The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.
Answers
Answer:
Volume of frustum is 1899.33 cm³ , slant height of a frustum is 16.1 cm and Total surface area of the frustum is 860.475 cm²
Step-by-step explanation:
SOLUTION :
Given height of frustum(h)= 16 cm
Let the radius, r1 and radius, r2 be the radii of the circular ends of the frustum and h be its height.
Perimeter of one end of frustum = 44 cm (given)
2π r1= 44
r1= 44/2π
r1 = 22/π
r1 = 22/(22/7)
r1 = 22 × 7/22 = 7
r1 = 7 cm
Perimeter of other end of frustum = 33 cm
2πr2 = 33
r2 = 33/2π
r2 = 33/(2× 22/7) = 33/(44/7)
r2 = 33 × 7/44 = 21/4
r2 = 21/4 = 5.25 cm
Slant height of a frustum , l = √(r1 - r2)² + h²
l = √(7 - 5.25)² + 16²
l = √1.75² + 256
l = √3.0625 + 256
l = √259.0625
slant height of a frustum,l = 16.1 cm
Volume of Frustum (V) = 1/3 π {(r1)²+ (r2)² + (r1r2)} h
V = ⅓ π {(7)² + (5.25)² + (7 x 5.25)} × 16
V = ⅓ × 22/7 {49 + 27.5625 + 36.75 }× 16
V = (22 × 16 × 113.3125)/21
V = 39866/21
V = 1899.33
Volume of frustum is 1899.33 cm³.
Curved surface area of a frustum = π(r1+ r2)l
= 22/7(7 + 5.25)× 16.1
= 22/7 × 12.25 × 16.1
= 22 × 1.75 × 16.1
= 619.85 cm²
Curved surface area of a frustum = 619.85 cm²
Total surface area of the frustum = π(r1+ r2)l + πr1² + πr2²
= 619.85 + π(r1² + r2²)
= 619.85 + 22/7 (7² + 5.25²)
= 619.85 + 22/7 (49 + 27.5625)
= 619.85 + 22/7× 76.5625
= 619.85 + 1,684.375/7
= 619.85 + 240.625
= 860.475 cm²
Total surface area of the frustum = 860.475 cm²
HOPE THIS ANSWER WILL HELP YOU...
Answer:
ANSWER
2πR
1
=44cm⇒R
1
=
44
44×7
=7cm
2πR
2
=33cm⇒R
2
=
44
33×7
=5.25cm
h=16cm
Volume =
3
1
πh[R
1
2
+R
2
2
+R
1
R
2
]=
3
1
×
7
22
×16[7
2
+5.25
2
+7×5.25]
=1899.33cm
3
≈1900cm
3
l=
(R
1
−R
2
)
2
+h
2
=
(7−5.25)
2
+16
2
=
259.0625
=16.10cm
CSA=π(R
1
+R
2
)l=
7
22
(7+5.25)×16.10=619.85cm
2
TSA=CSA+π(R
1
2
+R
2
2
)=619.85+
7
22
(7
2
+5.25
2
)=860.475cm
2