Question

A frustum of a right circular cone has a diameter of base 20 cm, of top 12 cm, and height 3 cm. Find the area of its whole surface and volume.

Answer 1

Answer:

The Whole Surface area of frustum is 678.85 cm²  and Volume of the of frustum = 616 cm³

Step-by-step explanation:

SOLUTION :  

Diameter of top of frustum = 12 cm  

Radius of top of frustum, r = 12/2 = 6 cm  

Diameter of Bottom of frustum = 20 cm

Radius of Bottom of frustum,R = 20/2 = 10 cm

Height of  frustum ,h = 3 cm  

Slant height of frustum,l = √ (R - r)² + h²  

= √(10 - 6)² + (3)²

= √4² + 9 = √16 + 9  

= √25

= 5 cm

slant height ,l =  5 cm

Whole Surface area of frustum = π(R + r)l + πR² + πr²

= 22/7 (10+6)×5 + 22/7 × 10² + 22/7 × 6²  

= 22/7(16×5  + 100 + 36)

= 22/7 (80 + 136)

= 22/7 × 216 = 4752/7  

Whole Surface area of frustum = 678.85 cm²

Volume of  frustum = 1/3πh(R² + r² + Rr)

= ⅓ × 22/7 ×3 (10² + 6² + 10×6)

= 22/7 (100 + 36 + 60)

= 22/7 × 196

= 22 × 28 = 616 cm³

Volume of the of frustum = 616 cm³

Hence, the Whole Surface area of frustum is 678.85 cm²  and Volume of the of frustum = 616 cm³

HOPE THIS ANSWER WILL HELP YOU....

Answer 2

ANSWER:

Given base diameter of cone

 (d1)(d1)=20cmRadius (r1)

(r1) = 202cm=10cm202cm=10c

mTop diameter of Cone (d2)(d2) = 12 cmRadius (r2)(r2) = 122

cm=6cm122cm=6cmHeight of the cone (h)=

3cmVolume of the frustum right circular cone

= Π3(r21+r22+r1r2)hΠ3(r21+r22+r1r2)

h= Π3(102+62+10×6)3Π3(102+62+10×6)3=616 cm3cm3

Let ‘L’ be the slant height of cone=>L

= √(r1−r2)2+h2(r1−r2)2+h2−−−−−−−−−−−−√=>L

= √(10−6)2+32(10−6)2+32−−−−−−−−−−−√=>L

= √2525−−√=>L = 5 cm

∴∴ Slant height of cone (L)

(L)= 5 cm Total surface area of the cone

=Π(r1+r2)×L+Π×r21+Pi×r22Π

(r1+r2)×L+Π×r21+Pi×r22

= Π(10+6)×5+Π×102+Pi×62Π(10+6)×5+Π×102+Pi×62

= Π(80+100+36)Π(80+100+36)

= Π(216)Π(216)= 678.85

cm 2678.85cm2∴∴Total surface area of the cone =678.85 cm2cm2

hope it helps:--