OpenStudy (anonymous):

Find the surface area of the pyramid shown to the nearest whole number.

4 years ago
OpenStudy (jhannybean):

the Surface area formula is \[\large SA =B+\frac{1}{2}(P)(S)\] Where B= Base, P =perimeter S = side length. In order to find the area of the base, we know the pentagon has 10 triangles. we need to take 1 of those triangles and find the area of it. Then multiply it by 10. |dw:1370134394122:dw|

4 years ago
OpenStudy (jhannybean):

and you know what... we dont even need to solve for individual triangles, since one base side is given. That is 10. So forget the drawing.

4 years ago
OpenStudy (anonymous):

Oh, okay XD

4 years ago
OpenStudy (jhannybean):

|dw:1370134617237:dw| we're going to take the Area of this, and multiply it by 5, instead of 10, because we have 5 BIG triangles that make up a pentagon. \[\large A =5[ (\frac{1}{2}) B*H]\]\[A_{B} = 5[(\frac{1}{2})(10)(5\sqrt{3})]\] \[\large A_{B}= 5[\frac{1}{2}(50 \sqrt{3})]\]\[\large A_{B}=125 \sqrt{3}\] Now weve found the Area of the Base. part 1 of our equation.

4 years ago
OpenStudy (anonymous):

I'm not sure how to simplify the square root

4 years ago
OpenStudy (jhannybean):

I base a mistake. the area of the base is actually \[\large A_{B}=6[\frac{1}{2}(10)(5\sqrt{3})]\]\[\large A_{B}=150 \sqrt{3}\]\[\large P_{B}= 60 m\]\[\large S=16 m\]\[\large SA = B+\frac{1}{2}(P)(S)\]\[\large SA = 150 \sqrt{3}+ \frac{1}{2}(60)(16)\]\[\large SA=150\sqrt{3} +480\] So \(\large 150\sqrt{3} \approx 259.8076\) and you add that to 480 \[\large SA = 259.8076 + 480 \approx 739.8076 = 740 \]

4 years ago
OpenStudy (jhannybean):

not base, made* xD you said pentagon, to i multiplied everything by 5, but it's actually a hexagon.... your problem confused me till i saw the figure.

4 years ago
OpenStudy (anonymous):

Thank you so much for taking your time to do this!! I understand the problem a lot better.

4 years ago
OpenStudy (jhannybean):

Awesome!!

4 years ago
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