Thread: Can someone please help me with these few problems?

So I've had to take home these math worksheets and a few of the problems really confuse me. Could someone please help me with these, I would really appreciate it. Thanks.


1. A diagonal of a square is 5 in. longer than a side of the square. Find the length of a side and the length of a diagonal.

2. The sum of the squares of two consecutive integers is 113. Find the integers.

3. One number is 6 less than 5 times another number. The product of the two numbers is 8. Find the numbers.

4. A new parking lot is twice the area of the old one. If the length of the old one is 100 ft more than the width and the width is the same as it was, what are the dimensions of the lot?

And now the really hard ones...

1. The area of a rectangular field is 1200 m^2. Two parallel sides are fenced with aluminum at $20/m. The remaining two sides are fenced with steel at $10/m. If the cost of the fencing is $2200, what is the length of each side fenced with aluminum? What is the length of each side fenced with steel?

I'll post the rest a little later.

1. Since a square can be divided into 2 right triangles the length of the diagonal can be calculated using the pythagorean theorem.

a^2+b^2=c^2

c is the hypotenuse, the diagonal of the square.

Since we know that the hypotenuse is 5 inches longer than a side of the square (and all sides are equal) we can use the equation

a^2+a^2=(a+5)^2
2a^2=a^2+10a+25
a^2=10a+25

Which can be further rearranged to make a quadratic function

a^2-10a-25=0

Which unfortunately cannot be factored so we have to use the quadratic formula

x=[-b±sqrt(b^2-4ac)]/2a
x=[10±sqrt([-10]^2-4[1][-25])]/2(1)
x=[10±sqrt(100+100)]/2
x=[10±sqrt(200)]/2

x=10+sqrt(200)/2 or x=10-sqrt(200)/2

since sqrt(200) is approximately 14 and a length cannot be negative

x=10+sqrt(200)/2 <--exact answer
x=12.0710678 <--rounded answer

So side length of the square is 12.0710678 inches and the length of the diagonal is 17.0710678




Please keep in mind I haven't done this type of problem for a month or two and may have made a mistake somewhere, but that is the general idea of how to do it.

2.
Let n represent the first integer
Let n+1 represent the second integer

n^2+(n+1)^2=113
n^2+n^2+2n+1=113
2n^2+2n=112

Which can be further rearranged to make the quadratic function

2n^2+2n-112=0

Which can be factored (to make it easier to work with) to

n^2+n-56=0

Which cannot be facotred so we use the quadratic formula

n=[-b±sqrt(b^2-4ac)]/2a
n=[-1±sqrt(1^2-4[1][-56])]/2(1)
n=[-1±sqrt(1+224)]/2
n=[-1±sqrt(225)]/2
n=(-1±15)/2

n=(-1+15)/2 or n=(-1-15)/2
n=14/2 or n=-16/2
n=7 or n=-8

Therfore the two consecutive integer are 7 and 8 or -8 and -7.

3.
Let n represent the first number
Let 5n-6 represent the second number

n(5n-6)=8
5n^2-6n=8
5n^2-6n-8=0
(5n+4)(n-2)=0

n=-4/5 or n=2

Sub these into 5n-6
Second number
n=-10 or n=4

The two numbers are -4/5 and -10 or 2 and 4

Thanks alot Philmac. This all very helpful and you explained these problems very nicely. I really appreciate this, thank you so much.

does anyone know how to do the other problems at all?

4.A new parking lot is twice the area of the old one. If the length of the old one is 100 ft more than the width and the width is the same as it was, what are the dimensions of the lot?

area of parking lot A = La * Wa.....eq1
area of parking lot B = Lb * Wb.....eq2

B = 2A...............eq3
La = Wa +100.....eq4
Wa = Wb...........eq5

It seems to me we have 6 unknowns and only 5 equations. Are we missing something. Did you write the problem correctly?

I'm sure that I wrote the problem correctly, its just a really confusing one.