Find the SURFACE AREA of this composite solid.
1 answer:
FINDING THE SURFACE AREA OF A COMPOSITE SOLID
About "Finding the surface area of a composite solid"
Finding the surface area of a composite solid :
A composite solid is made up of two or more solid figures.
To find the surface area of a composite solid, find the surface area of each figure. Subtract any area not on the surface.
Finding the surface area of a composite solid - Examples
Example 1 :
Daniel built the birdhouse shown below. What was the surface area of the birdhouse before the hole was drilled ?
Solution :
Step 1 :
Identify the important information.
• The top is a triangular prism with h = 24 cm. The base is a triangle with height 8 cm and base 30 cm.
• The bottom is a rectangular prism with h = 18 cm. The base is a 30 cm by 24 cm rectangle.
• One face of each prism is not on the surface of the figure.
Step 2 :
Find the surface area of each prism.
Add the areas. Subtract the areas of the parts not on the surface.
Step 3 :
Find the area of the triangular prism.
Perimeter = 17 + 17 + 30 = 64 cm
Base area = (1/2)(30)(8) = 120 sq.cm
Surface area = Ph + 2B
Surface area = 64(24) + 2(120)
Surface area = 1,776 sq.cm
Step 4 :
Find the area of the rectangular prism.
Perimeter = 2(30) + 2(24) = 108 cm
Base area = 30(24) = 720 sq.cm
Surface area = Ph + 2B
Surface area = 108(18) + 2(720)
Surface area = 3,384 sq.cm
Step 5 :
Add. Then subtract twice the areas of the parts not on the surface.
Surface area = 1,776 + 3,384 - 2(720) = 3,720 sq.cm
The surface area before the hole was drilled was 3,720 sq.cm.
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Problem 61
The nth triangular number is
T(n) = n(n+1)/2
I'll rewrite this into
T(n) = 0.5n(n+1)
The triangular number right after this is
T(n+1) = 0.5(n+1)(n+2)
I replaced every n with n+1 and simplified
Let's see what we get when we add up the two expressions
T(n) + T(n+1)
0.5n(n+1) + 0.5(n+1)(n+2)
0.5n^2+0.5n + 0.5(n^2+3n+2)
0.5n^2 + 0.5n + 0.5n^2 + 1.5n + 1
n^2+2n+1
(n+1)^2
This shows that the sum of any two consecutive triangular numbers results in a square number
Here's a few examples
- 0+1 = 1
- 1+3 = 4
- 3+6 = 9
- 6+10 = 16
- 10+15 = 25
Note each sum is a perfect square, which visually would plot out a square figure.
For quick reference, the set of the first few triangular numbers is {0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55,...}
<h3>Answer: Square number</h3>==========================================================
Problem 64
Let's say we go with n = 5.
This means,
T(n) = 0.5n(n+1)
T(n-1) = 0.5(n-1)(n-1+1)
T(n-1) = 0.5n(n-1)
T(5-1) = 0.5*5(5-1)
T(4) = 10
This says that when n = 5, the 4th triangular number is 10
Triple that result and add on n = 5
3*T(4) + n = 3*10+5 = 35
This result is beyond obvious which category of figurate number it belongs to. It's not a triangular number since it's not in the form n(n+1)/2. It's not a square number either.
Through a bit of trial and error, you should find it's a pentagonal number
Pentagonal numbers are of the form n(3n-1)/2
If you plugged n = 5 into that, it leads to 35
n(3n-1)/2 = 5*(3*5-1)/2 = 5*14/2 = 70/2 = 35
The diagram shown below represents the first few pentagonal numbers. The number of blue dots corresponds to the pentagonal number itself. Note the equal spacing when dealing with dots on each segment (eg: some interior blue dots are midpoints, others are quarter points, etc.)
<h3>Answer: Pentagonal number</h3>