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ISEE Upper Level Math : Parallelograms

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6 Diagnostic Tests 244 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #271 : Geometry

Parallelogram2

Give the area of the above parallelogram if BC = 10 .

Possible Answers:

$25\sqrt{3}$

$50\sqrt{2}$

$25\sqrt{2}$

Correct answer:

$25\sqrt{3}$

Explanation:

Multiply height by base to get the area.

By the 30-60-90 Theorem:

$BD = \frac{BC}{2} = \frac{10}{2}} = 5$

and

$CD =BD \cdot \sqrt{3} = 5\sqrt{3}$

The area is therefore

$BD \cdot CD = 5 \cdot 5\sqrt{3} = 25\sqrt{3}$

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Example Question #281 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Parallelogram1

Give the area of the above parallelogram if BC = 10 .

Possible Answers:

$50\sqrt{3}$

$25\sqrt{3}$

$25\sqrt{2}$

$50\sqrt{2}$

Correct answer:

Explanation:

Multiply height by base to get the area.

By the 45-45-90 Theorem,

$BD = CD = \frac{BC}{\sqrt{2}} = \frac{10}{\sqrt{2}}$ .

The area is therefore

$BD \cdot CD = \frac{10}{\sqrt{2}} \cdot \frac{10}{\sqrt{2}} = \frac{100}{2} = 50$

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

Three of the vertices of a parallelogram on the coordinate plane are (0,0), (9,12),(12,9) . What is the area of the parallelogram?

Possible Answers:

Insufficient information is given to answer the problem.

Correct answer:

Explanation:

As can be seen in the diagram, there are three possible locations of the fourth point of the parallelogram:

Axes_2

Regardless of the location of the fourth point, however, the triangle with the given three vertices comprises exactly half the parallelogram. Therefore, the parallelogram has double that of the triangle.

The area of the triangle can be computed by noting that the triangle is actually a part of a 12-by-12 square with three additional right triangles cut out:

Axes_1

The area of the 12 by 12 square is $12^{2} = 144$

The area of the green triangle is $\frac{1}{2} \times 12 \times 9 = 54$ .

The area of the blue triangle is $\frac{1}{2} \times 9 \times 12 = 54$ .

The area of the pink triangle is $\frac{1}{2} \times 3 \times 3 = 4 \frac{1}{2}$ .

The area of the main triangle is therefore

$144- 54-54-4\frac{1}{2} = 31\frac{1}{2}$

The parallelogram has area twice this, or $2 \times 31\frac{1}{2} = 63$ .

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Example Question #1 : Parallelograms

One of the sides of a square on the coordinate plane has an endpoint at the point with coordinates (4, 5 ) ; it has the origin as its other endpoint. What is the area of this square?

Possible Answers:

Correct answer:

Explanation:

The length of a segment with endpoints (0,0) and (4, 5 ) can be found using the distance formula with $x_{1} = y_{1} = 0$ , $x_{2} = 4$ , $y_{2} = 5$ :

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

$= \sqrt{(4-0)^{2}+(5-0 )^{2}}$

$= \sqrt{4^{2}+5^{2}}$

$= \sqrt{16+25 }$

$= \sqrt{41}$

This is the length of one side of the square, so the area is the square of this, or 41.

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Example Question #1 : How To Find The Perimeter Of A Parallelogram

Parallelogram1

The area of the Parallelogram ABCD is . Give its perimeter in terms of .

Possible Answers:

$4 \sqrt{2X}$

$\sqrt{6X}$

$2 \sqrt{3X}$

$2 \sqrt{X} + 2 \sqrt{2X}$

$4 \sqrt{X }$

Correct answer:

$2 \sqrt{X} + 2 \sqrt{2X}$

Explanation:

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, BD=CD . Since the product of the height and the base of a parallelogram is its area,

$BD \cdot CD = A$

$\left (BD \right )^{2} = X$

$BD = \sqrt{X}$

Also by the 45-45-90 Theorem,

$CD = AB = BD = \sqrt{X}$ , and

$AD = BC = \sqrt{X} \cdot \sqrt{2} = \sqrt{2X}$

The perimeter of the parallelogram is

$AB + CD + BC + AD = \sqrt{X} + \sqrt{X} + \sqrt{2X}+ \sqrt{2X} = 2 \sqrt{X} + 2 \sqrt{2X}$

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Example Question #1 : How To Find The Perimeter Of A Parallelogram

Parallelogram1

Calculate the perimeter of the above parallelogram if $BD = 8\sqrt{2}$ .

Possible Answers:

$48\sqrt{2}$

$16+ 16 \sqrt{2}$

$32+ 16 \sqrt{2}$

$32+ 32\sqrt{2}$

Correct answer:

$32+ 16 \sqrt{2}$

Explanation:

By the 45-45-90 Theorem,

$AB= CD = BD = 8\sqrt{2}$

$AD= BC = BD \cdot \sqrt{2} = 8\sqrt{2} \cdot \sqrt{2} = 8 \cdot 2 = 16$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 8 \sqrt{2} + 8 \sqrt{2} + 16+16 = 32+ 16 \sqrt{2}$

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Example Question #5 : Parallelograms

Parallelogram2

Calculate the perimeter of the above parallelogram if $BD = 30\sqrt{6}$ .

Possible Answers:

$360\sqrt{2}$

$180\sqrt{2}$

$240+120\sqrt{2}$

360

$180+180\sqrt{2}$

Correct answer:

$180\sqrt{2}$

Explanation:

By the 30-60-90 Theorem:

$AB = CD= \frac{BD}{\sqrt{3}} = \frac{30 \sqrt{6}}{\sqrt{3}}= 30\sqrt{2}$ , and

$BC = AD = 2\cdot AB = 2 \cdot 30\sqrt{2} =60\sqrt{2}$

The perimeter of the parallelogram is

$AB + CD + BC + AD = 30\sqrt{2}+ 30\sqrt{2}+60\sqrt{2}+60\sqrt{2}= 180\sqrt{2}$

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Example Question #284 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Find the perimeter of a parallelogram with a base of 6in and a side of length 8in.

Possible Answers:

$32\text{in}$

$48\text{in}$

$28\text{in}$

$36\text{in}$

$24\text{in}$

Correct answer:

$28\text{in}$

Explanation:

A parallelogram has 4 sides. A base (where the opposite side is equal) and a side (where the opposite side is equal). So, we will use the following formula:

P = b+b+s+s

where b is the base and s is the side of the parallelogram.

We know the base has a length of 6in. We also know the side has a length of 8in.

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

$P = 6\text{in} + 6\text{in} + 8\text{in} + 8\text{in}$

$P = 28\text{in}$

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Example Question #282 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Solve for :

Problem_10

Possible Answers:

x=95

x=89

x=109

x=19

Correct answer:

x=109

Explanation:

Find the sum of the interior angles of the polygon using the following equation where n is equal to the number of sides.

=180(n-2)

180(4-2)=180(2)=360

The sum of the angles must equal 360.

360=x+71+x+71=2x+142

360-142=2x+142-142

218=2x

109=x

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ISEE Upper Level Math : Parallelograms

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

All ISEE Upper Level Math Resources

Example Questions

Example Question #271 : Geometry

Example Question #281 : Isee Upper Level (Grades 9 12) Mathematics Achievement

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

Example Question #1 : Parallelograms

Example Question #1 : How To Find The Perimeter Of A Parallelogram

Example Question #1 : How To Find The Perimeter Of A Parallelogram

Example Question #5 : Parallelograms

Example Question #284 : Isee Upper Level (Grades 9 12) Mathematics Achievement

Example Question #282 : Isee Upper Level (Grades 9 12) Mathematics Achievement

All ISEE Upper Level Math Resources

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