(c) 9\(\frac{1}{11}\)%
Let the altitude of the triangle be h and corresponding base = b. Then, its area =\(\frac{1}{2}\)bh, Increased altitude = 1.1h,
Area remaining same =\(\frac{1}{2}\)bh
\(\therefore\) Reduced base = \(\frac{\frac{1}{2}bh}{1.1h}\) x 2 = \(\frac{b}{2.2}\) x 2 = \(\frac{10b}{11}\)
\(\therefore\) % reduction = \(\frac{b-\frac{10}{11}b}{b}\)x 100 = \(\frac{100}{11}\)%
= \(9\frac{1}{11}\)%