2
$\begingroup$

I have an isosceles trapezoid with an upper base length of $2.6$ units. I know that the length of the legs is $3$ units. I also know that the distance between the bottom of one leg and a point on the opposite leg is $5$ units. The point on the opposite leg is $1$ unit along the leg from the top ($2$ from the bottom). Given these parameters, how can I find the height (or the length of the lower base)?

The Trapezoid

I recognize that the fact that the trapezoid is isosceles ($\angle A=\angle B$), and the fact that $\overline{BE}=1$ is necessary to make both the lower base and height definite, but algebraically I cannot understand how to solve for either. I have attempted using law of cosines with two triangles ($\triangle ABC$ and $\triangle BCE$) and the algebraic relationships between the angles of points on the trapezoid to solve this, but all attempts have led to defining at least one angle as $0^\circ$ or $180^\circ$, always in a situation when not appropriate.

Using GeoGebra, I found that the lower base is approximately $5.7$ units, and the height is approximately $2.5$ units. However, I still need to algebraically solve for exact values.

$\endgroup$

1 Answer 1

1
$\begingroup$

Construction: Drop perpendiculars from points $A,B,E$, to base $CD$.

enter image description here

Let $$CH=GC=3x \;\; \text{and} \;\; AH=BG=3h$$ Since $\triangle BGC\sim \triangle EFC$, $$EF=2h\;,\;FD=2x \;,\; GF=x.$$ Using the Pythagorean Theorem, in $\triangle EFC$ and $\triangle EFD$, $$\begin{align*} x^2+h^2&=1\\ (4x+2.6)^2+4h^2&=25 \end{align*}$$ Solving the above equations, we have $(x,h)=\left(\dfrac{2\sqrt{13\sqrt{109}-95}}{15},\dfrac{2\sqrt{109}-13}{15}\right)$.
The base length is $6x+2.6$ and height is $3h$. $$\;\;\;\text{Base}\approx 5.752244$$ $$\text{Height}\approx 2.552613 $$

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .