The area of a triangle metal plate with base 88 cm and altitude 64 cm is to be reduced to one-fourth of its original area by making a hole of circular shape at the center. The radius of this hole will be:-

Area of triangular metal plate with base(b) = 88 cm and altitude(h) = 64 cm is given as A = $$\frac{1}{2} b\times h = \frac{1}{2} 88\times 64 = 2816 cm^2$$

Given this area is to be reduced to one-fourth by making a hole in the shape of circle

$$\Rightarrow$$ Reduction in the area of the triangle = Area of the circular hole

$$\Rightarrow\frac{3}{4}\times A=\pi r^2$$

$$\Rightarrow \frac{\frac{3}{4}\times 2816}{\pi} = r^2$$

$$\Rightarrow r = \sqrt672 = 4\sqrt42 $$

So, the radius of the circular hole = $$4\sqrt42$$ cm.