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ISEE Upper Level Math : Quadrilaterals

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Example Questions

Example Question #3 : How To Find The Area Of A Square

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

Possible Answers:

$\frac{5}{12}$

$\frac{2}{3}$

$\frac{7}{12}$

$\frac{1}{2}$

Correct answer:

$\frac{7}{12}$

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:

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

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Example Question #7 : How To Find The Area Of A Square

One of the sides of a square on the coordinate plane has its endpoints at the points with coordinates (6, 2) and (-5, 1) . What is the area of this square?

Possible Answers:

122

120

Correct answer:

122

Explanation:

The length of a segment with endpoints (6, 2) and (-5, 1) 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.

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Example Question #5 : How To Find The Area Of A Square

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

Possible Answers:

4AB

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

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

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

Correct answer:

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

Explanation:

The length of a segment with endpoints (A, B) and (-B, A) 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 $2A^{2} +2 B^{2}$ .

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Example Question #6 : How To Find The Area Of A Square

One of the vertices of a square is at the origin. The square has area 13. Which of the following could be the vertex of the square opposite that at the origin?

Possible Answers:

(6,0)

(3, 3)

(4, 2)

(5, 1)

Correct answer:

(5, 1)

Explanation:

Since a square is a rhombus, one way to calculate the area of a square is to take half the square of the length of a diagonal. If we let be the length of each diagonal, then

$\frac{1}{2} d^{2} = A$

$\frac{1}{2} d^{2} = 13$

$d^{2} =26$

$d = \sqrt{26}$

Therefore, we want to choose the point that is $\sqrt{26}$ units from the origin. Using the distance formula, we see that (5, 1) is such a point:

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

$= \sqrt{(5-0)^{2}+(1-0)^{2}}$

$= \sqrt{5^{2}+1^{2}}$

$= \sqrt{25+1}$

$= \sqrt{26}$

Of the other points:

(6,0) :

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

$= \sqrt{(6-0)^{2}+(0-0)^{2}}$

$= \sqrt{6^{2}+0^{2}}$

$= \sqrt{36+0}$

$= \sqrt{36 }$

(4, 2) :

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

$= \sqrt{(4-0)^{2}+(2-0)^{2}}$

$= \sqrt{4^{2}+2^{2}}$

$= \sqrt{16+4}$

$= \sqrt{20}$

(3, 3) :

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

$= \sqrt{(3-0)^{2}+(3-0)^{2}}$

$= \sqrt{3^{2}+3^{2}}$

$= \sqrt{9+9}$

$= \sqrt{18}$

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Example Question #21 : Quadrilaterals

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?

Possible Answers:

36 in^2

12in^2

18in^2

36 in

Correct answer:

36 in^2

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.

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

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Example Question #21 : Squares

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?

Possible Answers:

$60 \pi$

$120 \pi$

$\frac{60}{\pi}$

$\frac{120}{\pi}$

Correct answer:

$\frac{120}{\pi}$

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, 2 r , 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:

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

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Example Question #22 : Squares

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

Possible Answers:

$18\text{cm}^2$

$81\text{cm}^2$

$36\text{cm}^2$

$54\text{cm}^2$

$72\text{cm}^2$

Correct answer:

$81\text{cm}^2$

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

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

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

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Example Question #22 : Quadrilaterals

While out walking, you find a strange, square-shaped piece of metal. If the side length of the piece is 26 inches, what is the area of the square?

Possible Answers:

676in^2

54in^2

786in^2

104in^2

Correct answer:

676in^2

Explanation:

While out walking, you find a strange, square-shaped piece of metal. If the side length of the piece is 26 inches, what is the area of the square?

To find the area of a square, simply square the side length:

A=s^2=(26in)^2=676in^2

So, our answer is:

676in^2

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Example Question #24 : Squares

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

Possible Answers:

Correct answer:

Explanation:

Write the formula for the area of a square.

A=s^2

Substitute the side into the formula.

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

$A = 4\cdot 3 = 12$

The answer is:

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Example Question #23 : Quadrilaterals

What is the diagonal of a square with a side of 4?

Possible Answers:

$\sqrt{2}$

$4\sqrt{2}$

Correct answer:

$4\sqrt{2}$

Explanation:

Squares have all congruent sides. To find the diagonal, first recognize that you're dealing with an isoceles triangle when you draw the diagonal in the square. That means that two of the sides are congruent in the triangle. Thus, it's a special 45-45-90 triangle. In such triangles, the sides are x and the hypotenuse is $x\sqrt{2}$ . Since we know x is 4, we can plug in 4 to the expression $x\sqrt{2}$ . Thus, the answer is $4\sqrt{2}$ .

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All ISEE Upper Level Math Resources

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ISEE Upper Level Math : Quadrilaterals

Study concepts, example questions & explanations for ISEE Upper Level Math

All ISEE Upper Level Math Resources

Example Questions

Example Question #3 : How To Find The Area Of A Square

Example Question #7 : How To Find The Area Of A Square

Example Question #5 : How To Find The Area Of A Square

Example Question #6 : How To Find The Area Of A Square

Example Question #21 : Quadrilaterals

Example Question #21 : Squares

Example Question #22 : Squares

Example Question #22 : Quadrilaterals

Example Question #24 : Squares

Example Question #23 : Quadrilaterals

All ISEE Upper Level Math Resources

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