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C2 Question
Dont really know what to call this lol.
The diagram shows an open tank for storing water ABCDEF. The sides ABFE and CDEF are rectangles. The trianglular ends ADE and BCF are isoceles, and the <AED = <BFC = 90 degrees. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD = x metres
a) SHow that the area of triangle ADE is 1/4x^2 metres squared.
Given also that the capacity of the container is 4000m^3 and the the total area of the two triangular and two rectangular sides of the container is S m^2
b) Show taht s = x^2/2 + 16000root2/x
Thanks in advance.
The diagram shows an open tank for storing water ABCDEF. The sides ABFE and CDEF are rectangles. The trianglular ends ADE and BCF are isoceles, and the <AED = <BFC = 90 degrees. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD = x metres
a) SHow that the area of triangle ADE is 1/4x^2 metres squared.
Given also that the capacity of the container is 4000m^3 and the the total area of the two triangular and two rectangular sides of the container is S m^2
b) Show taht s = x^2/2 + 16000root2/x
Thanks in advance.
insparato
Dont really know what to call this lol.
The diagram shows an open tank for storing water ABCDEF. The sides ABFE and CDEF are rectangles. The trianglular ends ADE and BCF are isoceles, and the <AED = <BFC = 90 degrees. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD = x metres
a) SHow that the area of triangle ADE is 1/4x^2 metres squared.
Given also that the capacity of the container is 4000m^3 and the the total area of the two triangular and two rectangular sides of the container is S m^2
b) Show taht s = x^2/2 + 16000root2/x
Thanks in advance.
The diagram shows an open tank for storing water ABCDEF. The sides ABFE and CDEF are rectangles. The trianglular ends ADE and BCF are isoceles, and the <AED = <BFC = 90 degrees. The ends ADE and BCF are vertical and EF is horizontal.
Given that AD = x metres
a) SHow that the area of triangle ADE is 1/4x^2 metres squared.
Given also that the capacity of the container is 4000m^3 and the the total area of the two triangular and two rectangular sides of the container is S m^2
b) Show taht s = x^2/2 + 16000root2/x
Thanks in advance.
Could you scan in the diagram?
Doh i was gonna do a drawing forgot, i cant scan it in but ill have a bash at drawing it sorry.
Hello!
Diagram: it's a Toblerone box except it's a right-angled triangle instead of an equilateral triangle, tilted so that it's balancing on its spine (ie resting on the edge connecting the two right angles)
AED is a right-angled triangle, so for the first part you just use Pythagoras.
AE squared + ED squared = x squared. But AE=AD as it's an isosceles triangle,
so 2(AE squared) = x squared,
AE squared = 0.5xsquared
and AE = x/(root 2) also ED = x/(root 2)
Area of triangle = 0.5 x AE x ED
Area of triangle = 0.5 multiplied by x/(root 2) multiplied by x/(root 2)
which gives area of triangle = 0.5 (x squared)/2 = (x squared)/4
First part done.
Second part in a minute.
love Danniella
Diagram: it's a Toblerone box except it's a right-angled triangle instead of an equilateral triangle, tilted so that it's balancing on its spine (ie resting on the edge connecting the two right angles)
AED is a right-angled triangle, so for the first part you just use Pythagoras.
AE squared + ED squared = x squared. But AE=AD as it's an isosceles triangle,
so 2(AE squared) = x squared,
AE squared = 0.5xsquared
and AE = x/(root 2) also ED = x/(root 2)
Area of triangle = 0.5 x AE x ED
Area of triangle = 0.5 multiplied by x/(root 2) multiplied by x/(root 2)
which gives area of triangle = 0.5 (x squared)/2 = (x squared)/4
First part done.
Second part in a minute.
love Danniella
Hello!
Second part:
Call the length of the tank L.
The volume of this tank is therefore
(x squared/4) L = 4000 (given in the question)
Plan: rearrange to get L on its own, find an expression for the surface area of the tank and then substitute L into that to get an expression entirely in x)
The volume rearranges to L = 16000/(x squared)
Surface area of tank = 2 triangles + two rectangles (length L, width x/(root2)
Area = 2(x squared / 4) + 2(xL/(root 2)
area = x squared/2 + (root 2) xL (rearrange 2/(root 2) = root 2)
area = x squared/2 + (root 2)x multiplied by 16000/(x squared)
area = (x squared)/2 + 16000 (root 2)/x as required.
Presumably you then have to go on to differentiate this , then find dy/dx = 0 and then find the value of x for which the volume is a maximum?
love Danniella
Second part:
Call the length of the tank L.
The volume of this tank is therefore
(x squared/4) L = 4000 (given in the question)
Plan: rearrange to get L on its own, find an expression for the surface area of the tank and then substitute L into that to get an expression entirely in x)
The volume rearranges to L = 16000/(x squared)
Surface area of tank = 2 triangles + two rectangles (length L, width x/(root2)
Area = 2(x squared / 4) + 2(xL/(root 2)
area = x squared/2 + (root 2) xL (rearrange 2/(root 2) = root 2)
area = x squared/2 + (root 2)x multiplied by 16000/(x squared)
area = (x squared)/2 + 16000 (root 2)/x as required.
Presumably you then have to go on to differentiate this , then find dy/dx = 0 and then find the value of x for which the volume is a maximum?
love Danniella
@danniella although your method is correct. But you can just use A= 0.5 ab SinC to get to the answer.
Amusing that AE = y
sin90/x = sin45/y
from there y=xsqrt2/2
putting that in the formula of A= 0.5 ab SinC you would end up with A = 1/4 * x^2
Amusing that AE = y
sin90/x = sin45/y
from there y=xsqrt2/2
putting that in the formula of A= 0.5 ab SinC you would end up with A = 1/4 * x^2
Original post by dan94adibi
@danniella although your method is correct. But you can just use A= 0.5 ab SinC to get to the answer.
Amusing that AE = y
sin90/x = sin45/y
from there y=xsqrt2/2
putting that in the formula of A= 0.5 ab SinC you would end up with A = 1/4 * x^2
Amusing that AE = y
sin90/x = sin45/y
from there y=xsqrt2/2
putting that in the formula of A= 0.5 ab SinC you would end up with A = 1/4 * x^2
Around 6 years late, but better now than never.
Original post by tehforum
Around 6 years late, but better now than never.
yh lol xD but i'm doing C2 now so yh. I still don't kind of understand part B.
Original post by Sint
Where does the root 2 come from? 
This is a very old thread. If you require help with this question, please make a new thread about it as you're more likely to get your question answered there.
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