## Task B-2.7.

An isosceles trapezoid is inscribed in a semicircle of radius $4 \mathrm{~cm}$ such that its larger base is the diameter. Which of all such trapezoids has the maximum perimeter? Justify and determine its area.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_6dc7744662d888032875g-16.jpg?height=494&width=928&top_left_y=296&top_left_x=495)

We have $a=2r=8 \text{ cm}$.

The angle $\varangle A D B$ is a right angle because it is an inscribed angle over the diameter.

Also, $|A E|=\frac{a-c}{2}=4-\frac{c}{2}$.

According to the Euclidean theorem in the right triangle $A B D$:

$$
\begin{array}{rlr}
b^{2} & =|A E| \cdot a & 1 \text{ point } \\
& =\left(4-\frac{c}{2}\right) \cdot 8=32-4 c . &
\end{array}
$$

From this, we get $c=\frac{32-b^{2}}{4}$.

Then, the perimeter of the trapezoid as a function of the leg $b$ is:

$$
o=a+c+2 b=8+\frac{32-b^{2}}{4}+2 b=-\frac{b^{2}}{4}+2 b+16
$$

We see that the perimeter of the trapezoid is a quadratic function of $b$ which achieves its maximum for

$b=\frac{-2}{-\frac{2}{4}}=4$

Then the smaller base $c=\frac{32-b^{2}}{4}=4 \text{ cm}$.

Thus, the trapezoid will have the maximum perimeter if the legs and the smaller base are equal to the radius $r$.

From $|D E|^{2}=b^{2}-|A E|^{2}=r^{2}-\left(\frac{r}{2}\right)^{2}$ we get $|D E|=2 \sqrt{3}$.

The area of the given trapezoid is

$$
P=\frac{a+c}{2} \cdot|D E|=6 \cdot 2 \sqrt{3}=12 \sqrt{3} \text{ cm}^{2}
$$

Note: The area can also be calculated by observing that $\frac{a}{2}=b=c=r$ implies that the trapezoid is half of a regular hexagon with side length $r$. (After concluding $b=c=r$, the calculation of the area is worth 2 points.)

## COUNTY COMPETITION IN MATHEMATICS

3rd grade - high school - B variant
February 4, 2018.

## IF A STUDENT HAS A DIFFERENT APPROACH TO SOLVING THE PROBLEM, THE COMMITTEE IS OBLIGED TO GRADE AND EVALUATE THAT APPROACH APPROPRIATELY.