Question

what is the height of the cylindrical box of radius 10cm and total surface are 942sq.cm.​

Answer 1

Given :-

Radius of cylindrical box = 10 cm

Total Surface Area of cylindrical box = 942 cm²

To Find :-

• Height of the cylindrical box

Assumption:-

Let the height of the cylindrical box = h

Formula:-

T.S.A = 2πr (r + h)

Solution:-

As we know T.S.A = 942 cm²

So,

942 = 2πr (r + h)

Radius = 10 cm

⟹ 942 = 2 × 22/7 × 10 (10 + h)

⟹ 942 = 440/7 (10 + h)

⟹ 942 × 7 = 440 (10 + h)

⟹ 6594 = 440 (10 + h)

⟹ 6594/440 = (10 + h)

⟹ 3297/220 = (10 + h)

⟹ 3297/220 - 10/1 = h

⟹ 3297 - 2200/220 = h

⟹ 1097 / 220 = h

Height = 1097 / 220 cm

Verification:-

T.S.A = 2 × 22/7 × 10 (10 + 1097 / 220)

⟹ 942 = 440 / 7 ( 2200 + 1097 / 220)

⟹ 942 = 440 / 7 × 3097 / 220

⟹ 942 = 2 × 471

⟹ 942 = 942

$\therefore$L.H.S = R.H.S

✒ Required Answer :-

Height of the cylindrical box = 1097 / 220 cm

Answer 2

Given:

• Radius of cylindrical box(r)=10cm
• Total surface area(T.S.A) of the cylindrical box= 942sq.cm

To Find:

• Height(h) of the cylindrical box.

Solution:

According to formula,

$\sf{T.S.A \:of \:cylinder = 2 \pi r(h+r)}$

Given that,

T.S.A of cylindrical box = 942cm²

Therefore,

$\sf{2 \pi r(h+r) = 942cm^2}$

$\sf{\implies}{2 \times {\dfrac{22}{7}} \times 10(h+10) = 942cm^2}$

$\sf{\implies}{(h+10)= {\dfrac{942 \times 7}{2 \times 22 \times 10}}}$

$\sf{\implies}{(h+10)= {\dfrac{6594}{440}}}$

$\sf{\implies}{(h+10)=14.986}$

$\sf{\implies}{h=14.986-10}$

$\sf{\implies}{h=4.986cm}$

So, the height of the cylindrical box is 4.986cm[approx].

Verification:

$\sf{T.S.A \:of \:cylinder = 2 \pi r(h+r)}$

$\sf{942=2 \times {\dfrac{22}{7}}\times 10\times(10+4.986)}$

$\sf{\implies}{942=}{\dfrac{2\times \times 22 \times 10 \times 14.986}{7}}$

$\sf{\implies}{942=}{\dfrac{6594}{7}}$

$\sf{\implies}{942=942}$

Hence, Verified.