Question

answer this question ​

Answer 1

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hence,

➡A cylinder is hollowed out in coral cavity .

GIVEN

diameter of cylinder = diameter of conal cavity = 1.4

❇ Radius of cylinder = radius of conal cavity

$\frac{↪diameter }{2} = \frac{1.4}{2} = 0.7$

➡Hight of cylinder = hight of conal cavity = 2.4

↪slant height of conal cavity,

$↪l = \sqrt{ {h}^{2} + {r}^{2} }$

$↪ \sqrt{ {2.4}^{2} + {0.7}^{2} }$

$↪ \sqrt{5.76 + 0.49}$

$↪ \sqrt{6.25}$

$↪2.5$

Hence,

♠TSA of remaining solid = CSA of Coval cavity + CSA of cylinder + Area of the base of cylinder

$↪\pi \: rl \: + 2\pi \: rh + \pi \: {r}^{2}$

$↪\pi \: r(l + 2h + r)$

$↪ \frac{22}{7} \times 0.7 \times (2.5 + 2 \times 2.4 + 0.7)$

$↪22 \times 0.7 \times (2.5 \times 4.8 \times 0.7)$

$↪2.2 \times 8$

$↪17.6$

$\large\mathtt \green {Answer }$

✏ 17.6 CM ≈ 18 CM

Answer 2

Answer:

A cylinder is hollowed out in coral cavity .

GIVEN

diameter of cylinder = diameter of conal cavity = 1.4

❇ Radius of cylinder = radius of conal cavity

\frac{↪diameter }{2} = \frac{1.4}{2} = 0.7

2

↪diameter

=

2

1.4

=0.7

➡Hight of cylinder = hight of conal cavity = 2.4

↪slant height of conal cavity,

↪l = \sqrt{ {h}^{2} + {r}^{2} }↪l=

h

2

+r

2

↪ \sqrt{ {2.4}^{2} + {0.7}^{2} }↪

2.4

2

+0.7

2

↪ \sqrt{5.76 + 0.49}↪

5.76+0.49

↪ \sqrt{6.25}↪

6.25

↪2.5↪2.5

Hence,

♠TSA of remaining solid = CSA of Coval cavity + CSA of cylinder + Area of the base of cylinder

↪\pi \: rl \: + 2\pi \: rh + \pi \: {r}^{2}↪πrl+2πrh+πr

2

↪\pi \: r(l + 2h + r)↪πr(l+2h+r)

↪ \frac{22}{7} \times 0.7 \times (2.5 + 2 \times 2.4 + 0.7)↪

7

22

×0.7×(2.5+2×2.4+0.7)

↪22 \times 0.7 \times (2.5 \times 4.8 \times 0.7)↪22×0.7×(2.5×4.8×0.7)

↪2.2 \times 8↪2.2×8

↪17.6↪17.6

\large\mathtt \green {Answer }Answer

✏ 17.6 CM ≈ 18 CM