# Trapezoid

#### Joe_1234

##### New member
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
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

Staff member
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
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

Staff member

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

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:

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