Question

A men donates 10 aluminium buckets to an orphanage. A bucket made of aluminium is of height 20 cm and has its upper and lowest ends of radius 36 cm and 21 cm respectively. Find the cost of preparing 10 buckets if the cost of aluminium sheet is ruppes 42 per 100 cm^2. Write your comments on the act of the man

Answer 1

the cost of preparing 10 buckets is rupees 24608.8

Answer 2

Answer:

Step-by-step explanation:

Given the height of the bucket $(h)= 20\text{ cm}$

Radius of the upper end $R_{1}=36\text{ cm}$

Radius of the lower end $R_{2}=21\text{ cm}$

The shape of the bucket is frustum

The slant height of the frustum $S=\sqrt{h^{2}+(R_{1}-R{2})^{2}}\\\\\Rightarrow{S}=\sqrt{20^{2}+(36-21)^{2}}\\\\\Rightarrow{S}=\sqrt{20^{2}+15^{2}}\\\\\Rightarrow{S}=\sqrt{400+225}\\\\\Rightarrow{S}=\sqrt{625}\\\\\Rightarrow{S}=25$

The surface are of the bucket = Lateral Area of a flat surface + Circular area of bottom

therefore,

$A=\pi\times(R_{1}+R_{2})\times{S}+\pi\times({R}_{1})^{2}\\\\\Rightarrow{A}=\frac{22}{7}\times(36+21)\times25+\frac{22}{7}\times(21)^{2}\\\\\Rightarrow{A}=\frac{22}{7}\times57\times25+\frac{22}{7}\times(21)^{2}\\\\\Rightarrow{A}=\frac{22}{7}(1425+441)\\\\\Rightarrow{A}=\frac{22}{7}\times(1866)\\\\\Rightarrow{A}=5864.5$

therefore the total surface area of a bucket $= 5864.5\text{ cm}^{2}}$

Therefore the total surface area of 10 such bucket are $=58645\text{ cm}^{2}}$

Cost of aluminium sheet for 100 $\text{cm}^2$ is 42 rupees.

therefore, the Cost of aluminium sheet for $58645\text{ cm}^{2}}$ is $=\frac{42}{100}\times58645=24603.90$  

The total cost of the aluminium sheet is rupees 24604