289 + 36 = 325

so there are 2 answers, 15, 10 and 17, 6

Answer 4

Let the squares have lengths x and y, x being smaller.
now if you study 2 adjacent squares of different lengths, on the place where the 2 squares meet, the extra bits where the smaller square does not cover, and the adjacent side of the smaller square, add up to give the length of the larger square.

hence, perimeter= 3Y + Y + 2X
area= X^2 + Y^2

i am sure you can work out the rest.

Answer 5

Let x = length of side of smaller square
Let y = length of side of larger square
then Perimeter = 3x +3y+ y-x =80
So P = 2x+4y = 80, or x + 2y = 40, or x= 40 - 2y
Area = x^2 + y^2 = 325
So (40- 2y)^2 + y^2 =325
So 1600 -160y +4y^2 + y^2 = 325
So 5 y^2 - 160y + 1275 =0
So y^2 - 32y + 255 = 0
So (y-17)(y-15) = 0
So y = 17 or y = 15
Since x = 40 - 2y, then
x = 40 - 2(17) = 6, or
x = 40 -2(15) = 10

Answer 6

Interesting, I get two different answers:
(17,6)
(15,10)
and they both check out.

3*a + 4*b - a = 80 -> a + 2*b = 40
a^2 + b^2 = 325

Substituting the 1st equation into the second and simplifying gives,
b^2 - 32*b + 255 = 0

b = 16+-1 or b = 17,15 and a = 6,10.

Answer 7

the perimeter=80m=3a+3b+b-a=4b-2a=80
2b-a=40....................(1)
a=40-2b
a^2+b^2=325
(2b-40)^2+b^2=325
1600-160b+2b^2=325
5b^2-160b-1275=0
=>b^2-32b-255=0
(b-15)(b-17)=0
b=15 or 17
so the sides of the squares are
15 m and 17m repectively

Answer 8

25 and 48 meters