Unusual Quadrilateral
This quadrilateral has an unusual shape. Are you able to find its area?
Image
![Unusual Quadrilateral Unusual Quadrilateral](/sites/default/files/styles/large/public/thumbnails/content-id-6739-Unusual%252520quadrilateral%252520Stefania.png?itok=7-jcQSzl)
What is the area of this quadrilateral, in cm$^2$?
If you liked this problem, here is an NRICH task which challenges you to use similar mathematical ideas.
Image
![Unusual Quadrilateral Unusual Quadrilateral](/sites/default/files/styles/large/public/thumbnails/content-id-6739-Unusual%252520quadrilateral%252520sol%252520Stefania.png?itok=GippreIW)
Label the vertices of the quadrilateral $A$, $B$, $C$ and $D$ as shown.
By Pythagoras theorem, $AC^2=AD^2+DC^2=7^2+9^2=130$, so again by Pythagoras, $AB^2=130-BC^2=130-3^2=121$. Therefore $AB=11$cm.
The area of the quadrilateral is the area of the top right angled triangle plus the area of the bottom right angled triangle $=\frac{7\times 9}{2}+\frac{3\times 11}{2}=\frac{63+33}{2}=\frac{96}{2}=48$cm$^2$.
Image
![Unusual Quadrilateral Unusual Quadrilateral](/sites/default/files/styles/large/public/thumbnails/content-id-6739-Unusual%252520quadrilateral%252520sol%252520Stefania.png?itok=GippreIW)
By Pythagoras theorem, $AC^2=AD^2+DC^2=7^2+9^2=130$, so again by Pythagoras, $AB^2=130-BC^2=130-3^2=121$. Therefore $AB=11$cm.
The area of the quadrilateral is the area of the top right angled triangle plus the area of the bottom right angled triangle $=\frac{7\times 9}{2}+\frac{3\times 11}{2}=\frac{63+33}{2}=\frac{96}{2}=48$cm$^2$.