F.4 maths

1.
The ratio of the surface areas
= 4 : 25
= 2² : 5²

The ratio of their corresponding lengths
= 2 : 5

The ratio of their volume
= 2³ : 5³
= 8 : 125


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2.
Let 3k be the base radius of the cone.
Then, the height of the cone = 4k

slant height
= √[(3k)² + (4k)²]
= 5k

base circumference
= 2 x π x 3k
= 6kπ

curved surface area
= (1/2) x 5k x 6kπ
= 15k²π

base area
= π x (3k)²
= 9k²π

(curved surface area) : (total surface area)
= 15k²π : (15k²π +9k²π)
= 15 : 24
= 5 : 8

1. the ratio of the total surface areas of two similar solids is 4:25 = 2^2 : 5^2
so the ratio of the sides of two similar solids is 2:5
the ratio of their volumes is 2^3 : 5^3 = 8:125

2. let the base radius be 3k, the height of the cone be 4k, where k is a non-zero positive constant.
the curved surface area is π(3k)√((3k)2 + (4k)2) = π(3k)√(25k2)
= π(3k)(5k) = π(15k2)
the total surface = π(15k2) (π(3k)2)
so , the ratio is 1 : (π(3k)2)

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