Parallelograms - ISEE Upper Level Quantitative Reasoning

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Question

Give the area of the above parallelogram if .

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Answer

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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Question

Give the area of the above parallelogram if .

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Answer

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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Question

Three of the vertices of a parallelogram on the coordinate plane are . What is the area of the parallelogram?

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Answer

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

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:

The area of the 12 by 12 square is

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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Question

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

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Answer

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

$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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Question

Solve for :

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Answer

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

The sum of the angles must equal 360.

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Question

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

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Answer

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . 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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Question

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

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Answer

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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Question

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

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Answer

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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Question

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

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Answer

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:

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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Question

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

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Answer

The height of the parallelogram is , and the base is . By the 45-45-90 Theorem, . 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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Question

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

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Answer

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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Question

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

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Answer

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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Question

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

Tap to reveal answer

Answer

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:

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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Question

Solve for :

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Answer

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

The sum of the angles must equal 360.

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Question

Solve for :

Tap to reveal answer

Answer

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

The sum of the angles must equal 360.

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Question

Give the area of the above parallelogram if .

Tap to reveal answer

Answer

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}$$

← Didn't Know|Knew It →

Question

Give the area of the above parallelogram if .

Tap to reveal answer

Answer

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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Question

Three of the vertices of a parallelogram on the coordinate plane are . What is the area of the parallelogram?

Tap to reveal answer

Answer

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

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:

The area of the 12 by 12 square is

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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Question

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

Tap to reveal answer

Answer

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

$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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Question

Give the area of the above parallelogram if .

Tap to reveal answer

Answer

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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