The diameters of the top and the bottom portions of a bucket are 42 cm and 28 cm respectively. If the height of the bucket is 24 cm, then the cost of painting its outer surface at the rate of 50 paise / cm² is
(a)Rs. 1582.50
(b)Rs. 1724.50
(c)Rs. 1683
(d)Rs. 1642
Answers
Answer:
The Cost of painting its outer surface is ₹ 1683 .
Among the given options option (c) ₹ 1683 is the correct answer
Step-by-step explanation:
Given :
Diameter of top of bucket = 42
Radius of top of bucket ,R = 42/2 = 21 cm
Diameter of bottom of bucket = 28
Radius of bottom of bucket ,r = 28/2 = 14 cm
Height of the bucket (h) = 24 cm
Slant height of a frustum , l = √(R - r)² + h²
l = √(21 - 14)² + 24²
l = √7² + 24²
l = √49 + 576
l = √625
l = 25 cm
Slant height of the bucket ,l =25 cm
Outer Surface area of a bucket to be painted,S = curved surface area of bucket + area of bottom
S = π(R+ r)l + πr²
S = π(21 + 14) 25 + π × 14²
S = π × 875 + 196 π
S = π(875 + 196)
S = π × 1071
S = 22/7 × 1071
S = 22 × 153
S = 3366 cm²
Outer Surface area of a bucket to be painted = 3366 cm²
Rate of painting = 50 paise per cm²
Cost of painting its outer surface = 3366 × 50/100 = 3366 × ½ = ₹ 1683
Hence, the Cost of painting its outer surface is ₹ 1683 .
HOPE THIS ANSWER WILL HELP YOU….
SOLUTION ☺️❣️✌️
OPTION = (c)✓
Radius of the top bucket
=) r1= 42/2= 21cm
& Radius of the bottom
=) r2= 28/2= 14cm
Height of the bucket
=) h= 24cm
=) l= √h^2(r1-r2)
=) √576+(21-14)^2
=) √576+49
=) √625
=) 25
C.S.A of the bucket
=) π(r1+r2)l
=) π(21+14)×25
=) 22/7× 35×25
=) 2750 cm^2
Area of buttom
=) πr^2
=) 22/7× 196
=) 616cm^2
Now, the cost of painting its C.S
=) (2750+616)×1/2
=) 3366×1/2
=) Rs. 1683 [answer].
Hope it helps✌️