Mensuration

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For the Classic ACT exam:
The ACT Mathematics test is a timed exam...60 questions in 60 minutes
This implies that you have to solve each question in one minute.
Each of the first 20 questions (less challenging) will typically take less than a minute a solve.
Each of the next 20 questions (medium challenging) may take about a minute to solve.
Each of the last 20 questions (more challenging) may take more than a minute to solve.
The goal is to maximize your time.
You use the time saved on the questions you solve in less than a minute to solve questions that will take more than a minute.
So, you should try to solve each question correctly and timely.
So, it is not just solving a question correctly, but solving it correctly on time.
Please ensure you attempt all ACT questions.
There is no negative penalty for a wrong answer.
Also: please note that unless specified otherwise, geometric figures are drawn to scale. So, you can figure out the correct answer by eliminating the incorrect options.
Other suggestions are listed in the solutions/explanations as applicable.

These are the solutions to the ACT past questions on the topics in Mensuration.
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Formulas

Right Triangle

$ perpendicular\:\:height = height \\[3ex] Area = \dfrac{1}{2} * base * height \\[5ex] height = \dfrac{2 * Area}{base} \\[5ex] base = \dfrac{2 * Area}{height} \\[5ex] hypotenuse^2 = height^2 + base^2...Pythagorean\:\:Theorem \\[3ex] hypotenuse = \sqrt{height^2 + base^2} \\[3ex] height = \sqrt{hypotenuse^2 - base^2} \\[3ex] base = \sqrt{hypotenuse^2 - height^2} \\[3ex] Perimeter = hypotenuse + height + base \\[3ex] Area = \dfrac{1}{2} * height * base * \sin (hypotenuseAngle) \\[5ex] Area = \dfrac{1}{2} * height * hypotenuse * \sin (baseAngle) \\[5ex] Area = \dfrac{1}{2} * base * hypotenuse * \sin (heightAngle) \\[5ex] Semiperimeter = \dfrac{height + base + hypotenuse}{2} \\[5ex] Semiperimeter - height = firstdifference \\[3ex] Semiperimeter - base = seconddifference \\[3ex] Semiperimeter - hypotenuse = thirddifference \\[3ex] Area = \sqrt{Semiperimeter * firstdifference * seconddifference * thirddifference}...Hero's\:\:Formula\:\:or\:\:Heron's\:\:Formula \\[5ex] hypotenuse = {Perimeter^2 - 4 * Area}{2 * Perimeter} \\[5ex] base = \dfrac{(Perimeter - hypotenuse) \pm Math.sqrt((hypotenuse - Perimeter)^2 - 8 * Area)}{2} \\[5ex] height = \dfrac{2 * Area}{base} $

Triangle

$ Perimeter = firstside + secondside + thirdside \\[5ex] Area = \dfrac{1}{2} * firstside * secondside * \sin (thirdAngle) \\[5ex] Area = \dfrac{1}{2} * firstside * thirdside * \sin (secondAngle) \\[5ex] Area = \dfrac{1}{2} * secondside * thirdside * \sin (firstAngle) \\[5ex] Semiperimeter = \dfrac{firstside + secondside + thirdside}{2} \\[5ex] Semiperimeter - firstside = firstdifference \\[3ex] Semiperimeter - secondside = seconddifference \\[3ex] Semiperimeter - thirdside = thirddifference \\[3ex] Area = \sqrt{Semiperimeter * firstdifference * seconddifference * thirddifference}...Hero's\:\:Formula\:\:or\:\:Heron's\:\:Formula \\[5ex] \underline{Cosine\:\:Law} \\[3ex] firstside^2 = secondside^2 + thirdside^2 - 2 * secondside * thirdside * \cos (firstAngle) \\[3ex] secondside^2 = firstside^2 + thirdside^2 - 2 * firstside * thirdside * \cos (secondAngle) \\[3ex] thirdside^2 = firstside^2 + secondside^2 - 2 * firstside * secondside * \cos (thirdAngle) \\[5ex] firstAngle = \cos^{-1} \left(\dfrac{secondside^2 + thirdside^2 - firstside^2}{2 * secondside * thirdside}\right) \\[5ex] secondAngle = \cos^{-1} \left(\dfrac{firstside^2 + thirdside^2 - secondside^2}{2 * firstside * thirdside}\right) \\[5ex] thirdAngle = \cos^{-1} \left(\dfrac{firstside^2 + secondside^2 - thirdside^2}{2 * firstside * secondside}\right) \\[7ex] \underline{\text{Area of a Triangle given the vertices}} \\[3ex] \text{Let the vertices be:} \\[3ex] Vertex\;1:\;\;(x_1, y_1) \\[4ex] Vertex\;2:\;\;(x_2, y_2) \\[4ex] Vertex\;3:\;\;(x_3, y_3) \\[4ex] Area = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| $

Insribed Triangle (Circumscribed Circle)

$ Circumradius = \dfrac{firstside \cdot secondside \cdot thirdside}{4 \cdot Area} $

Square

$ side = length = width = height \\[3ex] Area = side^2 \\[3ex] side = \sqrt{Area} \\[3ex] Perimeter = 4 * side \\[3ex] side = \dfrac{Perimeter}{4} \\[5ex] diagonal = side * \sqrt{2} \\[3ex] side = \dfrac{diagonal * \sqrt{2}}{2} \\[5ex] Area = \dfrac{Perimeter^2}{16} \\[5ex] Perimeter = 4 * \sqrt{Area} \\[3ex] Area = \dfrac{diagonal^2}{2} \\[5ex] diagonal = \sqrt{2 * Area} \\[3ex] Perimeter = 2 * diagonal * \sqrt{2} \\[3ex] diagonal = \dfrac{Perimeter * \sqrt{2}}{4} $

Rectangle

$ Area = length * width \\[3ex] length = \dfrac{Area}{width} \\[5ex] width = \dfrac{Area}{length} \\[5ex] Area = \dfrac{(length * Perimeter) - (2 * length^2)}{2} \\[5ex] Area = \dfrac{(width * Perimeter) - (2 * width^2)}{2} \\[5ex] Perimeter = 2(length + width) \\[3ex] length = \dfrac{Perimeter - 2 * width}{2} \\[5ex] width = \dfrac{Perimeter - 2 * length}{2} \\[5ex] Perimeter = \dfrac{2(length^2 + Area)}{length} \\[5ex] Perimeter = \dfrac{2(width^2 + Area)}{width} \\[5ex] diagonal = \sqrt{length^2 + width^2} \\[4ex] length = \sqrt{diagonal^2 - width^2} \\[4ex] width = \sqrt{diagonal^2 - length^2} \\[4ex] diagonal = \dfrac{\sqrt{length^4 + Area^2}}{length} \\[5ex] diagonal = \dfrac{\sqrt{width^4 + Area^2}}{width} \\[5ex] diagonal = \dfrac{\sqrt{(Perimeter^2) + (5 * length^2) - (4 * Perimeter * length)}}{2} \\[5ex] diagonal = \dfrac{\sqrt{(Perimeter^2) + (5 * width^2) - (4 * Perimeter * width)}}{2} $

Circle

$ Area = A \\[3ex] Circumference = C \\[3ex] Radius = r \\[3ex] Diameter = d \\[3ex] d = 2r \\[3ex] r = \dfrac{d}{2} \\[5ex] A = \pi r^2 \\[3ex] A = \dfrac{\pi d^2}{4} \\[5ex] C = 2\pi r \\[3ex] C = \pi d \\[3ex] r = \dfrac{\sqrt{A\pi}}{\pi} \\[5ex] r = \dfrac{C}{2\pi} \\[5ex] d = \dfrac{2\sqrt{A\pi}}{\pi} \\[5ex] r = \dfrac{C}{\pi} \\[5ex] A = \dfrac{C^2}{4\pi} \\[5ex] C = 2\sqrt{A\pi} $

Cube

6 square faces
12 edges

$ edge = side = length = width = height \\[3ex] Surface\:\:Area = 6 * edge^2 \\[3ex] edge = \sqrt{\dfrac{Surface\:\:Area}{6}} \\[5ex] Volume = edge^3 \\[3ex] edge = \sqrt[3]{Volume} \\[3ex] Volume = \dfrac{edge * Surface\:\: Area}{6} \\[5ex] edge = \dfrac{6 * Volume}{Surface\:\:Area} \\[5ex] Surface\:\:Area = \dfrac{6 * Volume}{edge} \\[5ex] Volume = \dfrac{Surface\:\:Area * \sqrt{6 * Surface\:\:Area}}{36} \\[5ex] edge = \dfrac{diagonal * \sqrt{3}}{3} \\[5ex] diagonal = \sqrt{3} * edge \\[3ex] Surface\:\:Area = 2 * diagonal^2 \\[3ex] diagonal = \dfrac{\sqrt{2 * Surface\:\:Area}}{2} \\[5ex] Volume = \dfrac{diagonal^3 * \sqrt{3}}{9} \\[5ex] diagonal = \sqrt{3} * \sqrt[3]{Volume} $

Cuboid (Right Rectangular Prism)

$ Volume = Length \cdot Width \cdot Height \\[3ex] $

Right Cone

Curved Surface Area = Lateral Surface Area
Height = Perpendicular Height

$ Volume\:\:of\:\:Cone = \dfrac{1}{3} * Volume\:\:of\:\:Cylinder \\[5ex] Lateral\:\:Surface\:\:Area = LSA \\[3ex] Base\:\:Area = BA \\[3ex] Total\:\:Surface\:\:Area = TSA \\[3ex] Volume = V \\[3ex] Diameter = d \\[3ex] Radius = r \\[3ex] Height = h \\[3ex] Slant Height = l \\[3ex] r = \dfrac{d}{2} \\[5ex] d = 2r \\[3ex] l = \sqrt{h^2 + r^2} \\[3ex] l = \dfrac{\sqrt{4h^2 + d^2}}{2} \\[5ex] h = \sqrt{l^2 - r^2} \\[3ex] h = \dfrac{\sqrt{4l^2 - d^2}}{2} \\[5ex] r = \sqrt{l^2 - h^2} \\[3ex] d = 2 * \sqrt{l^2 - h^2} \\[3ex] BA = \pi r^2 \\[3ex] r = \dfrac{\sqrt{BA * \pi}}{\pi} \\[5ex] BA = \dfrac{\pi d^2}{4} \\[5ex] d = \dfrac{2\sqrt{BA * \pi}}{\pi} \\[5ex] LSA = \pi rl \\[3ex] LSA = \dfrac{\pi dl}{2} \\[5ex] l = \dfrac{LSA}{\pi r} \\[5ex] LSA = \pi r\sqrt{h^2 + r^2} \\[3ex] h = \dfrac{\sqrt{LSA^2 - \pi^2 r^4}}{\pi r} \\[5ex] TSA = BA + LSA \\[3ex] TSA = \pi r(r + l) \\[3ex] l = \dfrac{TSA - \pi r^2}{\pi r} \\[5ex] TSA = \dfrac{\pi d(d + 2l)}{4} \\[5ex] l = \dfrac{4 * TSA - \pi d^2}{2\pi d} \\[5ex] r = \dfrac{-\pi l \pm \sqrt{\pi^2 l^2 + 4\pi * TSA}}{2\pi} \\[5ex] TSA = \pi r(r + \sqrt{h^2 + r^2}) \\[3ex] h = \dfrac{\sqrt{TSA(TSA - 2\pi r^2)}}{\pi r} \\[5ex] V = \dfrac{BA * h}{3} \\[5ex] V = \dfrac{\pi r^2h}{3} \\[5ex] V = \dfrac{\pi hd^2}{12} \\[5ex] V = \dfrac{\pi h(l^2 - h^2)}{3} \\[5ex] h = \dfrac{3V}{\pi r^2} \\[5ex] r = \dfrac{\sqrt{3V\pi h}}{\pi h} $

Right Cylinder

Curved Surface Area = Lateral Surface Area
Height = Perpendicular Height

$ Volume\:\:of\:\:Cylinder = 3 * Volume\:\:of\:\:Cone \\[3ex] Lateral\:\:Surface\:\:Area = LSA \\[3ex] Base\:\:Area = BA \\[3ex] Total\:\:Surface\:\:Area = TSA \\[3ex] Volume = V \\[3ex] Diameter = d \\[3ex] Radius = r \\[3ex] Height = h \\[3ex] r = \dfrac{d}{2} \\[5ex] d = 2r \\[3ex] LSA = 2\pi rh \\[3ex] r = \dfrac{LSA}{2\pi h} \\[5ex] h = \dfrac{LSA}{2\pi r} \\[5ex] LSA = \pi dh \\[3ex] h = \dfrac{LSA}{\pi d} \\[5ex] d = \dfrac{LSA}{\pi h} \\[5ex] BA = \pi r^2 \\[3ex] r = \dfrac{\sqrt{\pi BA}}{\pi} \\[5ex] r = \dfrac{1}{\pi} * \sqrt{\dfrac{\pi(TSA - 2 * LSA)}{2}} \\[5ex] BA = \dfrac{\pi d^2}{4} \\[5ex] d = \dfrac{2\sqrt{\pi BA}}{\pi} \\[5ex] d = \dfrac{\sqrt{2\pi (TSA - LSA)}}{\pi} \\[5ex] TSA = 2\pi r(r + h) \\[3ex] h = \dfrac{TSA - 2\pi r^2}{2\pi r} \\[5ex] r = \dfrac{-\pi h \pm \sqrt{\pi(\pi h^2 + 2 * TSA)}}{2\pi} \\[5ex] TSA = 2BA + LSA \\[3ex] BA = \dfrac{TSA - LSA}{2} \\[5ex] LSA = TSA - 2BA \\[3ex] TSA = \pi d\left(\dfrac{d + 2h}{2}\right) \\[5ex] h = \dfrac{2 * TSA - \pi d^2}{2\pi d} \\[5ex] d = \dfrac{-\pi h \pm \sqrt{\pi(h^2 + 2 * TSA)}}{\pi} \\[5ex] h = \dfrac{LSA * \sqrt{\pi * BA}}{\pi * BA} \\[5ex] h = \dfrac{LSA}{\sqrt{2\pi(TSA - LSA)}} \\[5ex] BA = \dfrac{LSA^2}{\pi h^2} \\[5ex] BA = \dfrac{(4 * TSA + \pi h^2) \pm h\sqrt{\pi(\pi h^2 - 8 * TSA)}}{8} \\[5ex] LSA = h\sqrt{BA * \pi} \\[3ex] LSA = \dfrac{-\pi h^2 \pm h\sqrt{\pi(\pi h^2 + 8 * TSA)}}{4} \\[5ex] TSA = 2 * BA \pm h\sqrt{\pi * BA} \\[3ex] TSA = \dfrac{LSA(2 * LSA + \pi h^2)}{\pi h^2} \\[5ex] V = \pi r^2h \\[3ex] r = \dfrac{2V}{LSA} \\[5ex] d = \dfrac{4V}{LSA} \\[5ex] r = \dfrac{\sqrt{Vh\pi}}{h\pi} \\[5ex] V = BA * h \\[3ex] BA = \dfrac{V}{h} \\[5ex] h = \dfrac{V}{BA} \\[5ex] h = \dfrac{V}{\pi r^2} \\[5ex] h = \dfrac{4V}{\pi d^2} \\[5ex] V = \dfrac{\pi d^2h}{4} \\[5ex] d = \dfrac{\sqrt{Vh\pi}}2{h\pi} \\[5ex] V = \dfrac{LSA^2}{h\pi} \\[5ex] LSA = \sqrt{Vh\pi} \\[3ex] h = \dfrac{LSA^2}{4V\pi} \\[5ex] V = \dfrac{(h^3\pi + 4 * TSA * h) \pm h^2\sqrt{\pi(h^2\pi + 8 * TSA)}}{8} \\[5ex] TSA = \dfrac{2V + h\sqrt{Vh\pi}}{h} \\[5ex] TSA = \dfrac{2V + 2\pi rh^2}{h} \\[5ex] r = \dfrac{TSA * h - 2V}{2\pi h^2} \\[5ex] d = \dfrac{TSA * h - 2V}{\pi h^2} \\[5ex] h = \dfrac{TSA \pm \sqrt{TSA^2 - 16\pi rV}}{4\pi r} $

Trapezoid

Trapezoid's Midpoint Segment Theorem states that the line segment connecting the nonparallel sides of a trapezoid is parallel to the bases, and it's length is the average of the lengths of the bases.

$ Midline = \dfrac{short\;\;base + long\;base}{2} $

(1.) Yesterday, there was no snow on the ground at 8:00 a.m. when snow began to fall.
Snow fell from 8:00 a.m. to 8:00 p.m. at a constant rate of $\dfrac{3}{4}$ inch per hour.
Kate built a snowman made of 2 spherical snowballs, with the smaller placed on top of the larger.
When she built the snowman, the diameter of the larger snowball was 3 times the diameter of the smaller snowball.
Today, the day after she built the snowman, there is a 50% chance of rain.
If it rains today, then there is a 60% chance that the snowman will melt.
If it does not rain today, then there is a 10% chance the snowman will melt.

When Kate built the snowman, the circumference of the larger snowball was 36 π inches.
When Kate built the snowman, what was the diameter, in inches, of the smaller snowball?

$ F.\;\; 4 \\[3ex] G.\;\; 6 \\[3ex] H.\;\; 12 \\[3ex] J.\;\; 54 \\[3ex] K.\;\; 108 \\[3ex] $

Let the:
diameter of the larger snowball = D
circumference of the larger snowball = C
diameter of the smaller snowball = d

$ \underline{\text{Larger Snowball}} \\[3ex] C = \pi D ...formula \\[3ex] C = 36\pi ...given \\[3ex] C = C \implies \\[3ex] \pi D = 36\pi \\[3ex] D = 36\;inches \\[3ex] \underline{\text{Smaller Snowball}} \\[3ex] 3d = D \\[3ex] d = \dfrac{D}{3} \\[5ex] d = \dfrac{36}{3} \\[5ex] d = 12\;inches $

(2.) A certain rectangle has a width of 6km and a length that is 4km more than 2 times the width.
What is the area, in square kilometers, of this rectangle?

$ F.\;\; 16 \\[3ex] G.\;\; 24 \\[3ex] H.\;\; 32 \\[3ex] J.\;\; 60 \\[3ex] K.\;\; 96 \\[3ex] $

$ width = 6\;km \\[3ex] length = 4 + 2\cdot width \\[3ex] = 4 + 2(6) \\[3ex] = 4 + 12 \\[3ex] = 16\;km \\[3ex] Area = length \cdot width \\[3ex] = 16 \cdot 6 \\[3ex] = 96\;km^2 $

(3.) The table below gives the lengths, widths, and heights, in feet, of 3 right rectangular prisms.
The difference between the volume of Prism A and the volume of Prism B is how many times the volume of Prism C?

Prism	Length	Width	Height
A	4	4	3
B	3	3	4
C	4	3	1

$ F.\;\; 0 \\[3ex] G.\;\; 1 \\[3ex] H.\;\; 5 \\[3ex] J.\;\; 8 \\[3ex] K.\;\; 12 \\[3ex] $

Prism	Length	Width	Height	Volume = Length × Width × Height
A	4	4	3	48
B	3	3	4	36
C	4	3	1	12

$ 48 - 36 = what \cdot 12 \\[3ex] 12 = what \cdot 12 \\[3ex] what = 1 $

(4.) A circle with radius 4 centimeters is inscribed in a square, as shown below.
Which of the following represents the area, in square centimeters, of the shaded region?

$ A.\;\; 32 - 8\pi \\[3ex] B.\;\; 32 - 16\pi \\[3ex] C.\;\; 32 + 16\pi \\[3ex] D.\;\; 64 - 16\pi \\[3ex] E.\;\; 64 + 16\pi \\[3ex] $

Area of the shaded region = Area of the Square − Area of the Circle

$ \underline{Square} \\[3ex] apothem,\;a = 4\;cm \\[3ex] Length,\;L = 2a = 2(4) = 8\;cm \\[3ex] Area,\;A = L^2 \\[3ex] A = 8^2 \\[3ex] A = 64\;cm^2 \\[5ex] \underline{Circle} \\[3ex] radius,\;r = 4\;cm \\[3ex] Area,\;A = \pi r^2 \\[4ex] A = \pi(4)^2 \\[4ex] A = 16\pi\;cm^2 \\[5ex] \implies \\[3ex] \text{Area of the shaded region} = 64 - 16\pi $

(5.) Josie has decided to make a new lampshade for her bedroom lamp.
She will order materials from a website that offers 4 different print designs and 4 different color schemes that can be used with each design.
The top and bottom edges of Josie’s current lampshade are parallel circles with diameters of length 6 inches and 8 inches, as pictured below.
The centers of the 2 circles are directly above and below one another.

Number 5

(I.) What is the ratio of the area of the circle modeled by the top edge of the lampshade to the area of the circle modeled by the bottom edge of the lampshade?

$ A.\;\; 3:7 \\[3ex] B.\;\; 3:4 \\[3ex] C.\;\; 9:16 \\[3ex] D.\;\; 4:3 \\[3ex] E.\;\; 16:9 \\[5ex] $ (II.) Which of the following 2-dimensional shapes, when rotated about its vertical symmetry line, will form the shape of this lampshade?

A. Circle
B. Octagon
C. Pentagon
D. Rectangle
E. Isosceles trapezoid

(I.) Let:
r be the radius of the circle modeled by the top edge of the lampshade
R be teh radius of the circle modeled by the bottom edge of the lampshade

$ radius = \dfrac{diameter}{2} \\[5ex] Area = \pi radius^2 \\[4ex] r = \dfrac{6}{2} = 3\;in \\[5ex] R = \dfrac{8}{2} = 4\;in \\[5ex] \text{Ratio of the areas} \\[3ex] = \dfrac{\pi (3)^2}{\pi (4)^2} \\[5ex] = \dfrac{9}{16} \\[5ex] = 9 : 16 \\[3ex] $ (II.) The shape of Josie's lampshade is a truncated cone.
When rotated about its vertical symmetry line, the 2-dimensional shape that will form the shape of a truncated cone is the Isosceles trapezoid.

(6.) The base of a right square pyramid has a side length of 20 ft.
The pyramid’s slant height is 15 ft.
What is the total surface area, in square feet, of the pyramid?

$ A.\;\; 550 \\[3ex] B.\;\; 600 \\[3ex] C.\;\; 1,000 \\[3ex] D.\;\; 1,600 \\[3ex] E.\;\; 2,000 \\[3ex] $

The base of a right square pyramid is a square
The base has a side length of 20 ft.
This implies that:
The square has a side length of 20 ft.
This implies that the base area is the area of the square
Base Area, BA = Area of the square = side² = (20)² = 400 square feet

The slant height of the pyramid = 15 ft.
Lateral Surface Area, LSA = 2 × base × slant height
LSA = 2 × 20 × 15 = 600 square feet

Total Surface Area, TSA = BA + LSA
TSA = 400 + 600
TSA = 1000 square feet.

(7.) Tisha is building a wooden drying rack.
The top of the rack is in the shape of a trapezoid.
The trapezoid's bases are 24 inches long and 38 inches long, respectively.
Tisha will add a brace that joins the midpoints of the trapezoid's 2 nonparallel sides to support her drying rack, as shown in the figure below.
How many inches long will the brace be?

Number 7

$ A.\;\; 26 \\[3ex] B.\;\; 28 \\[3ex] C.\;\; 31 \\[3ex] D.\;\; 32 \\[3ex] E.\;\; 33.5 \\[3ex] $

$ Midline = \dfrac{short\;\;base + long\;base}{2} ...Midpoint\;\;Segment\;\;Theorem \\[5ex] = \dfrac{24 + 38}{2} \\[5ex] = \dfrac{62}{2} \\[5ex] = 31\;inches $

(8.) A rectangular field that measures 300 meters by 175 meters is to be completely fenced along its perimeter.
Given that fencing sells for $2.05 per meter, what will the fencing for the field cost?

$ A.\;\; \$973.75 \\[3ex] B.\;\; \$1,332.50 \\[3ex] C.\;\; \$1,405.00 \\[3ex] D.\;\; \$1,588.75 \\[3ex] E.\;\; \$1,947.50 \\[3ex] $

$ \underline{Rectangular\;\;Field} \\[3ex] Length,\;L = 300\;m \\[3ex] Width,\;W = 175\;m \\[3ex] Perimeter = P \\[3ex] P = 2(L + W) \\[3ex] = 2(300 + 175) \\[3ex] = 2(475) \\[3ex] = 950\;m \\[5ex] \underline{Fencing\;\;Cost} \\[3ex] 950\meters\;\;@\;\;\$2.05\;\;per\;\;meter \\[3ex] = 950(2.05) \\[3ex] = \$1947.50 $

(9.) The length of Rectangle A is equal to the length of Rectangle B.
The width of Rectangle A is 3 inches less than the width of Rectangle B.
If it can be determined, how many inches less is the perimeter of Rectangle A than the perimeter of Rectangle B?

$ F.\;\; 3 \\[3ex] G.\;\; 6 \\[3ex] H.\;\; 9 \\[3ex] J.\;\; 12 \\[3ex] K.\;\;\text{Cannot be determined from the given information} \\[3ex] $

Rectangle A	Rectangle B
$ Length = L_A \\[4ex] Width = W_A \\[4ex] Perimeter = P_A \\[5ex] L_A = L_B \\[4ex] W_A = W_B - 3 \\[4ex] P_A = 2(L_A + W_A) \\[4ex] = 2(L_B + W_B - 3) \\[4ex] = 2L_B + 2W_B - 6 $	$ Length = L_B \\[4ex] Width = W_B \\[4ex] Perimeter = P_B \\[5ex] P_B = 2(L_B + W_B) \\[4ex] = 2L_B + 2W_B $
How many inches less is the perimeter of Rectangle A than the perimeter of Rectangle B? $ P_B - P_A \\[4ex] = (2L_B + 2W_B) - (2L_B + 2W_B - 6) \\[4ex] = 2L_B + 2W_B - 2L_B - 2W_B + 6 \\[4ex] = 6 $

(10.) The figure below shows 3 rectangles where a, b, c, d, e, and f are side lengths measured in meters.
Which of the following expressions must give the area of the shaded region, in square meters?

Number 10

$ A.\;\; ab - cd \\[3ex] B.\;\; ab - ef \\[3ex] C.\;\; cd - ef \\[3ex] D.\;\; cd + ef \\[3ex] E.\;\; ab - cd + ef \\[3ex] $

$ \text{Area of a rectangle} = Length \cdot Width \\[3ex] \text{Area of the 1st (inner) rectangle},A_1 = f \cdot e = ef \\[4ex] \text{Area of the 2nd rectangle}, A_2 = d \cdot c = cd \\[4ex] \text{Area of the shaded region} = A_2 - A_1 \\[4ex] = cd - ef $

(11.) The vertices of $\triangle PQR$ are given in the standard (x, y) coordinate plane below.
What is the area, in square coordinate units, of $\triangle PQR$?

Number 11

$ F.\;\; 6 \\[3ex] G.\;\; 10 \\[3ex] H.\;\; 12 \\[3ex] J.\;\; 20 \\[3ex] K.\;\; 48 \\[3ex] $

$ P:\;Vertex\;1:\;\;(x_1, y_1) = (8, 12) \\[4ex] Q:\;Vertex\;2:\;\;(x_2, y_2) = (10, 8) \\[4ex] R:\;Vertex\;3:\;\;(x_3, y_3) = (8, 2) \\[4ex] x_1 = 8 ~~~~~~~~~~~~~~~~ y_1 = 12 \\[4ex] x_2 = 10 ~~~~~~~~~~~~~~~ y_2 = 8 \\[4ex] x_3 = 8 ~~~~~~~~~~~~~~~~~ y_3 = 2 \\[4ex] Area = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \\[4ex] = \dfrac{1}{2}|8(8 - 2) + 10(2 - 12) + 8(12 - 8)| \\[5ex] = \dfrac{1}{2}|8(6) + 10(-10) + 8(4)| \\[5ex] = \dfrac{1}{2}|48 - 100 + 32| \\[5ex] = \dfrac{1}{2}|-20| \\[5ex] = \dfrac{1}{2} \cdot 20 \\[5ex] = 10\;square\;\;units $

(12.) Which of the following equations expresses a relationship between D, the length of the diagonal of the square shown below, and S, the length of a side of the square?

Number 12

$ A.\;\; D = S \\[3ex] B.\;\; D = \sqrt{2}S \\[3ex] C.\;\; D = \sqrt{3}S \\[3ex] D.\;\; D = 2S \\[3ex] E.\;\; D = 3S \\[3ex] $

The four sides of the square are congruent
Each interior angle in a square is a right angle

Number 12

$ \underline{\triangle ABC} \\[3ex] hyp^2 = leg^2 + leg^2 ...\text{Pythagorean Theorem} \\[4ex] D^2 = S^2 + S^2 \\[4ex] D^2 = 2S^2 \\[4ex] D = \sqrt{2S^2} \\[4ex] D = \sqrt{2} \cdot \sqrt{S^2} \\[4ex] D = \sqrt{2}S $

(13.) The dimensions of a hallway floor are shown below.
Each pair of adjacent walls meets at a right angle.
Teresa plans to completely carpet this entire hallway floor.
At a cost of $2 per square foot, how much will carpeting this hallway floor cost?

Number 13

$ A.\;\; \$ 95 \\[3ex] B.\;\; \$ 150 \\[3ex] C.\;\; \$ 190 \\[3ex] D.\;\; \$ 285 \\[3ex] E.\;\; \$ 300 \\[3ex] $

15 feet − 4 feet = 11 feet
Hence we have two regions: Region 1 and Region 2

NUmber 13

$ \text{Area of Region 1 } = 4(10) = 40\;feet^2 \\[4ex] \text{Area of Region 2 } = 11(5) = 55\;feet^2 \\[4ex] \text{Area of Hallway Floor} = \text{Area of Region 1} + \text{Area of Region 2} \\[3ex] = 40 + 55 \\[3ex] = 95\;feet^2 \\[5ex] \text{Cost of Carpeting Entire Hallway Floor } @\;\;\$2\;\; \text{per square feet } \\[3ex] = 95\;square\;\;feet \cdot \dfrac{\$2}{1\;square\;\;foot} \\[5ex] = \$ 190 $

(14.) Each side of square ABCD has a length of 48 m.
A certain rectangle whose area is equal to the area of ABCD has a width of 12 m.
What is the length, in meters, of the rectangle?

$ F.\;\; 36 \\[3ex] G.\;\; 48 \\[3ex] H.\;\; 60 \\[3ex] J.\;\; 144 \\[3ex] K.\;\; 192 \\[3ex] $

$ \underline{Square\;\;ABCD} \\[3ex] Area = Length^2 = 48^2 \;m^2 \\[5ex] \underline{Rectangle} \\[3ex] Area = Length * Width \\[3ex] \text{Let the length of the rectangle } = L_R \\[4ex] Area = L_R \cdot 12 \\[5ex] \text{Area of the Rectangle = Area of the Square} \\[3ex] \implies \\[3ex] L_R \cdot 12 = 48^2 \\[4ex] L_R = \dfrac{48 \cdot 48}{12} \\[5ex] = 4 \cdot 48 \\[3ex] = 192\;m $

(15.) Hoshi is building a circular pond in a square section of her backyard.
The square section measures 20feet on each side.
The pond touches all 4 sides of the square section.
Approximately what is the total area, in square feet, of the region inside this square section but outside the circular pond?

$ A.\;\; 86 \\[3ex] B.\;\; 126 \\[3ex] C.\;\; 314 \\[3ex] D.\;\; 337 \\[3ex] E.\;\; 714 \\[3ex] $

Let us represent this information diagrammatically

Number 15

O is the center of the circle.
The region inside this square section but outside the circular pond is the shaded region.
The question is asking us to calculate the area of the shaded region.

$ \underline{\text{Square Section}} \\[3ex] \text{Length of the square} = 20\;feet \\[3ex] Area = Length^2 \\[4ex] = 20^2 \\[3ex] = 400\;feet^2 \\[5ex] \underline{\text{Circular Pond}} \\[3ex] \text{radius of the circle} = \dfrac{1}{2} \cdot 20 = 10\;feet ...\text{radius of an inscribed circle = half the length of the square} \\[5ex] Area = \pi * radius^2 \\[4ex] = \pi * 10^2 \\[4ex] = 100\pi \\[5ex] \underline{\text{Area of the shaded region}} \\[3ex] = \text{Area of the Square Section} - \text{Area of the Circular Pond} \\[3ex] = 400 - 100\pi \\[3ex] = 85.84073464 \\[3ex] \approx 86\;feet^2 ...\text{to the nearest whole number} $

(16.) A piece of chocolate candy enclosing an almond in the center is made in a rectangular mold that has inside dimensions 4 cm, 3 cm, and 2 cm.
If the volume of the almond is 2 cubic centimeters, what is the maximum volume, in cubic centimeters, of the chocolate in the piece of candy?

$ A.\;\; 7 \\[3ex] B.\;\; 12 \\[3ex] C.\;\; 16 \\[3ex] D.\;\; 22 \\[3ex] E.\;\; 24 \\[3ex] $

The rectangular mold is a rectangular prism.
The chocolate candy is made in the rectangular prism.
The almond is in the center of the chocolate candy.
This implies that the volume of the chocolate candy is the volume of the almond subtracted from the volume of the rectangular prism

$ \text{Volume of the Rectangular Prism} \\[3ex] = Length \cdot Width \cdot Height \\[3ex] = 4 \cdot 3 \cdot 2 \\[3ex] = 24\;cm^3 \\[5ex] \text{Volume of the Almond} = 2\;cm^3 \\[5ex] \text{Volume of the Chocolate Candy} = 24 - 2 \\[3ex] = 22\;cm^3 $

(17.) Rafael wrapped a gift box l inches long, w inches wide, and h inches high.
He tied a decorative rope around it and used an extra 6 inches for a tie at the top, as shown below.
On both the top and the bottom, he formed right angles where the rope crossed.
About how many inches of decorative rope did Rafael use?

Number 17

$ F.\;\; 2l + 2w + 2h \\[3ex] G.\;\; 2l + 2w + 2h + 6 \\[3ex] H.\;\; 2l + 2w + 4h + 6 \\[3ex] J.\;\; 2l + 4w + 2h + 6 \\[3ex] K.\;\; 4l + 2w + 2h + 6 \\[3ex] $

$ \text{Decorative Rope} \\[5ex] \underline{\text{Top and Bottom of Box}} \\[3ex] Length:\;\; 2 \cdot l = 2l\;inches \\[3ex] Width:\;\; 2 \cdot w = 2w\;inches \\[5ex] \underline{\text{4 Sides of the Box}} \\[3ex] Height:\;\; 4 \cdot h = 4h\;inches \\[5ex] \underline{\text{Extra 6 inches for a tie at the top}} \\[3ex] Tie:\;\; 6\;inches \\[5ex] \text{Perimeter of Decorative Rope} = (2l + 2w + 4h + 6)\;inches $

(18.) In a certain triangle that has an area of 12 square inches, the length of one altitude is $\dfrac{2}{3}$ the length of its corresponding base.
What is the length of that base, in inches?

$ A.\;\; 2 \\[3ex] B.\;\; 3 \\[3ex] C.\;\; 6 \\[3ex] D.\;\; 9 \\[3ex] E.\;\; \sqrt{18} \\[3ex] $

Assume the altitude to be the perpendicular height, h
base = b
Area = A

$ A = 12\;inches^2 \\[4ex] b = b \\[3ex] h = \dfrac{2}{3}b \\[5ex] A = \dfrac{1}{2}\cdot b \cdot h \\[5ex] 12 = \dfrac{1}{2} \cdot b \cdot \dfrac{2}{3}b \\[5ex] \dfrac{b^2}{3} = 12 \\[5ex] b^2 = 3(12) \\[3ex] b = +\sqrt{36}...\text{The negative square root is discarded} \\[3ex] \text{This is because the length cannot be negative} \\[3ex] b = 6\;inches $

(19.) Rectangular pyramids A (left) and B (right) are shown below with dimensions given in feet.

Number 19

Let a and b be the volumes of A and B, respectively.
Which of the following equations is true?

$ F.\;\; a = \dfrac{1}{3}b \\[5ex] G.\;\; a = \dfrac{2}{3}b \\[5ex] H.\;\; a = b \\[3ex] J.\;\; a = \dfrac{3}{2}b \\[5ex] K.\;\; a = 3b \\[3ex] $

$ \underline{Rectangular\;\;Pyramid} \\[3ex] Volume = \dfrac{1}{3} \cdot Length \cdot Width \cdot \perp Height \\[5ex] a = \dfrac{1}{3} \cdot \dfrac{12}{x} \cdot x \cdot 12 \\[5ex] a = 48\;cubic\;feet \\[5ex] b = \dfrac{1}{3} \cdot 12 \cdot x \cdot \dfrac{12}{x} \\[5ex] b = 48\;cubic\;feet \\[5ex] a = b $

(20.) A restaurant currently has an outdoor rectangular dining section measuring 30 feet by 40 feet.
The shorter sides will be increased by 10 feet each, resulting in a larger rectangular dining section.
What is the positive difference, in square feet, between the areas of the resulting and current dining sections?

$ F.\;\; 100 \\[3ex] G.\;\; 300 \\[3ex] H.\;\; 400 \\[3ex] J.\;\; 500 \\[3ex] K.\;\; 600 \\[3ex] $

$ \underline{\text{Current Dining Section}} \\[3ex] \text{Rectangular: 30 feet by 40 feet} \\[3ex] Area = 30(40) = 1200\;square\;\;feet \\[5ex] \underline{\text{Resulting Dining Section}} \\[3ex] \text{shorter sides} = 30 feet \\[3ex] \text{increase by 10 feet each} = 30 + 10 = 40\;feet \\[3ex] \implies \\[3ex] \text{Square: 40 feet by 40 feet} \\[3ex] Area = 40(40) = 1600\;square\;\;feet \\[5ex] \text{Positive Difference Between the Areas} \\[3ex] = 1600 - 1200 \\[3ex] = 400\;square\;\;feet. $

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(21.) What is the area, in square millimeters, of a triangle whose side lengths are 5 millimeters, 5 millimeters, and 6 millimeters?

A. 6
B. 12
C. 15
D. 25
E. 30

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \text{1st Approach: Hero's Formula} \\[3ex] side1, a = 5\;mm \\[3ex] side2, b = 5\;mm \\[3ex] side3, c = 6\;mm \\[3ex] semiperimeter, s = \dfrac{a + b + c}{2} \\[5ex] = \dfrac{5 + 5 + 6}{2} \\[5ex] = \dfrac{16}{2} \\[5ex] = 8\;mm \\[5ex] difference1 = semiperimeter - side1 \\[3ex] = s - a \\[3ex] = 8 - 5 \\[3ex] = 3\;mm \\[5ex] difference2 = semiperimeter - side2 \\[3ex] = s - b \\[3ex] = 8 - 5 \\[3ex] = 3\;mm \\[5ex] difference3 = semiperimeter - side3 \\[3ex] = s - c \\[3ex] = 8 - 6 \\[3ex] = 2\;mm \\[5ex] Area, A = \sqrt{s(s - a)(s - b)(s - c)} \\[4ex] = \sqrt{8(3)(3)(2)} \\[4ex] = \sqrt{144} \\[4ex] = 12\;mm^2 \\[5ex] \text{2nd Approach: Formula and Theorem} \\[3ex] $ To do the 2nd approach, let us represent the information diagrammatically.
Draw a perpendicular line from Vertex B to Midpoint D

Number 21

base = b = 6mm
perpendicular height = h

$ 5^2 = h^2 + 3^2 ...\text{Pythagorean Theorem} \\[4ex] 25 = h^2 + 9 \\[4ex] h^2 + 9 = 25 \\[4ex] h^2 = 25 - 9 \\[4ex] h^2 = 16 \\[4ex] h = \sqrt{16} \\[4ex] h = 4\;mm \\[5ex] Area, A = \dfrac{1}{2} \cdot b \cdot h \\[5ex] = \dfrac{1}{2} \cdot 6 \cdot 4 \\[5ex] = 12\;mm^2 $

(22.) A square vegetable garden is built in a rectangular 50-meter-by-40-meter lawn.
The lengths of the sides of the garden are 5 meters.
What area of the lawn, in square meters, is outside of the vegetable garden?

$ F.\;\; 1,575 \\[3ex] G.\;\; 1,750 \\[3ex] H.\;\; 1,800 \\[3ex] J.\;\; 1,975 \\[3ex] K.\;\; 2,475 \\[3ex] $

$ \underline{\text{Outer: Rectangular Lawn}} \\[3ex] Area = Length \cdot Width \\[3ex] = 50(40) \\[3ex] = 2000\;m^2 \\[5ex] \underline{\text{Inner: Square Garden}} \\[3ex] Area = Length^2 \\[4ex] = 5^2 \\[4ex] = 25\;m^2 \\[5ex] \underline{\text{Middle: Shaded Area: Outside Square Garden}} \\[3ex] Area = \text{Area of Rectangular Lawn} - \text{Area of Square Garden} \\[3ex] = 2000 - 25 \\[3ex] = 1975\;m^2 $

(23.) The vertices of $\triangle PQR$ are given in the standard (x, y) coordinate plane below.
What is the area, in square coordinate units, of $\triangle PQR$?

Number 23

$ A.\;\; 18 \\[3ex] B.\;\; 21 \\[3ex] C.\;\; 36 \\[3ex] D.\;\; 40 \\[3ex] E.\;\; 42 \\[3ex] $

$ P:\;Vertex\;1:\;\;(x_1, y_1) = (8, 10) \\[4ex] Q:\;Vertex\;2:\;\;(x_2, y_2) = (8, 4) \\[4ex] R:\;Vertex\;3:\;\;(x_3, y_3) = (1, 2) \\[4ex] x_1 = 8 ~~~~~~~~~~~~~~~~ y_1 = 10 \\[4ex] x_2 = 8 ~~~~~~~~~~~~~~~ y_2 = 4 \\[4ex] x_3 = 1 ~~~~~~~~~~~~~~~~~ y_3 = 2 \\[4ex] Area = \dfrac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| \\[4ex] = \dfrac{1}{2}|8(4 - 2) + 8(2 - 10) + 1(10 - 4)| \\[5ex] = \dfrac{1}{2}|8(2) + 8(-8) + 1(6)| \\[5ex] = \dfrac{1}{2}|16 - 64 + 6| \\[5ex] = \dfrac{1}{2}|-42| \\[5ex] = \dfrac{1}{2} \cdot 42 \\[5ex] = 21\;square\;\;units $

(24.) Rectangle P has an area of 24 square inches.
Rectangle Q has a perimeter of 24 inches.
The ratio of the area of Rectangle P to the area of Rectangle Q is 2:3.
Which of the quantities below can be determined from just the given information?

I. The perimeter of Rectangle P
II. The area of Rectangle Q
III. The ratio of the perimeter of Rectangle P to the perimeter of Rectangle Q

F. I only
G. II only
H. III only
J. I and II only
K. I, II, and III

Which of the quantities below can be determined from just the given information?
This is a tricky question.
The question is asking us to use only the information given.

Given:
Area of Rectangle P
Perimeter of Rectangle Q
Ratio of the area of Rectangle P to the area of Rectangle Q is 2:3.

$ \dfrac{\text{Area of Rectangle P}}{\text{Area of Rectangle Q}} = \dfrac{2}{3} \\[5ex] 2 \cdot \text{Area of Rectangle Q} = 3 \cdot \text{Area of Rectangle P} \\[3ex] 2 \cdot \text{Area of Rectangle Q} = 3 \cdot 24 \\[3ex] $ So, we can determine the Area of Rectangle Q directly from the given information.
We are not given the ratio of the perimeters so we cannot determine the perimeter of Rectangle P directly without doing some algebraic substitutions.
The same applies to the ratio of the perimeters.
Hence, the correct answer is Option G.

(25.) In kite ABCD, shown below, the diagonals intersect at E at a right angle.
The given lengths are in centimeters.
Which of the following values is closest to the perimeter, in centimeters, of ABCD?

Number 25

$ F.\;\; 32 \\[3ex] G.\;\; 40 \\[3ex] H.\;\; 65 \\[3ex] J.\;\; 91 \\[3ex] K.\;\; 182 \\[3ex] $

Considering the fact that you have to solve this question in 1 minute, it is recommended that you figure out the correct answer.
This is because the figure is drawn to scale
1st Approach:
In Figure ABE, the hypotenuse, |AB| is greater than 24
Assume that |AB| is about 25 cm
This also means that |AD| is about 25 cm

In Figure CBE, the hypotenuse, |CB| is greater than 7
Assume that |CB| is about 8 cm
This also means that |CD| is about 8 cm

So, the perimeter is about: 25(2) + 8(2) = 50 + 16 = 66 cm.
The closest answer in the option is 65 cm.
So, the correct answer is Option H.

$ \underline{\text{2nd Approach}} \\[3ex] \underline{\text{Right Triangles and the Pythagorean Theorem}} \\[3ex] \underline{\triangle ABE} \\[3ex] hyp^2 = 24^2 + 7^2 \\[4ex] hyp^2 = 576 + 49 \\[4ex] hyp = \sqrt{625} \\[3ex] |AB| = hyp = 25\;cm \\[5ex] \underline{\triangle ADE} \\[3ex] |AD| = hyp = 25\;cm ...\text{congruent triangle to triangle ABE} \\[5ex] \underline{\triangle BCE} \\[3ex] hyp^2 = 7^2 + 2^2 \\[4ex] hyp^2 = 49 + 4 \\[3ex] hyp = \sqrt{53} \\[3ex] |BC| = hyp = 7.280109889\;cm \\[5ex] \underline{\triangle DCE} \\[3ex] |DC| = hyp = 7.280109889\;cm ...\text{congruent triangle to triangle BCE} \\[5ex] Perimeter = |AB| + |AD| + |BC| + |DC| \\[3ex] = 25 + 25 + 7.280109889 + 7.280109889 \\[3ex] = 64.56021978 \\[3ex] \approx 65\;cm $

(26.) What is the volume, in cubic inches, of a cylinder that has a diameter of 16 inches and a height of 25 inches?
(Note: The volume, V, of a cylinder with radius r and height h is given by $V = \pi r^2h$.)

$ A.\;\; 200\pi \\[3ex] B.\;\; 400\pi \\[3ex] C.\;\; 1.600\pi \\[3ex] D.\;\; 5,000\pi \\[3ex] E.\;\; 6,400\pi \\[3ex] $

$ Let\;\;diameter = d \\[3ex] r = \dfrac{d}{2} = \dfrac{16}{2} = 8\;inches \\[5ex] h = 25\;inches \\[3ex] V = \pi r^2 h \\[3ex] = \pi \cdot 8^2 \cdot 25 \\[3ex] = 1,600\pi $

(27.) The dimensions, in feet, of a standard tennis court are shown in the figure below.
All lines that meet in the figure do so at right angles.
Which of the following values is closest to the area, in square feet, of the 1 service box shown shaded?

Number 27

F. 284
G. 527
H. 567
J. 1,053
K. 1,134

Dimensions of shaded box = 21 feet by 13.5 feet
Length = 21 feet
Width = 13.5 feet

$ Area = Length \cdot Width \\[3ex] = 21 \cdot 13.5 \\[3ex] = 283.5 \\[3ex] \approx 284\;square\;feet $

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