Question

An isosceles trapezium has an area of `300cm^2`, height `h=12cm` and base `b= 4/3 h`. Find: • The perimeter • The surface area obtained by revolving it around the major base • The volume & mass (`7.5g/(cm^3)`)?

Answer 1

The perimeter of the trapezium is`=80`
The surface area is `=744piu^2`. The volume is `=1088piu^3`. The mass is `=25.635kg`

Explanation:

`BE=CF=h=12cm`

`BC=b=4/3h=4/3*12=16cm`

`AD=a`

The area of the trapezium is

`area=1/2(BC+AD)*BE`

`1/2(a+b)h=300`

`a+b=600/12=50`

`a=50-16=34`

`AE=FD=(34-16)/2=9`

By Pythagoras,

`l=AB=CD=sqrt(BE^2+AE^2)=sqrt(144+81)=sqrt225=15`

The perimeter of the trapezium is

`P=AB+BC+CD+AD=15+16+15+34=80`

The surface area is made of 2 cones and a cylinder

Surface area is `S=2*pirl+2pirh`

`=2*pi*h*l+2*pi*h*b`

`=2pi*h(15+16)`

`=744pi`

The volume is made of 2 cones and a cylinder

`V=2*1/3pih^2*AE+pi*h^2*b`

`=2/3*pi*144*9+pi*144*16`

`=1088piu^3`

THe mass is

`m=V*rho=1088pi*7.5=25.635kg`