Finding the Length of a Dividing Line in a Trapezoid

[Joe_1234]

A trapezoid with a base of 100m and 160m is divided into 2 equal parts by a line parallel to the base. Find the length of dividing line.

 

[anemone]

Hi Joe_1234, can you check and confirm if the question is correctly typed?

Do you have any idea of how to begin tackling the problem?

 

[MarkFL]

Let's let the larger base be \(B\), the smaller base be \(b\) and the height be \(h\).

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?

 

[Joe_1234]

Let's let the larger base be \(B\), the smaller base be \(b\) and the height be \(h\).

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?

Yes. Tnx

 

[MarkFL]

Let's follow up...we left off with the quadratic:

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

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:

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

 

[MarkFL]

Here is a graph:

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Scroll down to the bottom of the expressions on the left and move the sliders. The red dashed line divides the trapezoid into two equal areas. :)