Geometry 2160 Exam 2 prep

Question 18 A: Suppose you are looking down a road and you see a person ahead of you. You hold out your arm and "sight" the person with your thumb, finding that the person appears to be as tall as your thumb is long. Assume that your thumb is 2 inches long and that the distance from your sighting eye to your thumb is 22 inches. If the person is 5 feet 4 inches tall, then how far away are you from the person? a. Make a math drawing showing that the thumb sighting problem involves similar triangles. Explain why the triangles are similar

The triangles are similar because the angle at the eye is the same for both triangles and the thumb and person are parallel, so the corresponding angles at the thumb and the person are congruent (by the Parallel postulate).

Question 1: What's the difference between a square and a cube?

A cube is a solid three-dimensional shape that has square faces. A square is a flat two-dimensional shape.

Question 10b:Your students have an open-top box that has a 2-in.-by-4-in. rectangular base and is 3 in. high. They also have a bunch of cubic-inch blocks and some rulers. What is a more advanced way for your students to determine the volume of the box, and why does this method work?

A more advanced way to determine the volume of the box is to multiply the height (3 in.) by the area of the base (8 in2) to obtain the volume, 24 in3. If you fill the box with 1-unit-by-1-unit cubes, then the box will be made of 3 layers. Each layer has 8 cubes in it—1 cube for each square unit of area in the base. So, the whole box is made of 3 groups (layers), with 8 cubes in each group, and therefore there are 3 × 8 cubes in the box. Each cube has a volume of 1 cubicinch; therefore, the volume of the cube is 3 × 8 cubic units

Question 13: What is the difference between a reflection and reflection symmetry?

A reflection is a transformation of the plane. Reflection symmetry is a property that certain shapes or designs have

Question 15: Explain why isosceles triangles can be decomposed into two congruent right triangles

Assume that AB and AC are the same length. Let M be the midpoint between B and C, so that BM and CM are the same length. Therefore, the triangles AMB and AMC have the same side lengths. According to the SSS congruence criterion, the triangles AMB and AMC must therefore be congruent. Since these triangles are congruent, all their angles must match, too. Therefore, angles AMB and AMC are the same size. Since they add to 180°, they must each be 90°. Therefore, triangles AMB and AMC are congruent right triangles that compose to make triangle ABC.

Question 4: That is the surface area of a closed box (rectangular prism) that is 4 ft wide, 3 ft deep, and 5 ft tall?Draw a small version of a pattern for the box to help you determine its surface area.

By subdividing the pattern into rectangles and using the additivity principle about areas, we see that the surface area of the box, in square feet, i s2 ∙ (3 ∙ 4) + 2 ∙ (5 ∙ 4) + 2 ∙ (5 ∙ 3) =94 ft 2.

Question 11: A cone without a base is made from 3 4 of a circle of radius 20 cm. Determine the volume of the cone. Explain your reasoning.

The arc length of the pattern is 3/4 ∙ 2 ∙ 𝜋 ∙ 20 = 30𝜋 cm. cm. This length is also the circumference of the base of the cone. So, the radius of the base of the cone is found from solving 2𝜋𝑟 = 30𝜋, 𝑟 = 15 cm. The radius of the original pattern is also the slant height of the cone. So, the height of the cone is the unknown leg of a right triangle and can be found by solving 𝑥2 + 152 = 202, 𝑥 =√175 ≈ 13.2 cm. Therefore, the volume of the cone is 13 (𝜋(152)) ∙ 13.2 ≈ 3110 cm3.

Question 18 B: Solve the thumb sighting problem in two different ways. In both cases, explain the logic behind the method you use.

5 ft 4 in = 5 x 12 inches + 4 inches = 64 inches. Internal factor method: the arm length is 11 times the thumb length. So, the distance is 11 times the person's height. Since the person is 64 inches tall, then the person is 11 x 64 = 704 inches (or 58 ft 8 in) away. Scale factor method: Using the thumb and person to find the scale factor, we have k ∙ 2 = 64 so the scale factor is k = 32. Since the distance from my sighting eye to my thumb is 22 inches, then the distance from me to the person is 32 x 22 = 704 inches (or 58 ft 8 in.)

Question 8:A pool will be made in the shape of an isosceles right triangle with legs measuring 4 meters each with a half circle attached to one of the legs, as shown below. (In a right triangle, the legs are the sides that are not the hypotenuse.)If the pool is 2 meters deep, how many liters of water are needed to fill the pool up to its edge?Show all your work. (1 m3 = 1,000 liters

First, we need to determine the volume of the pool. The pool can be thought of as a combination of a half-cylinder and a triangular prism. Therefore, we can use formula for volume:Volume = (Area of the base) ∙ (height) in which the height is 2 meters.Area of the base = Area of half circle + Area of triangle= 12 (𝜋(2)2) + 12 (4 ∙ 4) = 12 (3.14 ∙ 4) + 12 (16) = 6.28 + 8= 14.28 m2Volume = 14.28 x 2 = 28.56 m3 Since 1 m3 = 1,000 liters, we will need 28.56 x 1000 = 28,560 liters of water to fill the pool.

Question 16: Use a triangle congruence criterion and a fact about angles that we established in class to prove that every rhombus is also a parallelogram

Given a rhombus ABCD, we want to prove that its opposite sides are parallel. Draw the diagonal AC. Sides AB, BC, CD, and DA are all the same length by definition of rhombus. By SSS congruence, triangles ABC and CDA are congruent. Therefore, angles a and b (see figure above) are the same size. Viewing the diagonal AC as a transversal to lines AB and DC, angles a and b are alternate interior angles, so the converse of the Parallel Postulate tells us that sides AB and DC are parallel. The same way of reasoning explains why the other two sides of the rhombus are parallel.

Question 10a. Your students have an open-top box that has a 2-in.-by-4-in. rectangular base and is 3 in. high. Theyalso have a bunch of cubic-inch blocks and some rulers.a. What is the most primitive way for your students to determine the volume of the box

The most primitive way for your students to determine the volume of the box is to count how many cubes it takes to fill the box. That number of cubes is the volume of the box in cubic inches.

Question 2: Try to visualize a (right) prism that has hexagonal bases. How many faces, edges, and vertices does such a prism have?

The prism has 8 faces: 6 rectangular faces and 2 hexagonal faces. The prism has 18 edges and 12 vertices.

Question 3: Try to visualize a pyramid that has an octagonal base. How many faces, edges, and vertices does such a pyramid have?

The pyramid has 9 faces: 8 triangular faces and 1 octagonal face. The pyramid has 16 edges and 9 vertices.

Question 19: Suppose a large cylinder has twice the radius and twice the height of a small cylinder. Use a formula for the surface area of a cylinder to explain why the surface area of the large cylinder is 4 times the surface area of the small cylinder. Use the formula for the volume of a cylinder to explain why the volume of the large cylinder is 8 times the volume of the small cylinder

Let's call the radius of the small cylinder r units and the height of the small cylinder h units. Then the volume of the small cylinder is ℎ𝜋𝑟2 cubic units, according to the (height) ∙ (area of the base) volume formula. The larger cylinder has twice the radius and twice the height of the smaller cylinder; therefore, the larger cylinder has radius 2𝑟 units and height 2ℎ units. So, the larger cylinder has volume (2ℎ)𝜋(2𝑟)2 = 8(ℎ𝜋𝑟2) cubic units. Because 8ℎ𝜋𝑟2 is 8 times ℎ𝜋𝑟2, the volume of the large cylinder is 8 times the volume of the small cylinder. The surface area of the small cylinder of radius r and height h is 2𝜋𝑟2 + 2𝜋𝑟ℎ square units. (See answer to problem 7). Using the same formula again, but now with 2𝑟 substituted for 𝑟 and 2ℎ substituted for ℎ, we obtain the surface area of the big cylinder as 2𝜋(2𝑟)2 + 2𝜋(2𝑟)(2ℎ) = 8𝜋𝑟2 + 8𝜋𝑟ℎ = 4(2𝜋𝑟2 + 2𝜋𝑟ℎ) square units. Because 4(2𝜋𝑟2 + 2𝜋𝑟ℎ) is 4 times 2𝜋𝑟2ℎ + 2𝜋𝑟ℎ, the surface area of the large cylinder is 4 times the surface area of the small cylinder.

Question 20: According to one description, King Kong was 19 feet 8 inches tall and weighed 38 tons. Typical male gorillas are about 5 feet 6 inches tall and weigh between 300 and 500 pounds. Assuming that King Kong was proportioned like a typical male gorilla, does his given weight of 38 tons agree with what you would expect? Explain. (1 ton = 2000 pounds)

Since Kong's height is 236 in. and a typical gorilla's height is 66 in., the scale factor from a gorilla to Kong is 𝑘 = 236 ÷ 66 = 3.6. Since weight is directly related to volume, and since we can assume that Kong and a gorilla have the same weight per volume, we can apply the scale factor directly to the weight. Starting with the gorilla's weight and scaling it up: (3.6)3 ∙ 400 = 18,662 pounds = 9.331 tons. The 38-ton figure is much too large.

Question 14: Determine all the symmetries of the two designs below. Consider each design as a whole. For each design, describe all lines of symmetry (if the design has reflection symmetry), and determine whether the design has 2-fold, 3-fold, 4-fold, or other rotation symmetry. Explain your answers.

The design in the left has 6 lines of reflection symmetry and 6-fold rotation (also a 3-fold and a 2-fold rotation). The design in the right has 3 lines of reflection symmetry and 3-fold rotation.

Question 7:Find a formula for the surface area of a cylinder of radius r units and height h units by reasoning about a pattern for a cylinder. (Include the bases of the cylinder.)

The surface of the cylinder consists of two circles of radius r units (the bases) and a tube. Imagine slitting the tube open along its length and unrolling it, as indicated in the figure below.The tube then becomes a rectangle. The height, h, of the tube becomes the length of two sides of the rectangle. The circumference of the tube 2𝜋𝑟, becomes the length of two other sides of the rectangle. Therefore, the rectangle has area 2𝜋𝑟ℎ. According to the moving and additivity principles about area, the surface area of the cylinder is equal to the sum of the areas of the two circles (from the top and bottom) and the area of the rectangle (from the tube), which is (2𝜋𝑟2 + 2𝜋𝑟ℎ) units^2.

Question 5: A right pyramid has a square base with sides 60 m long. The distance from one vertex on the base to the apex of the pyramid (along an edge) is 50 m. Determine the surface area of the pyramid ( not including the base)

The surface of the pyramid (not including the base) consists of 4 isosceles triangles, each of which has a base of 60 m and 2 sides of length 50 m, as shown in the figure below. To find the area of one of these triangles, we must determine the triangle's height. If h stands for the height of the triangle, then, by the Pythagorean theorem, 302 + ℎ2 = 502.Therefore, ℎ2 = 502 − 302 = 2500 − 900 = 1600 and ℎ = 40, so the height of the triangle is 40m. The area of each of the 4 triangles making the surface of the pyramid is therefore 12 (60 ∙ 40) =1200 square meters. So, the surface area of the pyramid (not including the base) is 4 ∙ 1200 =4800 square meters.

Question 6: Suppose you take a rectangular piece of paper, roll it up, and tape two ends together, without overlapping them, to make a cylinder. If the cylinder is 12 inches long and has a diameter of 2 12 in.then what were the length and width of the original rectangular piece of paper?

The two edges of paper that are rolled up make circles of diameter 2.5 in. Therefore, the lengths of these edges are 𝜋 ∙ 2.5 in. (the circumference of the circles), which is about 8 in. The other two edges of the paper run along the length of the cylinder, so they are 12 in. long. Thus, the original piece of paper was about 8 in. by 12 in

Question 9: A cone is to be made from a circle of radius 3 cm (for the base) and a quarter-circle (for the lateral portion). Determine the radius of the quarter-circle.

Think about how the lateral portion of the cone would be attached to the base if you made the cone out of paper. The 14 portion of the circumference of the circle making the cone must wrap completely around the base. Since the base has radius 3 cm, the circumference of the base is 2𝜋 ∙ 3cm. So, if 𝑟 is the radius of the larger circle, 14 of 2𝜋𝑟 must be equal to 2𝜋 ∙ 3. In other words,14 ∙ 2𝜋𝑟 = 2𝜋 ∙ 3 Therefore, 𝑟 = 12 and so the radius of the quarter-circle making the cone is 12 cm.4 m4 m2 m

Question 17: Tyler has designed his own flag on a rectangle that is 3 inches tall and 6 inches wide. Now he wants to draw a larger version of his flag on a rectangle that is 9 inches tall. How wide should Tyler make his larger flag? Solve this problem in two ways: with the scale factor method and with the internal factor method. In each case explain clearly the idea and the reasoning of that method

Using the internal factor method, the width of the flag is 6 3 = 2 times its height. So, if the larger flag will be 9 in. tall, then its width is 2 ∙ 9 = 18 inches. Using the scale factor method, we can use the heights of the flags to find the scale factor from the smaller to the larger flag: 𝑘 ∙ 3 = 9, 𝑘 = 3. Applying the same scale factor to the widths, 3 ∙ 6 = 18 inches.