SED 100
Name: Unit 11: Volume & Surface Area
Date: Bell: Homework 6: Surface Area of Pyramids & Cones
** This is a 2-page document! **
Directions: Find the surface area of each figure. Round to the nearest hundredth when necessary.
20.7 in
8.2 yd
5.2 yd
11 in
6 yd
13.6 mm
8 mm
111
3.5 m
12.8 ft
23.6 km
9 ft
17 km
19.5 in
20 ft
12 in
30 ft
B – 247.75 in
Gina Wilson All Things Algebra
15.4 cm
17 cm
20 mm
If a cone with a diameter of 10 meters has a surface area of 290.6 square meters, find its slant height.
A regular hexagonal pyramid has a base area of 166.28 square feet, a slant height of 9.8 feet, and a surface area of 401.48 square feet. Find the side length of the base.
A triangular pyramid with an equilateral base has a side length of 10 centimeters and a surface area of 214.5 square centimeters. Find its slant height.
Find the surface area of the figure below.
10.3 mm
5 mm
13 mm
13 mm
Gina Wall Things 2015
Sure! Let’s work through one of the specific problems from the list you mentioned to provide the correct answer and a 300-word explanation.
Problem:
Find the surface area of a cone
A cone has a diameter of 10 meters and a total surface area of 290.6 square meters. Find its slant height.
Solution:
To solve this, we use the surface area formula for a cone: Surface Area=πr2+πrl\text{Surface Area} = \pi r^2 + \pi r l
Where:
- rr = radius
- ll = slant height
- π≈3.14\pi \approx 3.14
- Surface Area is given = 290.6 m²
- Diameter = 10 meters → Radius r=102=5r = \frac{10}{2} = 5 meters
Substitute into the formula: 290.6=π(5)2+π(5)(l)290.6 = \pi (5)^2 + \pi (5)(l) 290.6=3.14(25)+3.14(5)(l)290.6 = 3.14(25) + 3.14(5)(l) 290.6=78.5+15.7l290.6 = 78.5 + 15.7l
Now solve for ll: 290.6−78.5=15.7l290.6 – 78.5 = 15.7l 212.1=15.7l212.1 = 15.7l l=212.115.7≈13.5 metersl = \frac{212.1}{15.7} \approx 13.5 \text{ meters}
✅ Final Answer:
Slant height = 13.5 meters
✍️ Explanation:
The surface area of a cone is made up of two parts: the base and the curved surface. The base of a cone is a circle, so its area is calculated using the formula πr2\pi r^2. The curved surface, or lateral area, is calculated using πrl\pi r l, where ll is the slant height — the distance from the top of the cone to a point on the edge of the base, measured along the side.
In this problem, we are given the total surface area of a cone and its diameter, and we’re asked to find the slant height. First, we convert the diameter to the radius by dividing by two. That gives us a radius of 5 meters. We then substitute the known values into the surface area formula: Surface Area=πr2+πrl\text{Surface Area} = \pi r^2 + \pi r l
This becomes: 290.6=3.14(25)+3.14(5)(l)290.6 = 3.14(25) + 3.14(5)(l)
We simplify the equation step by step and isolate ll, eventually finding that the slant height is approximately 13.5 meters. This approach shows how algebra is used to solve geometry problems involving unknown values. Understanding the components of 3D shapes like cones helps in a variety of real-world applications, from construction to design, where precise measurements are critical.
By breaking the problem into small steps, we ensure accuracy while strengthening our understanding of how dimensions and formulas relate to real shapes.
