1. A frustum of a right circular cone of height 16cm with radii of its circular ends as 8cm and 20cm has its slant height equal to?
2. The perpendicular distance of A(5,12) from the y axis is
3. Sum of the area of two squares is 468m2 . If the difference of their perimeters is 24m, Find the sides of the two squares.
Answers
as we know that
l^2 = (r1 -r2)^2 + h^2
here r1 =20 cm ,r2 =8 cm and h =16 cm
now
l^2 = (20-8)^2 + 16^2
= 12^2 + 16^2
= 144 + 256
l^2 = 400
i.e. l= 20
Slant height = 20 cm
Step-by-step explanation:
radii of its circular ends as 8cm and 20cm
let say AB = 20 cm Radius of bottom
CD = 8 cm Radius of top
BD = height = 16 cm
AC is slant height
BD & AC extend to meet at O
now Δ AOB ≈ Δ COD
=> AO/CO = AB/CD = OB/OD
=> AO/CO = 20/8 = OB/OD
20/8 = OB/OD
=> 5/2 = (OD + 16)/OD
=> 5OD = 2OD + 32
=> 3OD = 32
=> OD = 32/3
OB = 16 + OD = 16 + 32/3 = 80/3
OA² = OB² + AB²
=> OA² = (80/3)² + (20)²
=> OA² = 20² ( (4/3)² + 1)
=> OA² = 20² (5/3)²
=> OA = 100/3
OA/OC = 5/2
=> 100/(3 * OC) = 5/2
=> OC = 40/3
AC = OA - OC = 100/3 - 40/3 = 60/3
=> AC = 20
Slant height = 20 cm
The perpendicular distance of A(5,12) from the y axis is 5
a² + b² = 468
4a - 4b = 24
=> a - b = 6
Squaring both sides
=> a² + b² -2ab = 36
=> 468 -2ab = 36
=> 2ab = 432
(a + b)² = a² + b² + 2ab
=> (a + b)² = 468 + 432
=> (a + b)² = 900
=> a + b = 30
a - b = 6
=> a = 18 , b = 12
Sides of two squares are 18 & 12 m
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