Volume and Surface Areas Quiz Set 008

Let the dimensions of the cuboid be L, 2L and L. Its total surface area is 2(L × 2L + 2L × L + L × L) = 10L². When it is placed on the table, the bottom face is hidden, so net visible area is 10L² - 2L² = 8L². When the cubical block of side L is placed on this block its(i.e., cube's) own L² and L² of the cuboidal block is hidden. So it hides 2L². Thus, the net visible is 8L² + 6L² - 2L² = 12L² = 12 × 6² = 432 sq. units.

One of the right triangles of the isosceles triangle has its base = 16/2 = 8. By Pythagorean theorem, the height = $√{10^2 - 8^2}$ = $√{100 - 64}$ = 6. The radius of the base of the cone r = 8 cm, and height h = 6 cm. The volume is $1/3$π$(r^2 × h)$ = $1/3$π$(8^2 × 6)$ = 128π.

Let the sides be L, B and H. Then LB = 3, BH = 4 and LH = 12. Multiplying all three of them L²B²H² = (3 × 4 × 12) = 144 = 12², which gives LBH = 12 = volume of the cuboid.

The volume of water to be displaced is 2 × 4 × 1 cu. m. This should be equal to the volume of the boxes to be dropped. So N × 1 × 1 × 1 = 2 × 4 × 1, which gives N = 2 × 4 = 8.

Let the side of a cube be L. Then L³ = 343, which gives L = 7. The resulting cuboid has L = L, H = L, and B = 2L. The surface area is 2 × (LB + BH + HL) = 2 × (2L² + 2L² + L²) = 10 × L² = 10 × 7² = 490 sq. cm.