Quadrilaterals - ISEE Upper Level Mathematics Achievement

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Question

Solve for :

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

Consider the rhombus below. Solve for .

Answer

The total sum of the interior angles of a quadrilateral is degrees. In this problem, we are only considering half of the interior angles:

$\frac{360}{2}=180$

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Question

Note: Figure NOT drawn to scale.

The above depicts a rhombus and one of its diagonals. What is ?

Answer

The diagonals of a rhombus bisect the angles.

The angle bisected must be supplementary to the $122^{\circ }$ angle since they are consecutive angles of a parallelogram; therefore, that angle has measure $\left (180 - 122 \right ) ^{\circ } = 58^{\circ }$ , and is half that, or $29^{\circ }$ .

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Question

Three of the interior angles of a quadrilateral measure $51^{\circ }$ , $120 ^{\circ }$ , and $77^{\circ}$ . What is the measure of the fourth interior angle?

Answer

The measures of the angles of a quadrilateral have sum $360^{\circ }$ . If is the measure of the unknown angle, then:

The angle measures $112^{\circ }$ .

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Question

The angles of a quadrilateral measure $x^{\circ }, x^{\circ }, \left ( x+45 \right ) ^{\circ }, \left ( x+65 \right ) ^{\circ }$ . Evaluate .

Answer

The sum of the degree measures of the angles of a quadrilateral is 360, so we can set up and solve for in the equation:

$x + x + \left ( x+45 \right ) + \left ( x+65 \right ) = 360$

$4x\div 4 = 250 \div 4$

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Question

The four angles of a quadrilateral have the following value: 79 degrees, 100 degrees, 50 degrees, and degrees. What is the value of ?

Answer

Given that there are 360 degrees when all the angles of a quadrilateral are added toghether, this problem can be solved with the following equation:

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Question

In a quadrilateral, the angles have the following values:

$94^{\circ},27^{\circ},81^{\circ},\textup{and }(4x)^{\circ}$ .

What is the value of ?

Answer

Given that there are 360 degrees when the angles of a quadrilateral are added together, it follows that:

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Question

Cassie is making a kite for her little brother. She has two plastic tubes to use as the skeleton, measuring inches and inches. If these two tubes represent the diagnals of the kite, how many square inches of paper will she need to make the kite?

Answer

To find the area of a kite, use the formula $\frac{1}{2}ab$ , where represents one diagnal and represents the other.

Since Cassie has one tube measuring inches, we can substitute for . We can also substitute the other tube that measures inches in for .

$\frac{1}{2}(7)(12)=42\ in^{2}$

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Question

Two diagonals of a kite have the lengths of $10\ in$ and $6\ in$ . Give the area of the kite.

Answer

The area of a kite is half the product of the diagonals, i.e.

$Area=\frac{d_{1}d_{2}}{2}$ ,

where $d_{1}$ and $d_{2}$ are the lengths of the diagonals.

$Area=\frac{d_{1}d_{2}}{2}=\frac{6\times 10}{2}=30\ in^2$

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Question

In the following kite, $a=10\ cm$ , $b=15\ cm$ and $\angle C=120^{\circ}$ . Give the area of the kite. Figure not drawn to scale.

Answer

When you know the length of two unequal sides of a kite and their included angle, the following formula can be used to find the area of a kite:

$Area=ab\times sinC$ ,

where __are the lengths of two unequal sides, is the angle between them and is the sine function.

$Area=ab\times sinC=10\times 15\times sin120^{\circ}=10\times 15\times \frac{\sqrt{3}}{2}\Rightarrow Area=75\sqrt{3}\ cm^2$

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Question

Find the area of a kite with one diagonal having length 18in and the other diagonal having a length that is half the first diagonal.

Answer

To find the area of a kite, we will use the following formula:

$A = \frac{pq}{2}$

where p and q are the lengths of the diagonals of the kite.

Now, we know the length of one diagonal is 18in. We also know the other diagonal is half of the first diagonal. Therefore, the second diagonal has a length of 9in.

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

$A = \frac{18\text{in} \cdot 9\text{in}}{2}$

$A = \frac{162\text{in}^2}{2}$

$A = 81\text{in}^2$

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Question

Give the area of the above parallelogram if .

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 .

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?

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?

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

A rectangle and a square have the same perimeter. The area of the square is square centimeters; the length of the rectangle is centimeters. Give the width of the rectangle in centimeters.

Answer

The sidelength of a square with area square centimeters is $\sqrt{225} = 15$ centimeters; its perimeter, as well as that of the rectangle, is therefore $15 \times 4 = 60$ centimeters.

Using the formula for the perimeter of a rectangle, substitute and solve for as follows:

$2 \cdot 21 + 2W = 60$

$2W \div 2 = 18\div 2$

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Question

A rectangle on the coordinate plane has its vertices at the points .

What percent of the rectangle is located in Quadrant I?

Answer

The total area of the rectangle is

$\left [ 3- (-6) \right ] \times\left [ 4- (-2) \right ] = 9 \times 6 = 54$ .

The area of the portion of the rectangle in Quadrant I is

$\left ( 3- 0 \right ) \times\left ( 4- 0 \right) = 3 \times 4 = 12$ .

Therefore, the portion of the rectangle in Quadrant I is

$\frac{12}{54} = \frac{2}{9} = \frac{2}{9} \times 100 % = 22 \frac{2}{9} %$ .

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Question

In a rectangle, the width is while the length is . If , what is the area of the rectangle?

Answer

In a rectangle in which the width is x while the length is 4x, the first step is to solve for x. If , the value of x can be found by dividing each side of this equation by 3.

Doing so gives us the information that x is equal to 3.

Thus, the area is equal to:

$Area=x\cdot4x$

$Area=3\cdot4\cdot3$

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Question

Your geometry book has a rectangular front cover which is 12 inches by 8 inches.

What is the area of your book cover?

Answer

Your geometry book has a rectangular front cover which is 12 inches by 8 inches.

What is the area of your book cover?

To find the area of a rectangle, use the following formula:

$A_{rectangle}=l*w$

Plug in our knowns and solve:

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Question

Find the area of a rectangle with a width of 5in and a length that is three times the width.

Answer

To find the area of a rectangle, we will use the following formula:

$\text{area of rectangle} = l \cdot w$

where l is the length and w is the width of the rectangle.

Now, we know the width of the rectangle is 5in. We also know the length is three times the width. Therefore, the length is 15in.

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

$\text{area of rectangle} = 15\text{in} \cdot 5\text{in}$

$\text{area of rectangle} = 75\text{in}^2$

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