For a complete understanding of the question please find the diagram in the file attached.
In the diagram a perpendicular DP is dropped from D on AB as shown.
It is given that [tex] \angle A=45^{\circ} [/tex] and that [tex] \angle B=90^{\circ} [/tex]. Thus, the [tex] \angle C=45^{\circ} [/tex]. This is because we know that the interior angles of any triangle add up to make [tex] 180^{\circ} [/tex].
Thus, [tex] \Delta CDP [/tex] gives:
[tex] tan(\angle C)=\frac{DP}{CP} [/tex]
[tex] tan(45^{\circ})=\frac{5}{CP} [/tex]
[tex] \therefore CP=\frac{5}{tan(45^{\circ})}=\frac{5}{1}=5 [/tex]
Thus, PB=CB-CP=15-5=10
Now, since, [tex] DE\parallel CB [/tex], then by corresponding angles, [tex] \angle D=45^{\circ} [/tex]
Also, we note that DE=PB=10
Thus, in [tex] \Delta ADE [/tex],
[tex] tan(\angle D)=\frac{AE}{DE} [/tex]
[tex] tan(45^{\circ})=\frac{AE}{10} [/tex]
[tex] \therefore AE=10\times tan(45^{\circ})=10\times 1=10 [/tex]
Thus, now to find the area of the triangle [tex] \Delta ADE [/tex] all that we have to do is use the Area formula as:
Area=[tex] \frac{1}{2} [/tex] x base x height =[tex] \frac{1}{2}\times 10\times 10=50 [/tex] square units
