Q. 85.0( 4 Votes )

# A bucket is in th

Answer :

Given.

Volume of frustum = 12308.8 cm^{3}

Radii of frustum are 20cm and 12cm

Formula used/Theory.

CSA of frustum = π[R + r]l

Volume of frustum = h[R^{2} + r^{2} + Rr]

Volume of frustum = h[R^{2} + r^{2} + Rr]

= × h[(20cm)^{2} + (12cm)^{2} + 20cm × 12cm]

= × h[400cm^{2} + 144cm^{2} + 240cm^{2}]

= × h × 784cm^{2}

= × h × 22 × 112

= × 22 × 112 × h = 12308.8 cm^{3}

h = = 15 cm

In frustum

(Slant height)^{2} = (difference in radii )^{2} + (height)^{2}

(Slant height)^{2} = (R - r)^{2} + (height)^{2}

(Slant height)^{2} = (20cm – 12cm)^{2} + (15 cm)^{2}

(Slant height)^{2} = 64 cm^{2} + 225 cm^{2}

(Slant height)^{2} = 289 cm^{2}

Slant height = √ (289 cm^{2})

Slant height = 17 cm

CSA of cone = π[R + r]l

= [12cm + 20cm] × 17cm

= × 32cm × 17cm

= 1709.71 cm^{2}

Rate of 1 cm^{2} = Rs. 10

Rate of 1709.71 cm^{2} = Rs. 10 × 1709.71

= Rs. 17097.1

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