When the base of a triangle is increased $10\%$ and the altitude to this base is decreased $10\%$, the change in area is

$\text{(A)} \ 1\%~ \text{increase} \qquad \text{(B)} \ \frac12 \%~ \text{increase} \qquad \text{(C)} \ 0\% \qquad \text{(D)} \ \frac12 \% ~\text{decrease} \qquad \text{(E)} \ 1\% ~\text{decrease}$
1. Let the original base of the triangle be \( b \) and the original altitude (height) be \( h \). The area \( A \) of the original triangle is given by:
   \[
   A = \frac{1}{2} b h
   \]

2. When the base is increased by \( 10\% \), the new base \( b_1 \) is:
   \[
   b_1 = b + 0.1b = 1.1b
   \]

3. When the altitude is decreased by \( 10\% \), the new altitude \( h_1 \) is:
   \[
   h_1 = h - 0.1h = 0.9h
   \]

4. The area \( A_1 \) of the new triangle with base \( b_1 \) and altitude \( h_1 \) is:
   \[
   A_1 = \frac{1}{2} b_1 h_1 = \frac{1}{2} (1.1b) (0.9h)
   \]

5. Simplifying the expression for \( A_1 \):
   \[
   A_1 = \frac{1}{2} \times 1.1 \times 0.9 \times b \times h = \frac{1}{2} \times 0.99 \times b \times h = 0.99 \times \frac{1}{2} b h
   \]

6. Comparing the new area \( A_1 \) with the original area \( A \):
   \[
   A_1 = 0.99 A
   \]

7. The change in area is:
   \[
   \Delta A = A_1 - A = 0.99A - A = -0.01A
   \]

8. The percentage change in area is:
   \[
   \frac{\Delta A}{A} \times 100\% = \frac{-0.01A}{A} \times 100\% = -1\%
   \]

Thus, the area of the triangle decreases by \( 1\% \).

The final answer is \(\boxed{1\% \text{ decrease}}\).