Squares - PSAT Math

The diagonal of a square creates two special 45-45-90 triangles, meaning that the diagonal of a square is just the length of one side of the square multiplied by the square root of 2.

In this problem, you can figure out the length of one side of the square by finding the square root of the area (which is equal to a side length), then multiplying that number by the square root of 2.

Assign variables such that

One side of ABCD = a

and One side of EFGH = e

Note that all sides are the same in a square. Since the perimeter is the sum of all sides, according to the question:

4a = 3 x 4e = 12e or a = 3e

From that area of EFGH is 25,

e x e = 25 so e = 5

Substitute a = 3e so a = 15

We aren’t done. Since we were asked for the area of ABCD, this is a x a = 225.

If the area of the half circle is , then the area of a full circle is twice that, or .

Use the formula for the area of a circle to solve for the radius:

36π = πr2

r = 6

If the radius is 6, then the diameter is 12. We know that the sides of the square are the same length as the diameter, so each side has length 12.

Therefore the area of the square is 12 x 12 = 144.

Let S be the original side length. S*S would represent the original area. Doubling the side length would give you 2S*2S, simplifying to 4*(S*S), giving a new area of 4x the original, or 144.

Area of a square in terms of each of its sides:

Area = S x S

Perimeter of a square:

Perimeter = 4S

So if 'the perimeter of a square is equal to twice its area':

2 x Area = Perimeter

2 x \[S x S\] = \[4S\]; divide by 2:

S x S = 2S; divide by S:

S = 2

If Freddie uses x feet of fencing makes a square, each side must be x/4 feet long. The area of this square is (x/4)2 = _x_2/16 = p square feet.

If Freddie uses one third of x feet of fencing makes a square, each side must be x/12 feet long. The area of this square is (x/12)2 = _x_2/144 = 1/9(_x_2/16) = 1/9(p) = p/9 square feet.

Alternate method:

The scale factor between the small perimeter and the larger perimeter = 1 : 3. Since we're comparing area, a two-dimensional measurement, we can square the scale factor and see that the ratio of the areas is 12 : 32 = 1 : 9.

This is an isosceles right triangle, so the diagonal must equal $\sqrt{2}$ times the length of a side. Thus, one side of the square measures 16 cm, and the area is equal to (16 $cm)^{2}$ = 256 $cm^{2}$

The area of A is n, the area of B is 5.0625n. Therefore, you can fit 5.06 A's in B.

The perimeter of a square is given by P=4s=12 so the side length of the original square is 3 in. The side of the new square is enlarged by a factor of 3 to give s=9 in.

So the area of the new square is given by A = $s^{2}$ = $(9)^{2}$ = 81 $in^{2}$.

To inscribe means to draw inside a figure so as to touch in as many places as possible without overlapping. The circle is inside the square such that the diameter of the circle is the same as the side of the square, so the side is actually 4 in. The perimeter of the square = 4s = 4 * 4 = 16 in.

Find the area of Square Y, then calculate the area of Square X.

If the perimeter of Square Y is 24, then each side is 24/4, or 6.

A = 6 * 6 = 36 sq ft, for Square Y

If Square X has 3 times the area, then 3 * 36 = 108 sq ft.

The area of the given square is given by A = $s^{2}$ so the side must be 6 in. The side is reduced by a factor of two, so the new side is 3 in. The perimeter of the new square is given by .

Area of a square in terms of each of its sides:

Area = S x S

Perimeter of a square:

Perimeter = 4S

So if 'the perimeter of a square is equal to twice its area':

2 x Area = Perimeter

2 x \[S x S\] = \[4S\]; divide by 2:

S x S = 2S; divide by S:

S = 2

Let S be the original side length. S*S would represent the original area. Doubling the side length would give you 2S*2S, simplifying to 4*(S*S), giving a new area of 4x the original, or 144.

Assign variables such that

One side of ABCD = a

and One side of EFGH = e

Note that all sides are the same in a square. Since the perimeter is the sum of all sides, according to the question:

4a = 3 x 4e = 12e or a = 3e

From that area of EFGH is 25,

e x e = 25 so e = 5

Substitute a = 3e so a = 15

We aren’t done. Since we were asked for the area of ABCD, this is a x a = 225.

If the area of the half circle is , then the area of a full circle is twice that, or .

Use the formula for the area of a circle to solve for the radius:

36π = πr2

r = 6

If the radius is 6, then the diameter is 12. We know that the sides of the square are the same length as the diameter, so each side has length 12.

Therefore the area of the square is 12 x 12 = 144.

If Freddie uses x feet of fencing makes a square, each side must be x/4 feet long. The area of this square is (x/4)2 = _x_2/16 = p square feet.

If Freddie uses one third of x feet of fencing makes a square, each side must be x/12 feet long. The area of this square is (x/12)2 = _x_2/144 = 1/9(_x_2/16) = 1/9(p) = p/9 square feet.

Alternate method:

The scale factor between the small perimeter and the larger perimeter = 1 : 3. Since we're comparing area, a two-dimensional measurement, we can square the scale factor and see that the ratio of the areas is 12 : 32 = 1 : 9.

This is an isosceles right triangle, so the diagonal must equal $\sqrt{2}$ times the length of a side. Thus, one side of the square measures 16 cm, and the area is equal to (16 $cm)^{2}$ = 256 $cm^{2}$

The area of A is n, the area of B is 5.0625n. Therefore, you can fit 5.06 A's in B.