S.3 maths (measuration)答到幾多得幾多

4.
1/3 x base area x height
=1/3 x [ 12sqrt(3)/4] x sqrt 8 = 1/3[3sqrt(3)][sqrt(8)] = 2sqrt(6) cm ^3

9.
let the height of the cone be x cm, the height of the cylinder be 4x cm,
(x + 4x)^2 + (2.5)^2 = 9^2
x = sqrt(2.99) ,the height of the cone be 1.73 cm
the slant height of the cone
= sqrt[ (sqrt2.99)^2 + (2.5)^2]
=sqrt 9.24 = 3.04 cm
the total curved surface area
= (pie)rl + 2(pie)rh
= (pie)(2.5)(sqrt9.24) + 2(pie)(2.5)(4sqrt2.99)
= 42.2 cm^2

10.
the radiua of each sector R = sqrt[5^2 + 12^2] = 13cm
the base circumference of the new cone C = 2(5)(pie) + 2(5)(pie) = 20(pie)
the base radius (r) of the new cone = 20(pie) /2(pie) = 10 cm
the height = sqrt[ 13^2 - 10^2 ] = sqrt(69) = 8.31cm
the volume of the new cone C = 1/3 x 10 ^2 x sqrt(69) x (pie)
the total volume of A and B = 1/3 x 5^2 x 12 x (pie) x 2

14.
4/3 x (pie) x 3^3 = 1/3 x r^2 x(12)x (pie)
r = 3 cm

17.
let the original spherical balloon be V, then the new spherical balloon be (1 -48.8%)V = 0.512V
the radius of the original spherical balloon be R, the radius of the spherical balloon be r
[r/R]^3 = 0.512V/V
r/R = 0.8
let the surface area of original spherical balloon be A, the surface area of new spherical balloon be a
[r/R]^2 = a/A
0.8^2 = a/A
a = 0.64A
% decrease in its surface area :
[ A - a]/A x 100% = [A - 0.64A]/A x 100% = 36%