Question

The radius of the circular ends of a bucket of height 15 cm are 14 cm and r сm (r < 14 cm). If the volume of bucket is 5390 cm3, then find the value of r. [Use π = $\frac{22}{7}$ ] ​

Answer 1

GiveN :-

• The radii of Circular ends of a Bucket of height 15 cm are 14 cm and r cm

• Volume of the Bucket = 5390 cm3

To FinD :-

• The Value of ' r '

SoluTioN :-

We are Given with two radius of a circular ends of a Bucket of

• Height = 15 cm and 14 cm

• Volume = 5390cm3

Volume of Bucket

$\boxed{ \bold{ \pink{ \frac{1}{3}\pi \: r( {R}^{2} + {r}^{2} + Rr)}{} }{} }{}$

$5390 = \frac{1}{3} \times \frac{22}{7} \times 15(196 + {r}^{2} + 14r) \\ \\ \frac{5390 \times 7}{100} = 196 + {r}^{2} + 14 \\ \\ 49 \times 7 = 196 + {r}^{2} + 14r \\ \\ 343 = 196 + {r}^{2} + 14r \\ \\ {r}^{2} + 14r - 147 = 0 \\ \\ {r}^{2} + 21r - 7(r + 21) = 0 \\ \\ r(r - 7) - 7(r + 21) = 0 \\ \\ ( r- 7)(r + 21) \\ \\ r = 7 \: or \: - 21$

• R can never be Negative

Hence,

$\boxed{ \bold{ \red{★Hence \: r = 7}{} }{} }{}$

Answer 2

Answer:

volume of the bucket=1/3πh(R²+r²+Rr)

=5390=1/3*22/7*15(196+r²+14r)

=5390=22/7*5(196+r²+14r)

=5390*7/110=196+r²+14r

=49*7=196+r²+14r

=343=196+r²+14r

=0=196-343+r²+14r

=0=r²+14r-147

=r²+21r-7r-147=0

=r(r+21)-7(r+21)=0

=(r-7)(r+21)=0

r=7 or -21(rejected)

hence r=7

Step-by-step explanation:

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