Quadrilaterals - PSAT Math

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Question

In Rhombus , $m \angle H = 45 ^{\circ }$ . If $\overline{RO }$ is constructed, which of the following is true about $\Delta RHO$ ?

Answer

The figure referenced is below.

The sides of a rhombus are congruent by definition, so $\overline{RH} \cong \overline{HO}$ , making $\Delta RHO$ isosceles. It is not equilateral, since $m \angle H = 45 ^{\circ }$ , and an equilateral triangle must have three $60 ^{\circ}$ angles.

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

$m \angle HRM = 180^{\circ} - m \angle H = 180^{\circ} - 45 ^{\circ} = 135^{\circ}$

A diagonal of a rhombus bisects its angles, so

$m \angle HRO = \frac{1}{2} m \angle HRM = \frac{1}{2} \cdot 135 ^{\circ } = 67\frac{1}{2}^{\circ }$

Similarly, $m \angle HOR = 67\frac{1}{2}^{\circ }$

This makes $\Delta RHO$ acute.

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

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Question

Quadrilateral ABCD contains four ninety-degree angles. Which of the following must be true?

I. Quadrilateral ABCD is a rectangle.

II. Quadrilateral ABCD is a rhombus.

III. Quadrilateral ABCD is a square.

Answer

Quadrilateral ABCD has four ninety-degree angles, which means that it has four right angles because every right angle measures ninety degrees. If a quadrilateral has four right angles, then it must be a rectangle by the definition of a rectangle. This means statement I is definitely true.

However, just because ABCD has four right angles doesn't mean that it is a rhombus. In order for a quadrilateral to be considered a rhombus, it must have four congruent sides. It's possible to have a rectangle whose sides are not all congruent. For example, if a rectangle has a width of 4 meters and a length of 8 meters, then not all of the sides of the rectangle would be congruent. In fact, in a rectangle, only opposite sides need be congruent. This means that ABCD is not necessarily a rhombus, and statement II does not have to be true.

A square is defined as a rhombus with four right angles. In a square, all of the sides must be congruent. In other words, a square is both a rectangle and a rhombus. However, we already established that ABCD doesn't have to be a rhombus. This means that ABCD need not be a square, because, as we said previously, not all of its sides must be congruent. Therefore, statement III isn't necessarily true either.

The only statement that has to be true is statement I.

The answer is I only.

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Question

Note: Figure NOT drawn to scale.

In the above figure,

$\textup{Rect } ABCD \sim \textup{Rect } DBEF$ .

.

Give the perimeter of $\textup{Rect } DBEF$ .

Answer

We can use the Pythagorean Theorem to find :

$DB = \sqrt{(AD)^{2}+(AB)^{2}}$

$DB = \sqrt{5^{2}+10^{2}} = \sqrt{25+100} = \sqrt{125} = \sqrt{25} \cdot \sqrt{5} =5 \sqrt{5}$

The similarity ratio of $\textup{Rect } DBEF$ to $\textup{Rect } ABCD$ is

$\frac{DB}{AB} = \frac{5\sqrt{5}}{10} = \frac{ \sqrt{5}}{2}$

so $\frac{ \sqrt{5}}{2}$ multiplied by the length of a side of $\textup{Rect } ABCD$ is the length of the corresponding side of $\textup{Rect } DBEF$ . We can subsequently multiply the perimeter of the former by $\frac{ \sqrt{5}}{2}$ to get that of the latter:

$\frac{ \sqrt{5}}{2} (10 + 5 + 10 + 5) = \frac{ \sqrt{5}}{2} (30) = 15 \sqrt{5}$

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Question

Note: Figure NOT drawn to scale.

In the above figure,

$\textup{Rect } ABCD \sim \textup{Rect } DBEF$ .

.

Give the area of $\textup{Rect } DBEF$ .

Answer

Corresponding sidelengths of similar polygons are in proportion, so

$\frac{DF}{AD}= \frac{DB}{AB}$

, so

$\frac{DF}{4}= \frac{DB}{8}$

$DF= \frac{DB}{8} \cdot 4$

$DF= \frac{DB}{2}$

We can use the Pythagorean Theorem to find :

$DB = \sqrt{(AD)^{2}+(AB)^{2}}$

$DB = \sqrt{4^{2}+8^{2}} = \sqrt{16+64 } = \sqrt{80} = \sqrt{16} \cdot \sqrt{5} = 4 \sqrt{5}$

$DF= \frac{DB}{2}= \frac{ 4 \sqrt{5}}{2} = 2\sqrt{5}$

The area of $\textup{Rect } DBEF$ is

$DB \cdot DF = 4 \sqrt{5} \cdot 2\sqrt{5} = 8 \cdot 5 = 40$

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Question

Note: Figure NOT drawn to scale.

In the above figure,

$\textup{Rect } ABCD \sim \textup{Rect } DBEF$ .

.

Give the area of Polygon .

Answer

Polygon can be seen as a composite of right $\Delta ABD$ and $\textup{Rect } DBEF$ , so we calculate the individual areas and add them.

The area of $\Delta ABD$ is half the product of legs and :

$\frac{1}{2} \cdot AB \cdot AD = \frac{1}{2} \cdot 12 \cdot 6 = 36$

Now we find the area of $\textup{Rect } DBEF$ . We can do this by first finding using the Pythagorean Theorem:

$DB = \sqrt{(AD)^{2}+(AB)^{2}}$

$DB = \sqrt{6^{2}+12^{2}} = \sqrt{36+144} = \sqrt{180} = \sqrt{36} \cdot \sqrt{5} = 6 \sqrt{5}$

The similarity of $\textup{Rect } DBEF$ to $\textup{Rect } ABCD$ implies

$\frac{DF}{AD}= \frac{DB}{AB}$

so

$\frac{DF}{6}= \frac{6\sqrt{5}}{12}$

$DF= \frac{6 \cdot 6\sqrt{5}}{12} = \frac{3 6\sqrt{5}}{12} = 3\sqrt{5}$

The area of $\textup{Rect } DBEF$ is the product of and :

$DB \cdot DF = 6 \sqrt{5} \cdot 3 \sqrt{5}= 6\cdot 3 \cdot \sqrt{5} \cdot \sqrt{5}= 18 \cdot 5 = 90$

Now add: , the correct response.

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Question

Note: figure NOT drawn to scale.

Refer to the above figure.

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$

, , .

Give the area of $\textup{Rect }ABGF$ .

.

Answer

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$ , so the sides are in proportion - that is,

$\frac{AF}{AB}= \frac{AE}{AC}$

Set

, , and solve for :

$\frac{AF}{12}= \frac{10}{15}$

$AF = \frac{10}{15} \cdot 12 = 8$

$\textup{Rect }ABGF$ has area

$AB \cdot AF = 12 \cdot 8 = 96$

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Question

Note: Figure NOT drawn to scale.

Refer to the above figure.

$\textup{Rect }ABGF \sim \textup{Rect }ACDE$

and .

What percent of $\textup{Rect }ACDE$ has been shaded brown ?

Answer

and , so the similarity ratio of $\textup{Rect }ACDE$ to $\textup{Rect }ABGF$ is 10 to 7. The ratio of the areas is the square of this, or

$10^{2}:7^{2}$

or

Therefore, $\textup{Rect }ABGF$ comprises $\frac{49}{100} = 49 %$ of $\textup{Rect }ABGF$ , and the remainder of the rectangle - the brown region - is 51% of $\textup{Rect }ABGF$ .

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Question

A contractor is going to re-tile a rectangular section of the kitchen floor. If the floor is 6ft x 3ft, and he is going to use square tiles with a side of 9in. How many tiles will be needed?

Answer

We have to be careful of our units. The floor is given in feet and the tile in inches. Since the floor is 6ft x 3ft. we can say it is 72in x 36in, because 12 inches equals 1 foot. If the tiles are 9in x 9in we can fit 8 tiles along the length and 4 tiles along the width. To find the total number of tiles we multiply 8 x 4 = 32. Alternately we could find the area of the floor (72 x 36, and divide by the area of the tile 9 x 9)

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Question

If the area Rectangle A is larger than Rectangle B and the sides of Rectangle A are and , what is the area of Rectangle B?

Answer

$A = \text{length}: \times :\text{width} = 10 \times 6 = 60$

$B =\frac{60}{1.25} = 48$

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Question

The front façade of a building is 100 feet tall and 40 feet wide. There are eight floors in the building, and each floor has four glass windows that are 8 feet wide and 6 feet tall along the front façade. What is the total area of the glass in the façade?

Answer

Glass Area per Window = 8 ft x 6 ft = 48 ft2

Total Number of Windows = Windows per Floor * Number of Floors = 4 * 8 = 32 windows

Total Area of Glass = Area per Window * Total Number of Windows = 48 * 32 = 1536 ft2

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Question

The rectangular bathroom floor in Michael’s house is ten feet by twelve feet. He wants to purchase square tiles that are four inches long and four inches wide to cover the bathroom floor. If each square tile costs $2.50, how much money will Michael need to spend in order to purchase enough tiles to cover his entire bathroom floor?

Answer

The dimensions for the bathroom are given in feet, but the dimensions of the tiles are given in inches; therefore, we need to convert the dimensions of the bathroom from feet to inches, because we can’t compare measurements easily unless we are using the same type of units.

Because there are twelve inches in a foot, we need to multiply the number of feet by twelve to convert from feet to inches.

10 feet = 10 x 12 inches = 120 inches

12 feet = 12 x 12 inches = 144 inches

This means that the bathroom floor is 120 inches by 144 inches. The area of Michael’s bathroom is therefore 120 x 144 in2 = 17280 in2.

Now, we need to find the area of the tiles in square inches and calculate how many tiles it would take to cover 17280 in2.

Each tile is 4 in by 4 in, so the area of each tile is 4 x 4 in2, or 16 in2.

If there are 17280 in2 to be covered, and each tile is 16 in2, then the number of tiles we need is 17280 ÷ 16, which is 1080 tiles.

The question ultimately asks us for the cost of all these tiles; therefore, we need to multiply 1080 by 2.50, which is the price of each tile.

The total cost = 1080 x 2.50 dollars = 2700 dollars.

The answer is $2700.

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Question

Ron has a fixed length of wire that he uses to make a lot. On Monday, he uses the wire to make a rectangular lot. On Tuesday, he uses the same length of wire to form a square-shaped lot. Ron notices that the square lot has slightly more area, and he determines that the difference between the areas of the two lots is sixteen square units. What is the positive difference, in units, between the length and the width of the lot on Monday?

Answer

Let’s say that the rectangular lot on Monday has a length of l and a width of w. The area of a rectangular is the product of the length and the width, so we can write the area of the lot on Monday as lw.

Next, we need to find an expression for the area of the lot on Tuesday. We are told that the lot is in the shape of a square and that it uses the same length of wire. If the length of the wire used is the same on both days, then the perimeter will have to remain the same. In other words, the perimeter of the square will equal the perimeter of the rectangle. The perimeter of a rectangle is given by 2_l_ + 2_w_.

We also know that if s is the length of a side of a square, then the perimeter is 4_s_, because each side of the square is congruent. Let’s write an equation that sets the perimeter of the rectangle and the square equal.

2_l_ + 2_w_ = 4_s_

If we divide both sides by 4 and then simplify the expression, then we can write the length of the square as follows:

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Question

A rectangle has a width of 2_x_. If the length is five more than 150% of the width, what is the area of the rectangle?

Answer

Given that w = 2_x_ and l = 1.5_w_ + 5, a substitution will show that l = 1.5(2_x_) + 5 = 3_x_ + 5.

A = lw = (3_x_ + 5)(2_x_) = 6_x_2 + 10_x_

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Question

Rectangle ABCD is shown in the figure above. Points A and B lie on the graph of y = 64 – _x_2 , and points C and D lie on the graph of y = _x_2 – 36. Segments AD and BC are both parallel to the y-axis. The x-coordinates of points A and B are equal to –k and k, respectively. If the value of k changes from 2 to 4, by how much will the area of rectangle ABCD increase?

Answer

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Question

Two circles of a radius of each sit inside a square with a side length of . If the circles do not overlap, what is the area outside of the circles, but within the square?

Answer

The area of a square = $\dpi{100} \small side^{2}$

The area of a circle is $\dpi{100} \small \pi r^{2}$

Area = Area of Square $\dpi{100} \small -$ 2(Area of Circle) =

$8^{2}-2\cdot \pi (2)^{2}$

$64-8\pi$

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Question

George wants to paint the walls in his room blue. The ceilings are 10 ft tall and a carpet 12 ft by 15 ft covers the floor. One gallon of paint covers 400 $ft^{2}$ and costs $40. One quart of paint covers 100 $ft^{2}$ and costs $15. How much money will he spend on the blue paint?

Answer

The area of the walls is given by $A = 2wh + 2lh = 2(12)(10) + 2(15)(10) = 540 ft^{2}$

One gallon of paint covers 400 $ft^{2}$ and the remaining 140 $ft^{2}$ would be covered by two quarts.

So one gallon and two quarts of paint would cost

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Question

Daisy gets new carpet for her rectangluar room. Her floor is $21\ ft \times 24\ ft$ . The carpet sells for $5 per square yard. How much did she spend on her carpet?

Answer

Since $3\ ft=1\ yd$ the room measurements are 7 yards by 8 yards. The area of the floor is thus 56 square yards. It would cost $5\cdot 56=$280$ .

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Question

The length of a rectangular rug is five more than twice its width. The perimeter of the rug is 40 ft. What is the area of the rug?

Answer

For a rectangle, $P=2w+2l$ and $A=lw$ where $w$ is the width and $l$ is the length.

Let $x=width$ and $2x+5=length$ .

So the equation to solve becomes $40=2x+2(2x+5)$ or $40=6x+10$ .

Thus $x=5\ ft$ and $2x+5=15\ ft$ , so the area is $75\ ft^{2}$ .

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Question

Note: Figure NOT drawn to scale

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

Answer

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 $(100 - 2\times 8) = 84$ feet and $(60 - 2\times 8) = 44$ feet, respectively, so its area is

$A = 84 \times 44= 3,696$ square feet.

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

square feet.

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Question

Note: Figure NOT drawn to scale

What percent of Rectangle is pink?

Answer

The pink region is Rectangle . Its length and width are

so its area is the product of these, or

$A = 13 \times 9 = 117$ .

The length and width of Rectangle are

so its area is the product of these, or

$20 \times 12 = 240$ .

We want to know what percent 117 is of 240, which can be answered as follows:

$\frac{117}{240} \times 100 % = 48\frac{3}{4} %$

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