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

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

Insufficient information is given to answer the problem.

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 $2 \times 31\frac{1}{2} = 63$ .

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?

Explanation

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.

Parallelogram

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 .

Explanation

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.

Rhombus

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

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

Parallelogram A is below:

Rhombus_1

Parallelogram B is below:

Parallelogram

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

(A) and (B) are equal

It is impossible to determine which is greater from the information given

(A) is greater

(B) is greater

Explanation

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.

Parallelogram1

Give the area of the above parallelogram if .

$50\sqrt{2}$

$50\sqrt{3}$

$25\sqrt{2}$

$25\sqrt{3}$

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

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

Parallelogram2

Give the area of the above parallelogram if .

$25\sqrt{3}$

$25\sqrt{2}$

$50\sqrt{2}$

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

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

Parallelogram

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

(A) The area of the parallelogram

(B) 120 square inches

(B) is greater

(A) is greater

(A) and (B) are equal

It is impossible to determine which is greater from the information given

Explanation

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.