Unusual quadrilateral
This quadrilateral has an unusual shape. Are you able to find its area?
Age
14 to 16
Challenge level
Problem
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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.
Getting Started
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Student Solutions
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$^2$.
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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$^2$.