Let $t$ be TNYWR.

Quadrilateral $A B C D$ has vertices $A(0,3), B(0, k), C(t, 10)$, and $D(t, 0)$, where $k>3$ and $t>0$. The area of quadrilateral $A B C D$ is 50 square units. What is the value of $k$ ?

## The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca 2020 Canadian Team Mathematics Contest  

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Using the information given, the quadrilateral looks like

![](https://cdn.mathpix.com/cropped/2024_04_17_a164435ebfb99ea5f1c4g-23.jpg?height=688&width=707&top_left_y=1938&top_left_x=812)

We note that the diagram has been drawn with $k3$, so $k-3$ is positive and makes sense as a length.

The area of the trapezoid is

$$
\frac{t}{2}(A B+C D)=\frac{t}{2}(k-3+10)=\frac{t}{2}(k+7) .
$$

Since the area is 50 , this means $50=\frac{t}{2}(k+7)$ or $k+7=\frac{100}{t}$ so $k=\frac{100}{t}-7$.

The answer to part (b) is 5 , so $t=5$. This means $k=\frac{100}{5}-7=20-7=13$.

ANSWER: $6,5,13$