Another assignment left unanswered need cphil expertise

Here's the last one

Plus sir cphil if I have a funnel 20cm wide and 25 cm long connected to a cylinder 12 cm high and 10 cm base with how much is its liquid capacity?

Assuming the funnel is a cone, the volume is (1/3)pi r^2 * h =  (1/3)pi (10^2) (25) cm^3  = pi*(1/3)(2500)cm^3

And assuming that the 10cm base of the cylinder is the diameter [r = 5 cm], we have

Volume = pi(r^2)(h)  = pi(5^2)(12)  cm^3  = pi (300) cm^3

Adding both of these gives  :   pi [ (1/3)*2500 + 300]  = [3400/3]pi cm^3

Here's the first one.......

We need to  calculate the slant height (h)  of the sides

This is given by :

sqrt [ [( 12 - 6)/2]^2  + 4^2]   =  sqrt [3^2 + 4^2]   = sqrt[9 + 16] =  sqrt(25) = 5cm

The lateral area is formed by 4 trapezoids each having this area :

(1/2)h(b1 + b2]   =  (1/2)(5)(12 + 6]  = 5* 18/ 2 = 90/ 2 = 45 cm^2

So....4 times this =  180 cm^2

Here's the second one :  [it's a little difficult]

2.Find the volume of the frustum of a regular hexagonal pyramid if an edge of the lower base is 12 cm, a laterral edge is 10 cm and whose altitude is 4cm.

We need to find the length of the edge  of the upper base 

This is given by:   12 - 2sqrt(10^2 - 4^2)   =   [12 - 2sqrt(84)]

The area of the lower base - ( A1) =   [3*sqrt(3)/2] [12^2]  =  about  216sqrt(3) cm^2  = about 374.12 cm^2

The area of the upper base - (A2) = [3*sqrt(3)/2] [12- 2sqrt(84)]^2 =   about 104.11cm^2

And the volume is given by :   (1/3) h [ A1 + A2 + sqrt(A1 * A2)]   .....so we have

V = (1/3) (4) [ 374.12 + 104.11 +  sqrt(374.12 * 104.11) ]  =   about 900.72 cm^3

Here's the third one:

3.The reservoir is in the form of the frustum of an inverted square pyramid with upper base edge 24m, lower base edge 18m and altitude 3m. how many hours will it require for an inlet pipe to fill the reservoir if water flows in at the rate of 200 liters per min. 

The area of the upper base (A1) is  [24m]^2 = [2400cm]^2 

The area of the lower base  (A2) = [18m]^2   = [1800cm[^2

The total volume is :

(1/3)(altitude) [ A1 + A2 + sqrt( A1 * A2) ]

(1/3)(3)( 2400^2  + 1800^2  + sqrt [ 2400^2  * 1800^2] ) = 13,320,000 cm^3

And 200 liters = 200,000cm^3

So....it takes...... [13,200,000 / 200/000] =  66 min = 1.1 hr to fill the tank

At number 4 what about the 20cm widht at the top 

CierDreaj.....if the funnel is 20 cm wide.....its radius would be 1/2 of that = 10cm

So....the volume of the funnel =  (1/2)pi (r^2) h =  (1/2)pi (10^2)(25)  = [2500/3] pi cm^3