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Trapezoid

Joe_1234

New member
May 15, 2020
25
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.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,588
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

Administrator
Staff member
Feb 24, 2012
13,735
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

New member
May 15, 2020
25
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

Administrator
Staff member
Feb 24, 2012
13,735
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

Administrator
Staff member
Feb 24, 2012
13,735
Here is a graph:



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. :)