The shaded triangle has a base of length $10\text{ cm}.$ Since the triangle is enclosed in a rectangle of height $3\text{ cm},$ then the height of the triangle is $3\text{ cm}.$ (We know that the enclosing shape is a rectangle, because any figure with four sides, including two pairs of equal opposite sides, and two right angles must be a rectangle.) Therefore, the area of the triangle is $$\frac{1}{2}\times 3 \times 10 = \boxed{15\mbox{ cm}^2}.$$
Question:
In the diagram, what is the area of the shaded triangle? [asy]
fill((0,0)--(2,3)--(10,0)--cycle,gray);
draw((0,0)--(10,0)--(10,3)--(0,3)--cycle,linewidth(1));
draw((0,0)--(2,3)--(10,0),linewidth(1));
label("10 cm",(5,3),N);
label("10 cm",(5,0),S);
label("3 cm",(0,1.5),W);
label("3 cm",(10,1.5),E);
draw((0,2.5)--(.5,2.5)--(.5,3));
draw((10,2.5)--(9.5,2.5)--(9.5,3));
[/asy]
Answer:
Let's think step by step.

user
15\mbox{ cm}^2
In the diagram, what is the area of the shaded triangle? [asy]
fill((0,0)--(2,3)--(10,0)--cycle,gray);
draw((0,0)--(10,0)--(10,3)--(0,3)--cycle,linewidth(1));
draw((0,0)--(2,3)--(10,0),linewidth(1));
label("10 cm",(5,3),N);
label("10 cm",(5,0),S);
label("3 cm",(0,1.5),W);
label("3 cm",(10,1.5),E);
draw((0,2.5)--(.5,2.5)--(.5,3));
draw((10,2.5)--(9.5,2.5)--(9.5,3));
[/asy]
test