Squares, Rectangles, and Parallelograms - GED Math

Card 0 of 505

Use the following rectangle to answer the question:

Find the perimeter.

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

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

Now, given the rectangle

we can see the length is 9cm and the width is 7cm. So, we can substitute. We get

$P = 2(9$\text{cm}$) + 2(7$\text{cm}$)$

$P = 18$\text{cm}$ + 14$\text{cm}$$

$P = 32$\text{cm}$$

What is the perimeter of a square if the side length is ?

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A square has four congruent sides.

Multiply the quantity by four.

The answer is:

In Rhombus , $m \angle R = 60 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

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The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH}$ \cong $\overline{HO}$ , making $\Delta RHO$ isosceles (and possibly equilateral).

Also, consecutive angles of a rhombus are supplementary, as they are with all parallelograms, so

$m \angle H = $180^{\circ}$ - m \angle HRM = $180^{\circ}$ - 60 ^{\circ} = $120^{\circ}$$ .

$\angle H$ , having measure greater than , is obtuse, making $\Delta RHO$ an obtuse triangle. Also, the triangle is not equilateral, since such a triangle must have three $60 ^{\circ}$ angles.

The correct response is that $\Delta RHO$ is obtuse and isosceles, but not equilateral.

Given Quadrilateral , which of these statements would prove that it is a parallelogram?

I) $\angle A \cong \angle B$ and $\angle C \cong \angle D$

II) $\angle A \cong \angle C$ and $\angle B \cong \angle D$

III) $\angle A$ and $\angle B$ are supplementary and $\angle C$ and $\angle D$ are supplementary

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Statement I asserts that two pairs of consecutive angles are congruent. This does not prove that the figure is a parallelogram. For example, an isosceles trapezoid has two pairs of congruent base angles, which are consecutive.

Statement II asserts that both pairs of opposite angles are congruent. By a theorem of geometry, this proves the quadrilateral to be a parallelogram.

Statement III asserts that two pairs of consecutive angles are supplementary. While all parallelograms have this characteristic, trapezoids do as well, so this does not prove the figure a parallelogram.

The correct response is Statement II only.

You are given Parallelogram with . Which of the following statements, along with what you are given, would be enough to prove that Parallelogram is a rectangle?

I) $m \angle A = 90 ^{\circ }$

II) $m \angle B = 90 ^{\circ }$

III) $m \angle C = 90 ^{\circ }$

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A rectangle is defined as a parallelogram with four right, or $90 ^{\circ}$ , angles.

Since opposite angles of a paralellogram are congruent, if one angle measures $90 ^{\circ}$ , so does its opposite. Since consecutive angles of a paralellogram are supplementary - that is, their degree measures total $180 ^{\circ}$ - if one angle measures $90 ^{\circ}$ , then both of the neighboring angles measure $180 ^{\circ} - 90 ^{\circ} = 90 ^{\circ}$ .

In short, in a parallelogram, if one angle is right, all are right and the parallelogram is a rectangle. All three statements assert that one angle is right, so from any one, it follows that the figure is a rectangle. The correct response is Statements I, II, or III.

Note that the sidelengths are irrelevant.

If the rectangle has a width of 5 and a length of 10, what is the area of the rectangle?

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Write the area for a rectangle.

Substitute the given dimensions.

The answer is:

In the figure below, find the measure of the largest angle.

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Recall that in a quadrilateral, the interior angles must add up to .

Thus, we can solve for :

Now, to find the largest angle, plug in the value of into each expression for each angle.

The largest angle is .

Find the area of the trapezoid:

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The area of a trapezoid is calculated using the following equation:

$A=\frac{b_1+b_2}{2}$h=\frac{7+5}{2}$(3)=\frac{12}{2}$(3)=(6)(3)=18$

A rectangle has length 10 inches and width 5 inches. Each dimension is increased by 3 inches. By what percent has the area of the rectangle increased?

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The area of a rectangle is its length times its width.

Its original area is $10 \times 5 = 50$ square inches; its new area is $13 \times 8 = 104$ square inches. The area has increased by

$$\frac{104 - 50}{50 }$ \times 100 = $\frac{54}{50 }$ \times 100 = 108 %$ .

A rectangle has length 10 inches and width 8 inches. Its length is increased by 2 inches, and its width is decreased by 2 inches. By what percent has the area of the rectangle decreased?

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The area of a rectangle is its length times its width.

Its original area is $10 \times 8 = 80$ square inches; its new area is $12 \times 6 = 72$ square inches. The area has decreased by

$$\frac{80 -72}{80}$ \times 100 % = $\frac{8}{80}$ \times 100 % = 10 %$ .

The length of each side of a square is increased by 10%. By what percent has its area increased?

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Let be the original sidelength of the square. Increasing this by 10% is the same as adding 0.1 times that sidelength to the original sidelength. The new sidelength is therefore

The area of a square is the square of its sidelength.

The area of the square was originally ; it is now

$\left ( 1.1s\right $)^{2}$= $1.21s^{2}$$

That is, the area has increased by

.

The above figure depicts the rectangular swimming pool at an apartment. The apartment manager needs to purchase a tarp that will cover this pool completely. However, because of the cutting device the pool store uses, the length and the width of the tarp must each be a multiple of four yards. Also, the tarp must be at least one yard longer and one yard wider than the pool.

What will be the minimum area of the tarp the manager purchases?

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Three feet make a yard, so the length and width of the pool are $\frac{50}{3}$ = 16 $\frac{2}{3}$ yards and $\frac{35}{3}$ = 11 $\frac{2}{3}$ yards, respectively. Since the dimensions of the tarp must exceed those of the pool by at least one yard, the tarp must be at least $17 $\frac{2}{3}$$ yards by $12 $\frac{2}{3}$$ yards; but since both dimensions must be multiples of four yards, we take the next multiple of four for each.

The tarp must be 20 yards by 16 yards, so the area of the tarp is the product of these dimensions, or

$20 \times 16 = 320$ square yards.

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path five feet wide throughout. What is the area of that dirt path?

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The dirt path can be seen as the region between two rectangles. The outer rectangle has length and width 100 feet and 60 feet, respectively, so its area is

$A = 100 \cdot 60 = 6,000$ square feet.

The inner rectangle has length and width $(100 - 2\times 5) = 90$ feet and $(60 - 2\times 5) = 50$ feet, respectively, so its area is

$A = 90 \times 50 = 4,500$ square feet.

The area of the path is the difference of the two:

square feet.

Note: Figure NOT drawn to scale

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in brown). The dirt path is feet wide throughout. Which of the following polynomials gives the area of the garden?

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The length of the garden is feet less than that of the entire lot, or

.

The width of the garden is feet less than that of the entire lot, or

.

The area of the garden is their product:

$= 100 \cdot 60 - 100 \cdot 2y - 2y \cdot 60+2y \cdot 2y$

$= $6,000-200y-120y+4y^{2}$$

Note: Figure NOT drawn to scale.

Refer to the above figure, which shows a rectangular garden (in green) surrounded by a dirt path (in brown). The dirt path is five feet wide throughout. Which of the following polynomials gives the area of the garden in square feet?

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The length of the garden is $2 \times 5 = 10$ feet less than that of the entire lot, or

.

The width of the garden is $2 \times 5 = 10$ less than that of the entire lot, or

.

The area of the garden is their product:

$=2x \cdot x - 2x \cdot 10 - 10 \cdot x + 10 \cdot 10$

$= 2x ^{2} - 20x - 10 x + 100$

$= 2x ^{2} - 30 x + 100$

Note: Figure NOT drawn to scale

Refer to the above diagram.

Half of Rectangle is pink. .

Evaluate .

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Rectangle has length and width , so it has area

$LW = 20 \times 15 = 300$ .

The area of Rectangle is twice that of Rectangle , or 600. Its length is

.

Its width is

.

Plug in what we know and solve for :

$600 =30 \times \left (15 + DF \right )$

Note: Figure NOT drawn to scale

What percent of Rectangle is pink?

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The pink region is Rectangle . Its length and width are

so its area is the product of these, or

$14 \times 8 = 112$ .

The length and width of Rectangle are

so its area is the product of these, or

$20 \times 12 = 240$ .

So we want to know what percent 112 is of 240, which can be answered as follows:

$$\frac{112}{240}$ \times 100 % = 46$\frac{2}{3}$ %$

Note: Figure NOT drawn to scale

What percent of Rectangle is white?

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The pink region is Rectangle . Its length and width are

so its area is the product of these, or

$14 \times 7 = 98$ .

The length and width of Rectangle are

so its area is the product of these, or

$20 \times 10 = 200$ .

The white region is Rectangle cut from Rectangle , so its area is the difference of the two:

.

So we want to know what percent 102 is of 200, which can be answered as follows:

$$\frac{102}{200}$ \times 100 % = 51 %$

Note: Figure NOT drawn to scale.

Calculate the area of Rhombus in the above diagram if:

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The area of a rhombus is half the product of the lengths of diagonals $\overline{AC}$ and $\overline{BD}$ . This is

$$\frac{1}{2}$ \times 10 \times 24 = 120$ .

Find the area of a square with a side of .

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Write the formula for the area of a square.

Substitute the side into the equation.

Simplify the equation.

The answer is: