Parallelograms - ISEE Upper Level: Quantitative Reasoning

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Question

In the above parallelogram, $\angle 1$ is acute. Which is the greater quantity?

(A) The area of the parallelogram

(B) 120 square inches

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Answer

Since $\angle 1$ is acute, a right triangle can be constructed with an altitude as one leg and a side as the hypotenuse, as is shown here. The height of the triangle must be less than its sidelength of 8 inches.

The height of the parallelogram must be less than its sidelength of 8 inches.

The area of the parallelogram is the product of the base and the height - which is $A = 15 \cdot n$

Therefore,

$15 \cdot n <15 \cdot 8$

(B) is greater.

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Question

Parallelogram A is below:

Parallelogram B is below:

Note: These figures are NOT drawn to scale.

Refer to the parallelograms above. Which is the greater quantity?

(A) The area of parallelogram A

(B) The area of parallelogram B

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Answer

The area of a parallelogram is the product of its height and its base; its slant length is irrelevant. Both parallelograms have the same height (8 inches) and the same base (1 foot, or 12 inches), so they have the same areas.

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Question

Figure NOT drawn to scale

The above figure shows Rhombus ; and are midpoints of their respective sides. Rectangle has area 150.

Give the area of Rhombus .

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Answer

A rhombus, by definition, has four sides of equal length. Therefore, . Also, since and are the midpoints of their respective sides,

$AX = XB = \frac{1}{2 }AB=\frac{1}{2 } CD = CY = YD$

We will assign to the common length of the four half-sides of the rhombus.

Also, both $\overline{AY}$ and $\overline{XC}$ are altitudes of the rhombus; the are congruent, and we will call their common length (height).

The figure, with the lengths, is below.

Rectangle has dimensions and ; its area, 150, is the product of these dimensions, so

$A_{1} = hn = 150$

The area of the entire Rhombus is the product of its height and the length of a base , so

$A = h \cdot 2n = 2hn = 2 \cdot hn = 2 \cdot 150 = 300$ .

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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

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 $12^{2} = 144$

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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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 $x_{1} = y_{1} = 0$ , $x_{2} = 4$ , $y_{2} = 5$ :

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

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

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

$= \sqrt{16+25 }$

$= \sqrt{41}$

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

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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

In the above parallelogram, $\angle 1$ is acute. Which is the greater quantity?

(A) The area of the parallelogram

(B) 120 square inches

Tap to reveal answer

Answer

Since $\angle 1$ is acute, a right triangle can be constructed with an altitude as one leg and a side as the hypotenuse, as is shown here. The height of the triangle must be less than its sidelength of 8 inches.

The height of the parallelogram must be less than its sidelength of 8 inches.

The area of the parallelogram is the product of the base and the height - which is $A = 15 \cdot n$

Therefore,

$15 \cdot n <15 \cdot 8$

(B) is greater.

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Question

Parallelogram A is below:

Parallelogram B is below:

Note: These figures are NOT drawn to scale.

Refer to the parallelograms above. Which is the greater quantity?

(A) The area of parallelogram A

(B) The area of parallelogram B

Tap to reveal answer

Answer

The area of a parallelogram is the product of its height and its base; its slant length is irrelevant. Both parallelograms have the same height (8 inches) and the same base (1 foot, or 12 inches), so they have the same areas.

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Question

Figure NOT drawn to scale

The above figure shows Rhombus ; and are midpoints of their respective sides. Rectangle has area 150.

Give the area of Rhombus .

Tap to reveal answer

Answer

A rhombus, by definition, has four sides of equal length. Therefore, . Also, since and are the midpoints of their respective sides,

$AX = XB = \frac{1}{2 }AB=\frac{1}{2 } CD = CY = YD$

We will assign to the common length of the four half-sides of the rhombus.

Also, both $\overline{AY}$ and $\overline{XC}$ are altitudes of the rhombus; the are congruent, and we will call their common length (height).

The figure, with the lengths, is below.

Rectangle has dimensions and ; its area, 150, is the product of these dimensions, so

$A_{1} = hn = 150$

The area of the entire Rhombus is the product of its height and the length of a base , so

$A = h \cdot 2n = 2hn = 2 \cdot hn = 2 \cdot 150 = 300$ .

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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

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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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 $12^{2} = 144$

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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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 $x_{1} = y_{1} = 0$ , $x_{2} = 4$ , $y_{2} = 5$ :

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

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

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

$= \sqrt{16+25 }$

$= \sqrt{41}$

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

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