You need to use a ratio of height (H) to shadow length (L) to solve the first problem. It's basically a use of similar triangles, with two perpendicular sides, and with the shadow making the same angle with the vertical.
6 ft = 72 ins, so that rH/L = 72/16 = 9/2 for the player.
So the bleachers are 9/2 x 6 ft = 27 ft.
For the second problem, 9 ft = 108 in, so that the ratio of the actual linear dimensions to the plan's linear dimensions are 9ft/(1/2in) = 2 x 108 = 216.
So the stage will have dimensions 216 times larger than 1.75" by 3".
That would be 31ft 6ins x 54ft.
Live long and prosper.
Surface area of cone is a sum of surface of its base and surface of its mantle.
Surface of its base is a circle which surface we will calculate like this:
Sb = pi*r^2 where r is d/2
Sb = 201m^2
Surface of mantle we calculate by:
Sm = pi*l*r where l is length of side of cone.
l =
= 25.3
Sm = 635.8m^2
Total suface is:
Sb + Sm = 836.8m^2
Answer:
22 m^2
Step-by-step explanation:
The area of the bottom rectangle is A = l*w
The length is 8 and width is 2
A = 8*2 =16
The area of the top rectangle is A = l*w
The length is 3 and width is 2
A = 3*2 =6
Adding the areas together
A = 16+6 = 22
Tell Me if I'm Wrong But I Believe It's 198m3