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surface area..
I seem to get close to the answer, but it is not happening, could anyone help?..
A tray is made from sheet metal. The horizontal base is a rectangle measuring 8x cm by y cm and the two vertical sides are trapezia of height x cm with parallel edges of length 8x cm and 10x cm. The remainig two sides are rectangles inclined at 45degrees to the horizontal.
given that the capacşty of the tray is 900cm^3
a) find an expression for y in terms of x,
I get:
9 x^2 y = 900
x^2 y = 100
y = 100 / x^2
b) show that the area of metal used to make the tray, A cm^2 is given by
A = 18x^2 + [200(4+[sqroot]2)] / x
i keep on getting:
A = 18x^2 + 1000/x
Thanks in advance..
A tray is made from sheet metal. The horizontal base is a rectangle measuring 8x cm by y cm and the two vertical sides are trapezia of height x cm with parallel edges of length 8x cm and 10x cm. The remainig two sides are rectangles inclined at 45degrees to the horizontal.
given that the capacşty of the tray is 900cm^3
a) find an expression for y in terms of x,
I get:
9 x^2 y = 900
x^2 y = 100
y = 100 / x^2
b) show that the area of metal used to make the tray, A cm^2 is given by
A = 18x^2 + [200(4+[sqroot]2)] / x
i keep on getting:
A = 18x^2 + 1000/x
Thanks in advance..
Reply 1
Reply 2
n0b0dy
OP
1
Rabite
Er, here's what I make of it ~
where did you get the [sq root]2 from..
that's what i seem to be missing.. Reply 3
Ah.
Okay.
Um.
The height is x.
The angle is 45°, which means length = height.
So length of that triangle is x, as is height.
Use pythagoras (or trig) to see that the height is sqrt(x²+x²) = sqrt(2)x
Okay.
Um.
The height is x.
The angle is 45°, which means length = height.
So length of that triangle is x, as is height.
Use pythagoras (or trig) to see that the height is sqrt(x²+x²) = sqrt(2)x
Reply 4
n0b0dy
OP
1
but we are just trying to find the area of each side and add them together, right? i don't see what the angle has got to do with it..
Reply 5
you need to use right angled trig to find the length of the diagonal side in order to find it's area
Reply 6
Uhmm... *how to explain*
Okay, we have five "bits" of our tray.
The base (you're fine with this)
The trapeziums (you're fine with those two)
And whatever's left.
"Whatever's left" is two rectangles.
And the area of a rectangle is, of course, the product of the lengths of their two sides.
One side is 100/x², and the other side is the line at 45° to the horizontal.
Actually screw it, I'll draw it for you :P
Right. Here we go. So the area we want is (two times of) b*h.
b = 100/x² by looking at the previous diagram.
And h² = x²+x² by pythagoras.
How can it be found that the two sides (that are more like construction lines) are both x, you say?
Well, we know that the vertical line in the right triangle must be length x, because we are told (somewhere) that the height of the tray is 'x'.
And since there's a 45° angle, we know that the horizontal line and the vertical line in the right triangle are the same.
It takes a bit of visualising, but you'll get it ~
Okay, we have five "bits" of our tray.
The base (you're fine with this)
The trapeziums (you're fine with those two)
And whatever's left.
"Whatever's left" is two rectangles.
And the area of a rectangle is, of course, the product of the lengths of their two sides.
One side is 100/x², and the other side is the line at 45° to the horizontal.
Actually screw it, I'll draw it for you :P
Right. Here we go. So the area we want is (two times of) b*h.
b = 100/x² by looking at the previous diagram.
And h² = x²+x² by pythagoras.
How can it be found that the two sides (that are more like construction lines) are both x, you say?
Well, we know that the vertical line in the right triangle must be length x, because we are told (somewhere) that the height of the tray is 'x'.
And since there's a 45° angle, we know that the horizontal line and the vertical line in the right triangle are the same.
It takes a bit of visualising, but you'll get it ~
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