exercise_1_solution.md

Unit 3 Practice Solutions

Exercise 1 - Geometry

1.1

Solution

# Create three variables a, b and c
# Assign a number to each to represent the lengths of the sides of a triangle
a = 3
b = 4
c = 5

# Calculate the perimeter of the triangle
perimeter = a + b + c

# print using an f-string
print(f'A triangle with sides of {a}, {b}, and {c} has a perimeter of {perimeter}.')

Output

A triangle with sides of 3, 4, and 5 has a perimeter of 12.

---

1.2

Solution

# Create three variables a, b and c
# Assign a number to each to represent the lengths of the sides of a triangle
a = 3
b = 4
c = 5

# Select a, b or c to represent the base
base = c
# Create variables representing the height
height = 8


# Calculate the area
area = .5 * (base * height)

# print using an f-string
print(f'A triangle with a base of {base} and a height of {height} has an area of {area}.')

Output

A triangle with a base of 5 and a height of 8 has an area of 20.0.

---

1.3

Solution

import math # for pi constant

#Assign a number to a variable torepresent the radius of a circle.
radius = 5

# calculate the circumference and area
circumference = 2 * math.pi * radius

area = math.pi * (radius ** 2)

print(f'''
Circle

Radius: {radius}
Circumference: {circumference}
Area: {area}
''')

Output

Circle

Radius: 5
Circumference: 31.41592653589793
Area: 78.53981633974483

---

1.4

Solution

import math # for pi constant

#Assign a number to a variable torepresent the radius of a sphere.
radius = 5

# calculate spherical volume

volume = (4/3) * math.pi * (radius ** 3)

print(f'''
Sphere

Radius: {radius}
Volume: {volume}
''')

Output

Sphere

Radius: 5
Volume: 523.5987755982989

---

1.5

Solution

import math # for pi constant

# outer and inner radii
outer_radius = 5
inner_radius = 3

# calculate the area
annulus_area = math.pi * (outer_radius ** 2 - inner_radius ** 2)

print(f'''
Annulus

Outer radius: {outer_radius}
Inner radius: {inner_radius}

Area: {annulus_area}
''')

Output

Annulus

Outer Radius: 5
Inner Radius: 3
Area: 50.26548245743669

---

1.6

Solution

import math # for the square root function sqrt()

# Assign a number to each to represent two sides adjacent sides of a triangle
a = 3
b = 4

# calculate the hypotenuse with Pythagorean Theorem
hypotenuse = math.sqrt(a**2 + b**2)

# print using an f-string
print(f'A triangle adjacent sides of {a} and {b} has a hypotenuse of {hypotenuse}.')

Output

A triangle adjacent sides of 3 and 4 has a hypotenuse of 5.0.

---

1.7

The solution will be the same except with all numbers fulfilled by random.randint()

1.8

The solution will be the same except with all numbers collected from the user using input(). You can then convert the number to an integer using int() or a float using float().

Keep in mind that this is just one potential solution.

< Exercise 1

<< Back to Unit 3 Practice