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A net of a rectangular pyramid is shown. The rectangular base has length 25ft and width 22ft . The net of the pyramid has length 72.4 and width 67.8. Find the surface area of the pyramid.

Guest Mar 4, 2015

Best Answer 

 #7
avatar+93297 
+5

Oh, I see what you are saying Alan,  but I think it the lengths of the legs of the isosceles triangles are the same then the height in the middle would be the same also.  So you can do it either way i think.  

What are the sides of the isosceles triangles.

 

1)  $${\sqrt{{{\mathtt{22.9}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{12.5}}}^{{\mathtt{2}}}}} = {\mathtt{26.089\: \!461\: \!473\: \!936\: \!176\: \!3}}$$

and

 

2)   $${\sqrt{{{\mathtt{23.7}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{11}}}^{{\mathtt{2}}}}} = {\mathtt{26.128\: \!337\: \!107\: \!439\: \!501\: \!2}}$$

 

Again correct to the neares 0.1 of a foot.   Mmm

Melody  Mar 4, 2015
 #1
avatar+93297 
+5

Area of base = 25*22 = 550 ft square

  $$\frac{72.4-25}{2}=\frac{47.4}{2}=23.7$$

 

so 2 of the triangle sides have base 22 and height 23.7 Their area combined = 2* (1/2)*22*23.7

 

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{22}}{\mathtt{\,\times\,}}{\mathtt{23.7}} = {\frac{{\mathtt{2\,607}}}{{\mathtt{5}}}} = {\mathtt{521.4}}$$    square feet

 

$$\frac{67.8-22}{2}=\frac{45.8}{2}=22.9$$

 

so 2 of the triangle sides have base 25 and height 22.9 Their area combined = 2* (1/2)*25*22.9

 

$${\mathtt{2}}{\mathtt{\,\times\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right){\mathtt{\,\times\,}}{\mathtt{25}}{\mathtt{\,\times\,}}{\mathtt{22.9}} = {\frac{{\mathtt{1\,145}}}{{\mathtt{2}}}} = {\mathtt{572.5}}$$     square feet

 

total area =  $${\mathtt{550}}{\mathtt{\,\small\textbf+\,}}{\mathtt{521.4}}{\mathtt{\,\small\textbf+\,}}{\mathtt{572.5}} = {\mathtt{1\,643.9}}$$     square feet

 

If you want me to draw a picture to explain what i  have done I can do that but try and understand it by yourself first.   

Melody  Mar 4, 2015
 #2
avatar+88791 
+5

What is the "net" of the pyramid???

Can you provide a pic of this, Melody??

 

  

CPhill  Mar 4, 2015
 #3
avatar+93297 
0

Yes I will, but I will have to get back to it.  :/

Melody  Mar 4, 2015
 #4
avatar+26971 
+5

I think this is the net that Melody had in mind:

Net

 

A net is a pattern you can fold to make a solid shape.

.

Alan  Mar 4, 2015
 #5
avatar+93297 
+5

Thanks Alan, that  makes my life easier.    (๑‵●‿●‵๑)

 

I thought I would play around with it and determine is this really is  a rectangular pyramid.  :)

 

If it is then the height given by one triangle must be equal to the height given by the other one.

 

$${\sqrt{{{\mathtt{23.7}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{12.5}}}^{{\mathtt{2}}}}} = {\mathtt{20.135\: \!540\: \!717\: \!845\: \!150\: \!5}}$$

 

$${\sqrt{{{\mathtt{22.9}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{11}}}^{{\mathtt{2}}}}} = {\mathtt{20.085\: \!069\: \!081\: \!285\: \!232\: \!2}}$$

 

Mmm it is a little off but I suppose it is correct to 0.1 of a foot.    Is that near enough.   Not sure.  :/

Melody  Mar 4, 2015
 #6
avatar+26971 
+5

I think it's the sloping sides that have to be the same length Melody, but even here there is a small discrepancy.

.

Alan  Mar 4, 2015
 #7
avatar+93297 
+5
Best Answer

Oh, I see what you are saying Alan,  but I think it the lengths of the legs of the isosceles triangles are the same then the height in the middle would be the same also.  So you can do it either way i think.  

What are the sides of the isosceles triangles.

 

1)  $${\sqrt{{{\mathtt{22.9}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{12.5}}}^{{\mathtt{2}}}}} = {\mathtt{26.089\: \!461\: \!473\: \!936\: \!176\: \!3}}$$

and

 

2)   $${\sqrt{{{\mathtt{23.7}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{11}}}^{{\mathtt{2}}}}} = {\mathtt{26.128\: \!337\: \!107\: \!439\: \!501\: \!2}}$$

 

Again correct to the neares 0.1 of a foot.   Mmm

Melody  Mar 4, 2015

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