ISEE Upper Level Quantitative Reasoning - How to find the area of a square

A square is made into a rectangle by increasing the width by 20% and decreasing the length by 20%. By what percentage has the area of the square changed?

decreased by 4%

increased by 20%

the area remains the same

decreased by 10%

Explanation

The area decreases by 20% of 20%, which is 4%.

The easiest way to see this is to plug in numbers for the sides of the square. If we are using percentages, it is easiest to use factors of 10 or 100. In this case we will say that the square has a side length of 10.

10% of 10 is 1, so 20% is 2. Now we can just increase one of the sides by 2, and decrease another side by 2. So our rectangle has dimensions of 12 x 8 instead of 10 x 10.

The original square had an area of 100, and the new rectangle has an area of 96. So the rectangle is 4 square units smaller, which is 4% smaller than the original square.

Which of the following is equal to the area of a square with sidelength $1 \frac{1}{6}$ yards?

$1,764 \textrm{ in}^{2}$

$1,225 \textrm{ in}^{2}$

$2,304 \textrm{ in}^{2}$

$2,025 \textrm{ in}^{2}$

Explanation

Multiply the sidelength by 36 to convert from yards to inches:

$1 \frac{1}{6} \times36 = \frac{7}{6} \times \frac{36}{1} = \frac{7}{1} \times \frac{6}{1} = 42\ inches$

Square this to get the area:

$42^{2} = 1,764$ square inches

Find the area of a square with a base of 9cm.

$81\text{cm}^2$

$18\text{cm}^2$

$36\text{cm}^2$

$54\text{cm}^2$

$72\text{cm}^2$

Explanation

To find the area of a square, we will use the following formula:

$\text{area of square} = l \cdot w$

where l is the length and w is the width of the square.

Now, we know the base (or length) of the square is 9cm. Because it is a square, all sides are equal. Therefore, the width is also 9cm.

Knowing this, we can substitute into the formula. We get

$\text{area of square} = 9\text{cm} \cdot 9\text{ccm}$

$\text{area of square} = 81\text{cm}^2$

One of the sides of a square on the coordinate plane has its endpoint at the points with coordinates and , where and are both positive. Give the area of the square in terms of and .

$2A^{2} +2 B^{2}$

$2A^{2} +4AB +2 B^{2}$

$2A^{2} -4AB +2 B^{2}$

Explanation

The length of a segment with endpoints and can be found using the distance formula as follows:

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$d = \sqrt{((-B)-A)^{2}+(A-B)^{2}}$

$d = \sqrt{(-B-A)^{2}+(A-B)^{2}}$

$d = \sqrt{(-1)^{2}(B+A)^{2}+(A-B)^{2}}$

$d = \sqrt{1 \cdot (B^{2} +2AB +A^{2})+(A^{2}-2AB+B^{2})}$

$d = \sqrt{ (B^{2} +2AB +A^{2})+(A^{2}-2AB+B^{2})}$

$d = \sqrt{ 2A^{2} +2 B^{2}}$

This is the length of one side of the square, so the area is the square of this, or $2A^{2} +2 B^{2}$ .

One of the sides of a square on the coordinate plane has its endpoints at the points with coordinates and . What is the area of this square?

Explanation

The length of a segment with endpoints and can be found using the distance formula with $x_{1} = 6$ , $y_{1} = 2$ , $x_{2} = -5$ , $y_{2} = 1$ :

$d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}$

$= \sqrt{(6 - (-5))^{2}+(2-1)^{2}}$

$= \sqrt{11^{2}+1^{2}}$

$= \sqrt{121+1}$

$= \sqrt{122}$

This is the length of one side of the square, so the area is the square of this, or 122.

What is the area of the square with a side length of $2\sqrt{3}$ ?

Explanation

Write the formula for the area of a square.

Substitute the side into the formula.

$A=(2\sqrt{3})^2$

$A = 4\cdot 3 = 12$

The answer is:

A rectangle and a square have the same perimeter. The rectangle has length centimeters and width centimeters. Give the area of the square.

$4,900 \textrm{ cm}^{2}$

$19,600 \textrm{ cm}^{2}$

$3,600 \textrm{ cm}^{2}$

$4,000 \textrm{ cm}^{2}$

$8,000 \textrm{ cm}^{2}$

Explanation

The perimeter of the rectangle is

$P = 2L + 2W = 2\cdot100 +2\cdot40 = 200 + 80 = 280$ centimeters.

This is also the perimeter of the square, so divide this by to get its sidelength:

$s = \frac{P}{4} = 280 \div 4 = 70$ centimeters.

The area is the square of this, or $A = s^{2} = 70^{2} = 4,900$ square centimeters.

One of your holiday gifts is wrapped in a cube-shaped box.

If one of the edges has a length of 6 inches, what is the area of one side of the box?

Explanation

One of your holiday gifts is wrapped in a cube-shaped box.

If one of the edges has a length of 6 inches, what is the area of one side of the box?

We are asked to find the area of one side of a cube, in other words, the area of a square.

We can find the area of a square by squaring the length of the side.

$A_{square}=s^2=(6in)^2=36in^2$

What is the area of a square in which the length of one side is equal to $\sqrt{\frac{1}{4}+\frac{1}{3}}$ ?

$\frac{7}{12}$

$\frac{5}{12}$

$\frac{1}{2}$

$\frac{2}{3}$

Explanation

The area of a square is equal to the product of one side multiplied by another side. Therefore, the area will be equal to:

$\sqrt{\frac{1}{4}+\frac{1}{3}}\cdot \sqrt{\frac{1}{4}+\frac{1}{3}}=\frac{1}{4}+\frac{1}{3}$

The next step is to convert the fractions being added together to a form in which they have a common denominator. This gives us:

$\frac{3}{12}+\frac{4}{12}=\frac{7}{12}$

Inscribed circle

In the above diagram, the circle is inscribed inside the square. The circle has area 30. What is the area of the square?

$\frac{120}{\pi}$

$\frac{60}{\pi}$

$120 \pi$

$60 \pi$

Explanation

Inscribed circle

In terms of , the area of the circle is equal to

$A_{1} = \pi r^{2}$ .

Each side of the square has length equal to the diameter, , so its area is the square of this, or

$A_{2} = (2r)^{2} = 4 r^{2}$

Therefore, the ratio of the area of the square to that of the circle is

$\frac{A_{2} }{A_{1} } = \frac{4 r^{2}}{\pi r^{2}}= \frac{4}{\pi }$

Therefore, the area of the circle is multiplied by this ratio to get the area of the square:

$A_{1} \cdot \frac{A_{2} }{A_{1} } =A_{1} \cdot \frac{4}{\pi }$

$A_{2} =A_{1} \cdot \frac{4}{\pi }$

Substituting:

$A_{2} = 30 \cdot \frac{4}{\pi } = \frac{120}{\pi }$

How to find the area of a square

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