Question

The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at 100 per sq. m.

Answer 1

Answer:

CSA= 48m^2

COST= 4800 RS

Step-by-step explanation:

2πr1=18

r1=9/π

2πr2=6

r2=3/π

csa= πl(r1+r2)

= π×4(9/π+3/π)

= 4π × 12/π

=4×12

= 48cm^2

cost per sq.m = 100 RS

cost for 48 sq.= 100×48

= 4800 RS

Answer 2

Answer:

Let R, r be the radius of the top and bottom of the frustum. h, l be the height and slant height of the frustum.

GIVEN :

• slant height of the frustum l = 4m
• Perimeter of the top of the frustum = 18m

2πR = 18

• Radius of the top region of the

frustum R = 18/2π m

• Perimeter of the bottom of the

frustum = 16m

2πr = 16m

• Radius of the bottom region of the

frustum r = 16/2π m

C.S.A. of the frustum = π (R + r) l sq.units

= π (18/2π + 16/2π) 4

$= \pi \times 1 \div 2\pi \times (18 + 16) \times 4$

• C.S.A. of the frustum = 1/2 × 34 × 4 = 68m^2
• Cost for painting 1 sq.m = Rs. 100

∴ cost for painting 68 sq.m = 68 × 100

= ₹ 6,800

∴ Cost for painting the C.S.A. of the frustum = ₹ 6,800

Step-by-step explanation:

@GENIUS