A 12 year old probably should introduce point F and see, that ABD and DFE are congruent and ABD and EFC are similiar.
So we have 3 cm : 4 cm = EF : (1 cm-EF) and this yields EF=BD=3/7 cm.
Now the area of the three right angled triangles can easily be added up to 69/49 cm².
But if you are looking for a more elaborate Mathcad solution, you may take this one 😉
or this one
EDIT: Once we have determined that BD=3/7 cm by whatever method, its easier to calculate the area without using any trigonometry by calculating area of triangle ABC minus area of the half square ADE:
A 12 year old probably should introduce point F and see, that ABD and DFE are congruent and ABD and EFC are similiar.
So we have 3 cm : 4 cm = EF : (1 cm-EF) and this yields EF=BD=3/7 cm.
Now the area of the three right angled triangles can easily be added up to 69/49 cm².
But if you are looking for a more elaborate Mathcad solution, you may take this one 😉
or this one
EDIT: Once we have determined that BD=3/7 cm by whatever method, its easier to calculate the area without using any trigonometry by calculating area of triangle ABC minus area of the half square ADE: