- Thread starter
- #1
Welcome to our community
Be a part of something great, join today!
Triangular prism on top of rectangular prism in the shape of a house.....
- Thread starter Sting124
- Start date
- Admin
- #2
Re: triangular prism on top of rectangular prism in the shape of a house.....
Do you have any information regarding the dimensions?
edit: I recently was giving a student help on a practice worksheet containing 11 problems, and this was essentially problem 10, but a diagram was missing and the question unanswerable. Are you perhaps working on the same set of practice problems?
Do you have any information regarding the dimensions?
edit: I recently was giving a student help on a practice worksheet containing 11 problems, and this was essentially problem 10, but a diagram was missing and the question unanswerable. Are you perhaps working on the same set of practice problems?
Last edited:
- Thread starter
- #3
- Admin
- #4
- Mar 5, 2012
- 8,888
Re: triangular prism on top of rectangular prism in the shape of a house.....
If it is the shape of a house, it should still be doable.
We'd have a house with say length L and width W.
The block part of the house can have height H.
And the roof part can have height h for a total height of (H+h).
Then you'd have to calculate how much area the walls and roof have.
If it is the shape of a house, it should still be doable.
We'd have a house with say length L and width W.
The block part of the house can have height H.
And the roof part can have height h for a total height of (H+h).
Then you'd have to calculate how much area the walls and roof have.
- Thread starter
- #5
Re: triangular prism on top of rectangular prism in the shape of a house.....
The slant heights are already in the question I provided for you
- slant heights 15m
- length = 48m
- Width = 24m
- height = 32m
Mark could you help my sibling over at the other forum for question 10 if not help me here please
The slant heights are already in the question I provided for you
- slant heights 15m
- length = 48m
- Width = 24m
- height = 32m
Mark could you help my sibling over at the other forum for question 10 if not help me here please
- Thread starter
- #6
Geometry problem please help
Shape of a house, triangular prism on top a rectangular prism
- Slant heights 15m
- length = 48m
- Width = 24m
- height = 32m
What is the area (excluding the base)?
What is the volume?
If a scale model was made with the longest side being 2m in length. What would the volume of scale model to actual model as a ratio be.
Shape of a house, triangular prism on top a rectangular prism
- Slant heights 15m
- length = 48m
- Width = 24m
- height = 32m
What is the area (excluding the base)?
What is the volume?
If a scale model was made with the longest side being 2m in length. What would the volume of scale model to actual model as a ratio be.
- Admin
- #7
- Mar 5, 2012
- 8,888
Re: triangular prism on top of rectangular prism in the shape of a house.....
The area of a rectangle is length x width.
The area of a triangle is base x height / 2.
Can you find the total area with this?
Well, a house consists of 4 rectangular walls, 2 rectangular roof parts, and 2 triangular wall sections that are part of the roof.
The area of a rectangle is length x width.
The area of a triangle is base x height / 2.
Can you find the total area with this?
- Admin
- #8
Re: triangular prism on top of rectangular prism in the shape of a house.....
The total area consists of 6 rectangles and 2 triangles. All measures are in meters.
There are 4 vertical rectangles making up the outer wall of the rectangular prism, 2 of these are 24 X 32 and two are 48 X 32.
There are two rectangles making up the roof, and they are 15 X 48.
There are two triangles at each end of the roof, whose height $h$ may be found using the Pythagorean theorem:
\(\displaystyle 12^2+h^2=15^2\)
Once you find $h$ then the 2 triangles have a combined area of:
\(\displaystyle A=2\left(\frac{1}{2}\cdot24h \right)=24h\)
This should allow you to find the area. Now as for the model, its longest side is 2, and the longest side of the actual shed is 48 so the scale of the mode is 1:24. When the linear measure of a 3 dimensional solid is multiplied by some factor $k$, the ratio of the volume of the original to the volume of the scaled version is then \(\displaystyle \frac{1}{k^3}\). So, what would the ratio be in this case?
I just sent a PM at the other forum.
The total area consists of 6 rectangles and 2 triangles. All measures are in meters.
There are 4 vertical rectangles making up the outer wall of the rectangular prism, 2 of these are 24 X 32 and two are 48 X 32.
There are two rectangles making up the roof, and they are 15 X 48.
There are two triangles at each end of the roof, whose height $h$ may be found using the Pythagorean theorem:
\(\displaystyle 12^2+h^2=15^2\)
Once you find $h$ then the 2 triangles have a combined area of:
\(\displaystyle A=2\left(\frac{1}{2}\cdot24h \right)=24h\)
This should allow you to find the area. Now as for the model, its longest side is 2, and the longest side of the actual shed is 48 so the scale of the mode is 1:24. When the linear measure of a 3 dimensional solid is multiplied by some factor $k$, the ratio of the volume of the original to the volume of the scaled version is then \(\displaystyle \frac{1}{k^3}\). So, what would the ratio be in this case?
Last edited: