Name: Brisila Barros | Student Number: JM1500460 | Geometry

Namestudent Numberbrisila Barrosjm1500460geometry Quarter 4 Examansw

Answer the questions below. Make sure to show your work and provide complete geometric explanations for full credit. The questions encompass various topics in geometry, including triangle classifications, trigonometry applications, geometric transformations, surface area calculations, volume computations, properties of circles, and probability distributions. The assignment requires detailed step-by-step solutions, explanations of methods, and proper use of formulas and theorems.

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The problem set begins with a triangle measurement exercise, where the side lengths are 10, 15, and 7. To determine whether the triangle is acute, right, or obtuse, I will utilize the Pythagorean theorem. The first step involves identifying the longest side, which is 15, and assigning this as c, with the other sides 10 and 7 as a and b respectively. Calculating the squares:

• a² = 10² = 100
• b² = 7² = 49
• c² = 15² = 225

Next, sum a² and b²: 100 + 49 = 149. Comparing this sum to c², which is 225, shows that 149

Such classification follows the Pythagorean inequality that states: if c² > a² + b², the triangle is obtuse; if c² = a² + b², the triangle is right; and if c²

The next problem involves a highway making an angle of 6° with the horizontal and maintaining this angle over a horizontal distance of 5 miles. To find the height (rise) of the highway over this distance, I use the tangent function:

tan(6°) = height / 5 miles

Therefore, height = 5 tan(6°) ≈ 5 0.1051 ≈ 0.5255 miles, which rounds to approximately 0.53 miles.

Similarly, a forest ranger observes a fire from a 28-foot tall tower. The angle of depression from the top of the tower to the fire is 11°. Since the angle of depression equals the angle of elevation from the fire to the top of the tower, I set up the tangent function:

tan(11°) = distance / 28 ft

Calculating, distance = 28 tan(11°) ≈ 28 0.1944 ≈ 5.44 feet.

This is the horizontal distance from the base of the tower directly beneath the fire to the fire itself, rounded to the nearest foot, it remains 5 feet.

The question on transformations asks whether a certain transformation appears to be rigid. Rigid motions include rotations and translations that preserve shape and size. If a circle maintains its shape and size but is moved or rotated, the transformation is rigid. For example, if a circle appears unchanged after the transformation (only moved or rotated), then the transformation is rigid. If the circle is expanded or shrunk, then it is not rigid.

To verify Euler’s formula for a net of a cube, we can reconstruct the cube by folding the net and counting the number of vertices (V), edges (E), and faces (F). Euler’s formula states that V - E + F = 2 for convex polyhedra. Constructing or visualizing the folded cube confirms this relationship by counting:

• Vertices: 8
• Edges: 12
• Faces: 6

Checking, 8 - 12 + 6 = 2 confirms Euler’s formula holds.

The problem involving wrought iron for an arch consists of calculating the total length of iron used, given that 11 segments between the two concentric circles are each 1.25 meters long. First, find the circumference of the outer semicircle with diameter 15 m:

Radius = 15 / 2 = 7.5 m

Circumference of full circle = 2 π 7.5 ≈ 47.12 m

Semicircular outer arc length = 47.12 / 2 ≈ 23.56 m

Next, for the inner semicircle, assuming the rings are concentric and separated by 1.25 m segments, the diameter of the inner semicircle would be 15 - 2 1.25 = 12.5 m (assuming symmetrical separation). Radius = 12.5 / 2 = 6.25 m. Its circumference = 2 π * 6.25 ≈ 39.27 m. Then, the semicircular inner arc length = 19.63 m.

However, the problem states the total length of the wrought iron segments as 11 * 1.25 = 13.75 m, which likely refers to the connecting segments, not entire arcs. The total length of the wrought iron used is approximately 58 meters, considering additional segments connecting the arcs.

The surface area of a prism is calculated by summing the areas of its rectangular sides and its two bases. If the prism dimensions are known, such as length, width, and height, area calculations follow standard formulas:

Surface Area = 2 * base area + lateral surface area

e.g., for a rectangular prism: SA = 2(lw) + 2(hw) + 2(lh)

Details depend on the specific dimensions, which are not specified here, so the formula approach is emphasized.

For Janine’s cylindrical vase, with height 50 cm and a combined lateral area and one base about 3000 cm², the method involves using the formulas:

• Lateral area = 2πrh
• Area of one base = πr²

The sum is: lateral area + base area = 3000 cm². Plugging in the known height and rearranging the formula allows solving for r, the radius.

Regarding the balloon with a design width of 5 cm holding 71 cm³ of air, the volume of a sphere relates to its radius by V = (4/3)πr³. The diameter scales with the design width, so the ratio between volumes when the diameter changes from 5 cm to 10 cm is based on the cube of the ratio of radii:

Volume scales as the cube of the diameter since radius doubles, volume increases by a factor of 2³ = 8. Therefore, when the design width is 10 cm, the volume ≈ 71 * 8 = 568 cm³.

In circle-related problems involving tangents, the key is recognizing that tangent lines are perpendicular to radii at the point of tangency. For problem parts asking to prove or find angles, properties such as tangent-secant angle measures and the circle theorem are used. For example, if AB = 7.75, OB = 4, and AO = 8.75, checking whether line AB is tangent involves verifying if AB is perpendicular to radius OB at point B.

When problems present measurements like CD=44, OM=20, and ON=19, to find the radius, we use geometric relationships involving chords, radii, and Pythagoras theorem, especially if right triangles are formed within the circle. Expressing answers in radical form may involve square roots if the calculations produce non-integer results.

For the problem about a tangent at point B, if AB = 7.75 and OB = 4, then OA = 8.75, and verifying if AB is tangent involves confirming if ∠OBA = 90°, which is true if AB ⟂ OB. Since the radii are perpendicular to tangents at the point of contact, the property applies here.

Particularly, findings like x, y, and EF involve applying the Pythagorean theorem, circle theorem properties, and coordinate geometry as appropriate, often with radical expressions for precision.

In the probability distribution of finish times at a track meet, the frequencies are used to compute relative probabilities by dividing each count by the total number of runners. Summing all probabilities should equal 1, validating the distribution.

When discussing transformations, a similarity transformation involving scaling, rotation, or reflection can map one figure onto another. If a figure is scaled uniformly with respect to a point or axis, the transformation is a similarity transformation, which preserves shape ratios but not size unless the scale factor is 1.

For Janine’s painting problem of a cylindrical vase, the lateral surface area is calculated by:

Lateral Area = 2πrh. To find the total area painted, she needs the lateral area alone, which involves knowing the radius and height. With the height 45 cm and diameter 14 cm, radius r = 7 cm. Then, lateral area ≈ 2 π 7 * 45 ≈ 1980 cm², rounded to the nearest whole number.

The tangent line verification involves checking the perpendicularity property between radii and tangents; if a radius drawn to the point of tangency is perpendicular to the tangent line, the line is tangent. For example, when AB=7, OB=3.75, and AO=8, confirming perpendicularity involves calculating angles or slopes as needed.

The equation (x + 5) + (y + 3) = 169 indicates the position and range of a radio source. Solving for y in terms of x, or analyzing the sum's interpretation provides details about the source's coordinates and the signal's spatial extent.

The hourglass problem involves calculating the volume of salt needed to fill the bottom cone. Using the cone volume formula V = (1/3)πr²h, with radius 3 cm and height 6 cm (since the total height is 12 cm and the cones are identical), results in V = (1/3)π 9 6 ≈ 18π ≈ 56.55 cm³ of salt.

Finally, for the three-ball cylinder packaging, the volume of the cylindrical container is V = πr²h. With the diameter 13 cm, radius r = 6.5 cm, and height h=39 cm, the volume is V = π (6.5)² 39 ≈ 5176.6 cm³. The total volume of three spheres with radius 6.5 cm is V = 3 (4/3)πr³ ≈ 3451.04 cm³. The percentage of the container occupied by the balls is (3451.04 / 5176.6) 100 ≈ 66.7%.

References

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• Smith, J. (2018). Basic Geometric Constructions and Transformations. Oxford University Press.
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