Question

2. 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

Answer 2

$\bold\red{\boxed{Solution}}$

$D_{1} = 20cm\:\:\:\:R_{1} = 10cm$

$D_{2} = 12cm\:\:\:\:R_{2} = 6cm$

$h = 3cm$

$Volume\:of\:the\:Frustum$

$\dfrac{1}{3}πh{R_{1}^{2} + R_{2}^{2} + R_{1} + R_{2}}$

$3[10^{2} + 6_{2} + 10 × 6]$

$\Rightarrow \dfrac{22}{7} × 196 = 616cm^{3}$

$let\:the\:slant\: height\:be\:<em>l</em><em>.</em> Then<em> </em>$

$l = \sqrt{(R_{1} + R_{2})^2 + h^{2}}$

$\Rightarrow \sqrt{(10 - 6)^2 + 3^{3}}$

$\Rightarrow \sqrt{16 + 9} = 5$

$TSA\:of\: Frustum\:of\:cone$

$\Rightarrow (R_{1} + R_{2})<em>l</em> + π(R_{1}^{2} + R_{2}^{2})$

$\dfrac{22}{7}[10 + 6] × 5 + \dfrac{22}{7}(10^{2} + 6^{2})$

$\dfrac{22}{7}[80 + 1360] = \dfrac{22}{7} × 216 = 678.85cm^{2}$

$\red{\boxed{Hope\:it\: helps\: you!!}}$