Math Help for 4th Grader?

1. In a square, the length and the width are the same, so the area is equal to one side squared.

A = L*W

L = W = S (side)

A = S^2

We know that the area is 36, so we can solve for the length of one of the sides.

36 = S^2

root (36) = S

S = 6

Now, we know that the perimeter of a square is the four sides added together, or 4S. So,

P = 4S

P = 4*6

P = 24 in.

2. To do this, you just have to draw to rectangles with lengths and widths that multiply to 12. For example one rectangle that you might have is one with a width of 3 m and a length of 4 m. Next, draw a rectangle that has a width of 2 m and a length of 6 m.

Now that you have to rectangles with areas of 12, you can calculate their perimeters.

The first rectangle has a perimeter of:

P = 2L + 2W

P = 2*4 + 2*3

P = 8 + 6 = 14 m

The second rectangle has a perimeter of:

P = 2L + 2W

P = 2*6 + 2*2

P = 12 + 4 = 16 m

Since 16 is greater than 14, you circle the second rectangle.

I hope this helped!

1.

Let each side of the square be represented by s.

Area of Square: s^2 = 36.

Parameter of Square: P = 4*s.

From the Area of Square, we get:

s = sqrt(36)

s = 6

Thus:

P = 4*s = 4*[6] = 24.

2.

I cannot draw on this computer.

Length will be l, width will be w.

However, you can solve this Mathematically using Calculus.

Area: A = l*w = 12

Para: P = 2*(l+w)

However, 4th Graders probably don't use Calculus.

I think this question is meant for intuition.

Since the question is only asking for two rectangles, we can choose any two.

Let's pick easy numbers.

Let the first one have: l=6 and w=2 [Area is 12]

Let the 2nd one have: l=4 and w=3 [Area is 12]

The Parameter, however, will be different.

The first one will be 2*(6+2) = 16

The 2nd one will be 2*(4+3) = 14

So draw those two rectangles and circle the one with P = 16.

In terms of explaining this one, tell him that P=2*(l+w) and A = l*w.

Explain how l and w can change even if A is constant.

So it certainly is possible for P to change while A is constant.

Just so you know, there is no exact correct answer for this question.

I can draw a rectangle with l=3.998 and w=3.001500750375...

So my point here is that you are not trying to arrive at a "correct" answer for this question, but instead the "correct" idea.

This "correct" idea is that P can change even while A is constant.

Think of P as a string. If you have two different lengths string, you can still use these two strings to encircle the same area. It sounds kind of crazy, but it's true.

Perhaps you can explain to him by using a demo with string of length 14 and 16, showing how both strings can produce the same area, but not the same length.

Anyway, I hope I helped.

1) The area of a square is side squared, or s^2

Since the area of the square is 36, just set it equal to the area formula

36=s^2

s=+/-6

Since length cannot be negative (Because you can't), 6 is the length of a side of the square. To find the perimeter, just multiply a side by 4, or 4s.

4(6)=24.

2) Area of a rectangle= l*w. (Similar to a square, but a square has congruent sides)

Simply draw two rectangles and you need to know multiples of 12.

1*12=12

2*6=12

3*4=12

So say I took the bottom to to draw. Rectangles have a perimeter of (2*length)+(2*width)

or 2l+2w

for the 2*6=12 rectangle, it has a width of 2 and a length of 6. Width is the shorter side of the rectangle.

And this rectangle will have a perimeter of 2(6)+2(2)=16 I believe.

And for the 3*4 rectangle, you will get

2(3)+2(4)=6+8=14

So in this situation, the 2*6 rectangle has the largest perimeter, so your kid would want to circle it.

Good Luck!