find the lengths of the hypotenuse
whose sides are 12 cm and 16 cm
Answers
Answer:
The volume of the double cone so formed = 1929.21 cm^3
Step-by-step explanation:
From the figure attached with this answer consists of two cone.
The slant height of each cone is 16 cm and 12 cm respectively.
Let x be the base radius of each cone.
h be the height of large cone and 20 - h be the height of small cone
To find h and (20 -h)
From the figure we get,
x^2 = 16 ^2 - h^2 ------(1)
x^2 = 12^2 - (20 - h)^2 ----(2)
16 ^2 - h^2 = 12^2 - (20 - h)^2
256 = 144 -400 -40h
h = 64/5 = 12.8 cm
there fore height of small cone 20 - h = 20 - 12.8 = 7.2cm
To find x
x^2 = 16^2 - h^2 = 256 - 12.8^2 = 92.16
x =9.6 cm
Volume of cone
V = 1/3[πr^2h]
To find the volume of large cone
V1 = 1/3[πr^2h] = 1/3[3.14 x 9.6 x 9.6 x 12.8] = 1234.7 cm^3
To find the volume of large cone
V2 = 1/3[πr^2h] = 1/3[3.14 x 9.6 x 9.6 x 7.2] = 694.5 cm^3
Total volume
Volume V = v1 +v2 = 1234.7 + 694.5 =1929.21 cm^3
Step-by-step explanation:
Answer:
20
Step-by-step explanation:
as it is a right angled triangle
the sum of squares of the given two sides is equal to the square of the hypotenuse
12^2=144
16^2=256
144+256=400
root of 400=20