7. Given two lines with slopes of $2$ and $\frac{1}{2}$ intersecting at the point $(2,2)$. The area of the triangle formed by these lines and a third line $x+y=10$ is ( ).
(A) 4
(B) $4 \sqrt{2}$
(C) 6
(D) 8
(E) $6 \sqrt{2}$
7. C.

From the problem, we can set the equations of the two lines as
$$
l_{1}: y=2 x+b_{1}, l_{2}: y=\frac{1}{2} x+b_{2} \text {. }
$$

Substituting the point $(2,2)$, we get
$$
l_{1}: y=2 x-2, l_{2}: y=\frac{1}{2} x+1 \text {. }
$$

Thus, the intersections with the line $l_{3}: y=-x+10$ are at points
$$
B(4,6), C(6,4) \text {. }
$$

Therefore, the area of the triangle formed by points $A(2,2) 、 B(4,6) 、 C(6,4)$ is 6.