On these or one of need a. s.a. p !

Problem 5

Work Shown:

The lateral faces of this pyramid are triangles. Focus on half of one of the triangles. We have a right triangle with a leg of 8 cm and a hypotenuse of 17 cm. The dashed line is the other leg, which we'll find using the pythagorean theorem

a^2 + b^2 = c^2

8^2 + b^2 = 17^2

64 + b^2 = 289

b^2 = 289 - 64

b^2 = 225

b = sqrt(225)

b = 15

Each triangle face has a base of 2*8 = 16 and a height of 15. So the area of one triangle is A = b*h/2 = 16*15/2 = 240/2 = 120

There are four of these triangles, so 4*120 = 480 is the lateral surface area. Basically the surface area for everything but the base.

Problem 6

Work Shown:

L = slant height = 12.5

h = height of cone = 12

To find the radius, we form a right triangle such that the height 12 is one leg and the hypotenuse is 12.5; the unknown radius is the other leg of this right triangle.

So,

a^2 + b^2 = c^2

(12)^2 + b^2 = (12.5)^2

144 + b^2 = 156.25

b^2 = 156.25 - 144

b^2 = 12.25

b = sqrt(12.25)

b = 3.5

The radius is therefore 3.5 cm

Now let's use that formula given

LSA = lateral surface area of the cone

LSA = pi*r*L

LSA = pi*3.5*12.5

LSA = pi*43.75

LSA = 43.75pi <<-- exact lateral surface area in terms of pi

LSA = 137.444678594553 <<--- approx lateral surface area

LSA = 137.4 <<--- rounding to nearest tenth (aka 1 decimal place)

answer: the first graph

step-by-step explanation:

step-by-step explanation: