H3. The diagram shows squares $A B C D$ and $E F G D$. The length of $B F$ is $10 \mathrm{~cm}$. The area of trapezium $B C G F$ is $35 \mathrm{~cm}^{2}$.
What is the length of $A B$ ?
Solution
The point $F$ lies on the diagonal $B D$ of the square $A B C D$, so that $\angle F B C$ is equal to $45^{\circ}$. Let point $X$ lie on $B C$ so that $\angle F X B=90^{\circ}$, as shown in the diagram alongside. Then $\angle X F B=45^{\circ}$ from the angle sum of triangle $B X F$; it follows from 'sides opposite equal angles are equal' that $B X=X F$.
Now, using Pythagoras' Theorem in triangle $B X F$, we obtain $X F=5 \sqrt{2} \mathrm{~cm}$. But $C G F X$ is a rectangle, so that $C G=X F$. Let the length of $A B$ be $a \mathrm{~cm}$. Then $\frac{1}{2} \times 5 \sqrt{2} \times(2 a-5 \sqrt{2})=35$ and so $2 a-5 \sqrt{2}=\frac{14}{\sqrt{2}}$. Hence $2 a=5 \sqrt{2}+14 \frac{\sqrt{2}}{2}$, so that $a=6 \sqrt{2}$. Therefore the length of $A B$ is $6 \sqrt{2} \mathrm{~cm}$.