Find the surface area of the pyramid shown to the nearest whole number.
http://assets.openstudy.com/updates/attachments/4fcd07a2e4b0c6963ad7f294-cammo923-1338836941300-staffuserscodellmypicturesaab.jpg A. 740 m^2 B. 480 m^2 C. 1,480 m^2 D. 715 m^2
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|
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.
Oh, okay XD
|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.
I'm not sure how to simplify the square root
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 \[\large SA = 259.8076 + 480 \approx 739.8076 = 740 \]
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.
Thank you so much for taking your time to do this!! I understand the problem a lot better.
Awesome!!
Join our real-time social learning platform and learn together with your friends!