What is the lateral area of this regular octagonal pyramid? (picture one)

114.8 cm²

162.4 cm²

229.7 cm²

281.3 cm²

2.what is the slant height x of the square pyramid? (picture 2)

the figure shows a square pyramid. the slant height is shown as a dashed line perpendicular to the base edge and is labeled as x. the length of the lateral edge is 8 meters. the lateral edge makes a 60 degree angle with the base edge

express your answer in radical form.

3.what is the surface area of this square pyramid? (picture 3)

round your answer to the nearest tenth, if necessary.

43.8 yd²

66.6 yd²

105 yd²

171.6 yd²

Q1)
the lateral area of the pyramid is the total area of all the lateral faces excluding the base.
In this regular octagonal pyramid, the lateral sides are triangles. As there are 8 triangles we need to find the area of all 8 sides.
Area of one lateral triangle face = 1/2 * base * slant height 
slant height is the hypotenuse of the right angled triangle formed from the base of the pyramid with the perpendicular height.
slant height - l
l² = 7² + 7² = 49 *2
 l²  = 98 
l = √98
l = 9.9
Area = 1/2 * 5.8 cm * 9.9 cm 
         = 28.71 cm²
There are 8 sides 
total lateral area = 8 * 28.71 = 229.68 rounded off is 229.7 cm²
third option is correct - 229.7 cm²

Q2)
in the triangular face, the lateral edge makes a 60° angle with the base edge. Therefore 2 of the angles are 60° each, since the sum of the interior angles of a triangle is 180°, the third angle too is 60°. this makes the triangle an equilateral triangle with equal angles, hence equal sides. 
since lateral edge is 8 cm,base edge too is 8 cm. 
since this is an equilateral triangle, the perpendicular line cuts the base edge at its midpoint, bisecting the line forming 2 right angled triangles.
in the right angled triangle, height of triangle is x slant height ,
base = 8 /2 = 4 cm
hypotenuse = 8 cm
We need to find x, use Pythogoras' theorem 
4² + x² = 8²
16 + x² = 64
x² = 62 - 16
x = √48
x = √4x√4x√3
  = 2x2√3
  = 4√3 cm

Q3)
surface area of the square pyramid 
surface area of the base + surface area of triangular faces 
square area = length x length 
                    = 6.2 x 6.2 
                    = 38.44 yd²
triangular face area = 1/2 * length * height 
since the angle between lateral edge and base edge is 60°, its an equilateral triangle where all sides are equal. in this case each side is 6.2 yd. 
to find the perpendicular height, use pythogoras' theorem
the perpendicular line(slant height ) cuts the base edge at its midpoint, therefore length of the right angled triangle is = 6.2/2 = 3.1 yd

slant height - l
l² + 3.1² = 6.2²
l² = 38.44 -9.61
l²  = 28.83 
l = 5.37
area = 1/2 *length *height 
        = 1/2 * 6.2 * 5.37
        = 16.64 yd²
there are 4 triangles = 4 * 16.64 = 66.58 yd²
total area = 38.44 + 66.58 = 105 yd²
correct answer is 3rd option - 105 yd²

X^2 - 4x +(4/2^2) + y^2 + 8y + (8/2^2) = -4 + (4/2^2) + (8/2^2)(x + 2)^2 + (y + 4)^2 = -4 + 4 + 16your center is (-2, -4) and the radius is the sqrt of -4 + 4 + 16 which is 4