A square pyramid has side lengths of 4 and the height of 6 what is the surface area, rounded to the nearest tenth?

66.59

Step-by-step explanation:

The surface area can be calculated as follows

= a^2 + 2a √ a^2/4 + h^2

= 4^2 + 2(4) √ 4^2/4 + 6^2

= 16 + 8 √16/4 + 36

= 24 √ 4+36

= 24 √40

= 66.59

  b:   ∠x = 28.6°, ∠y = 31.2°, ∠z = 120.2°

step-by-step explanation:

this is one of those questions where "test taking skill" is all you need. the only answer that has angles that sum to 180° is the one shown above. the remaining choices do not describe the angles of a triangle.

if you want to do more work than simply adding the offered numbers, you can also check them against the law of sines. it should be true that

  x/sin(x) = y/sin(y) = z/sin(z)

and all of these ratios must be greater than 15, the longest side length.

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if you want to solve the triangle from scratch, i'd suggest solving for the largest angle using the law of cosines.

  z² = x² +y² -2xy·cos(z)

  z = arccos((x² +y² -z²)/(2xy)) = arccos(-75.11/149.4) = 120.182°

now you can find y from the law of sines

  y = arcsin(y/z·sin(z)) = 31.242°