Find the surface area of a pyramid if its height is 4 cm and its base is a rectangle with side lengths 4 cm and 6 cm. Round to the nearest whole number.

- anonymous

- schrodinger

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- jim_thompson5910

What is the area of the base?

- anonymous

24 cm^2

- jim_thompson5910

correct

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## More answers

- jim_thompson5910

|dw:1435548018859:dw|

- jim_thompson5910

|dw:1435548042782:dw|

- jim_thompson5910

|dw:1435548055051:dw|

- jim_thompson5910

do you see how I got that triangle I just added?

- anonymous

yes

- jim_thompson5910

it's a right triangle with legs 3 and 4
the hypotenuse is what?

- anonymous

5

- jim_thompson5910

|dw:1435548291256:dw|

- jim_thompson5910

we will have 2 triangles on the lateral face of the pyramid that look like this
|dw:1435548321950:dw|
base = 4
height = 5

- jim_thompson5910

what is the area of that triangle?

- anonymous

10cm^2

- jim_thompson5910

there are 2 of these triangles, so they add to 10+10 = 20 cm^2
the front and back faces
|dw:1435548418230:dw|

- jim_thompson5910

|dw:1435548440907:dw|

- jim_thompson5910

|dw:1435548484189:dw|

- anonymous

So for the other 2 sides would the area be 15cm^2.

- anonymous

15+15=30cm^2

- jim_thompson5910

how are you getting 15?

- anonymous

And the bottom area would be 6x4=24cm^2

- anonymous

I got 15 from multiplying 6x5=30/2=15cm^2. Would that not be correct?

- jim_thompson5910

that other hypotenuse isn't 5

- jim_thompson5910

|dw:1435548754459:dw|

- anonymous

4.47

- jim_thompson5910

so you have 2 more triangles, each looking like this
|dw:1435548837682:dw|

- anonymous

Would i make 4.47 into 4 because the problem says to round the nearest whole number.

- jim_thompson5910

round when you're done calculating

- jim_thompson5910

so it might be better to use more decimal digits
instead of 4.47 maybe use 4.47213595 or something
\[\Large \sqrt{20} \approx 4.47213595\]

- anonymous

Area=13.416407

- jim_thompson5910

that's the area of one triangle
there are 2 of them

- anonymous

26.832814cm^2

- jim_thompson5910

Summary:
you have a base that's 24 cm^2
the front and back triangles have a total area of 20cm^2
the left and right triangles have a total area of 26.832814 cm^2

- anonymous

The answer would be then 70.83 or 71cm^2 (the answer was in my book but i couldn't see what i was doing wrong THANK YOU SO MUCH :D)

- jim_thompson5910

you're welcome

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