3. In an isosceles trapezoid $\mathrm{ABCD}$, where $\overline{\mathrm{AD}}=\overline{\mathrm{BC}}=6 \mathrm{~cm}$, the diagonal divides the midline into segments of $2 \mathrm{~cm}$ and $5 \mathrm{~cm}$. Determine:

a) the perimeter of the trapezoid

b) the angles of the trapezoid
Solution. Let's denote the midline of the trapezoid as MN, and the intersection of MN with the diagonal $\mathrm{AC}$ as $\mathrm{S}$. Then, from the condition of the problem, $\overline{\mathrm{MS}}=2 \mathrm{~cm}$ and $\overline{\mathrm{SN}}=5 \mathrm{~cm}$. Since MS is the midline of triangle ACD, it follows that $\overline{\mathrm{DC}}=2 \cdot \overline{\mathrm{MS}}=4 \mathrm{~cm}$. Similarly, $\overline{\mathrm{AB}}=2 \cdot \overline{\mathrm{SN}}=10 \mathrm{~cm}$.

a) The perimeter of the trapezoid is $\mathrm{L}=26 \mathrm{~cm}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_02cd33546561a568c7abg-3.jpg?height=298&width=432&top_left_y=1884&top_left_x=1053)
b) Let's denote the height of the trapezoid as DE. Then $\overline{\mathrm{AE}}=\frac{10-4}{2}=3 \mathrm{~cm}$. Therefore, $\triangle \mathrm{AED}$ is a right triangle with the leg $3 \mathrm{~cm}$ and the hypotenuse $6 \mathrm{~cm}$, so $\triangle \mathrm{EAD}=60^{\circ}$ and $\square \mathrm{ADC}=180^{\circ}-60^{\circ}=120^{\circ}$.