1. The base edges of a frustum of a regular square pyramid whose base edges are 5cm and 10cm and whose alant height is 8cm and which makes with the lower base an angle of 60°. Find the total area of the frustum.
2 A milldam of earth, with planes slopping sides and rectangle bases 10m by 6m at the top and 16m by 12m. Find the volume in m³ if the altitude is 8meters
+130518 Here's the first one :
The area of the two bases is just (5cm)^2 + (10cm)^2 = 25cm^2 + 100cm^2 = 125cm^2
The sides are trapezoids that look like this :
The area of each trapezoid is (1/2)(8cm)(10cm + 5cm) = (4cm)(15cm) = 60cm^2
And we have 4 of these......so their total area is 240cm^2
So.......the total surface area = [125 + 240]cm^2 = 365cm^2
+130518 Here's the first one :
The area of the two bases is just (5cm)^2 + (10cm)^2 = 25cm^2 + 100cm^2 = 125cm^2
The sides are trapezoids that look like this :
The area of each trapezoid is (1/2)(8cm)(10cm + 5cm) = (4cm)(15cm) = 60cm^2
And we have 4 of these......so their total area is 240cm^2
So.......the total surface area = [125 + 240]cm^2 = 365cm^2
+130518 Here's the second one:
2 A milldam of earth, with planes slopping sides and rectangle bases 10m by 6m at the top and 16m by 12m. Find the volume in m³ if the altitude is 8meters
We need to apply this formula:
V = [h/6] [a*b + c*d + (a + c)(b + d)]
Where
a = 10m
b = 6m
c = 16m
d = 12
h = 8m
So we have
V = [8/6] [10*6 + 16*12 + (10 + 16) (6 + 12) ] = 960m^3
