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I would consider how long a line will be that cuts the trapezoid parallel to the bases. We know this length \(L\) will decrease linearly as we move from \(0\) to \(h\), and in fact, the line will contain the points:

\(\displaystyle L(0)=B\)

\(\displaystyle L(h)=b\)

And so:

\(\displaystyle L(y)=\frac{b-B}{h}y+B\)

Now, we require:

\(\displaystyle \frac{y}{2}(B+L(y))=\frac{h}{4}(B+b)\)

\(\displaystyle \frac{y}{2}\left(B+\frac{b-B}{h}y+B\right)=\frac{h}{4}(B+b)\)

\(\displaystyle 2y\left(B+\frac{b-B}{h}y+B\right)=h(B+b)\)

\(\displaystyle 2y(2Bh+(b-B)y)=h^2(B+b)\)

Arrange as quadratic in \(y\) in standard form:

\(\displaystyle 2(B-b)y^2-4Bhy+h^2(B+b)=0\)

Can you proceed?

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Yes. Tnx

I would consider how long a line will be that cuts the trapezoid parallel to the bases. We know this length \(L\) will decrease linearly as we move from \(0\) to \(h\), and in fact, the line will contain the points:

\(\displaystyle L(0)=B\)

\(\displaystyle L(h)=b\)

And so:

\(\displaystyle L(y)=\frac{b-B}{h}y+B\)

Now, we require:

\(\displaystyle \frac{y}{2}(B+L(y))=\frac{h}{4}(B+b)\)

\(\displaystyle \frac{y}{2}\left(B+\frac{b-B}{h}y+B\right)=\frac{h}{4}(B+b)\)

\(\displaystyle 2y\left(B+\frac{b-B}{h}y+B\right)=h(B+b)\)

\(\displaystyle 2y(2Bh+(b-B)y)=h^2(B+b)\)

Arrange as quadratic in \(y\) in standard form:

\(\displaystyle 2(B-b)y^2-4Bhy+h^2(B+b)=0\)

Can you proceed?

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\(\displaystyle 2(B-b)y^2-4Bhy+h^2(B+b)=0\)

Applying the quadratic formula, we find:

\(\displaystyle y=\frac{4Bh\pm\sqrt{(4Bh)^2-4(2(B-b))(h^2(B+b))}}{2(2(B-b))}=\frac{h(2B\pm\sqrt{2(B^2+b^2)})}{2(B-b)}\)

And so:

\(\displaystyle L\left(\frac{h(2B\pm\sqrt{2(B^2+b^2)})}{2(B-b)}\right)=\frac{b-B}{h}\left(\frac{h(2B\pm\sqrt{2(B^2+b^2)})}{2(B-b)}\right)+B=\pm\frac{\sqrt{2(B^2+b^2)}}{2}\)

Discarding the negative root, we obtain:

\(\displaystyle L(y)=\frac{\sqrt{2(B^2+b^2)}}{2}\)

As we should have expected, the result is independent of \(h\). Plugging in the given data, we find:

\(\displaystyle L=\frac{\sqrt{2(160^2+100^2)}}{2}\text{ m}=10\sqrt{178}\text{ m}\approx133.4\text{ m}\)

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