A4. For quadrilateral $A B C D$, it is given: $|A B|=3,|B C|=4,|C D|=5$, $|D A|=6$ and $\angle A B C=90^{\circ}$. ( $|A B|$ stands for the length of line segment $A B$, etc.) What is the area of quadrilateral $A B C D$?
A) 16
B) 18
C) $18 \frac{1}{2}$
D) 20
E) $6+5 \sqrt{11}$

![](https://cdn.mathpix.com/cropped/2024_04_17_17b311164fb12c8a5afcg-1.jpg?height=326&width=328&top_left_y=2328&top_left_x=1582)
A4. B) 18 According to the Pythagorean theorem, $|A C|=5$. Triangle $A C D$ is thus isosceles with apex $C$. In this triangle, the altitude from $C$ divides the triangle into two triangles with sides 3, 4, and 5. Quadrilateral $A B C D$ can thus be divided into three triangles with sides 3, 4, and 5. The required area is therefore $3 \times 6=18$.

![](https://cdn.mathpix.com/cropped/2024_04_17_7fefb8cd27a15e1d29a5g-1.jpg?height=318&width=312&top_left_y=1592&top_left_x=1573)