108. The base and the altitude of a triangular metal disc are 66 cm and 28 cm respectively. By drilling a circular hole through this metal disc, its area is reduced to one-third. Find the diameter of the hole. (Take only one side of the disc into consideration).

We are given a triangular disc with a base of 66 cm and an altitude of 28 cm. The area Aₜ of the triangle is   Aₜ = ½ × base × altitude = ½ × 66 × 28 = 66 × 14 = 924 cm². When a circular hole of area A_c is drilled into the disc, the remaining area becomes one-third of the original area:   Remaining area = 924/3 = 308 cm². Thus, the area removed (the area of the circular hole) is   A_c = Original area – Remaining area = 924 – 308 = 616 cm². Assuming that π = 22/7 (which is a common approximation in such problems), we can write the area of the circle as   A_c = (22/7) r² Set the equation:   (22/7) r² = 616 To solve for r², multiply both sides by 7/22:   r² = 616 × (7/22) Notice that 616 ÷ 22 = 28 (since 22 × 28 = 616), so   r² = 28 × 7 = 196 Taking the square root gives   r = √196 = 14 cm. The diameter d of the circular hole is twice the radius:   d = 2r = 2 × 14 = 28 cm. Thus, the diameter of the hole is 28 cm. (Note: “Take only one side of the disc into consideration” means we are only considering the area of one face for the removal of material.)