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Find the total surface area of the right square pyramid with

a=1
h=2
Round to the nearest tenth.

User Ostecke
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1 Answer

6 votes

Answer:


A</p><p></p><p>=</p><p></p><p></p><p> {a}^(2) </p><p></p><p>+</p><p></p><p>2a \sqrt{ \frac{ {a}^(2) }{4} + {h}^(2) } </p><p> \\ </p><p></p><p>=</p><p></p><p> {1}^(2) </p><p></p><p>+</p><p></p><p>2</p><p></p><p>·</p><p></p><p>1</p><p></p><p>· \sqrt{ \frac{ {1}^(2) }{4}+ {2}^(2) } </p><p> \\ </p><p></p><p>≈</p><p></p><p>5.12311

User Jdm
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