Question

$\large\red{ \bf \implies{Question}}$
Find the total surface area and lateral surface area of a shoe box whose length, breath, and height are 30CM, 25CM and 15 cm respectively.
$\large\blue{\bf{this \: is \: for \: mathdude}}$
​

Answer 1

Given that,

=> Length = 30cm

=> Breadth = 25cm

=> Height = 15cm

To find :-

=> Total surface area (TSA) and Lateral surface area (LSA)

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Solution :-

=> TSA = 2(lb+bh+hl)

=> 2( (30×25)+(25×15)+(15×30)

=> 2(750+375+450)

=> 2 ( 1575 )

=> 3150

★ Therefore, Total surface area (TSA) is 3150cm² .

=> LSA = 2h(l+b)

=> 2×15(30+25)

=> 30(55)

=> 1650

★ Therefore, Lateral surface area (LSA) is 1650cm².

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Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

• Length of shoe box, l = 30 cm

• Breadth of shoe box, b = 25 cm

• Height of shoe box, h = 15 cm

We know,

Total Surface Area of cuboid of length l, breadth b and height h is given by

$\boxed{ \rm{ \:TSA_{(Cuboid)} \: = \: 2(lb \: + \: bh \: + \: hl )\: \: }}$

So, on substituting the values, we get

$\rm \: TSA_{(Cuboid)} = 2(30 \times 25 + 25 \times 15 + 15 \times 30)$

$\rm \: TSA_{(Cuboid)} = 2(750 + 375 + 450)$

$\rm \: TSA_{(Cuboid)} = 2 \times 1575$

$\rm\implies \:\boxed{ \rm{ \:TSA_{(Cuboid)} \: = \: 3150 \: {cm}^{2} \: }}$

Now, We know

Curved Surface Area of cuboid of length l, breadth b and height h is given by

$\boxed{ \rm{ \:CSA_{(Cuboid)} \: = \: 2(l + b) \times h \: }}$

So, on substituting the values, we get

$\rm \: CSA_{(Cuboid)} \: = \: 2 \times (30 + 25) \times 15$

$\rm \: CSA_{(Cuboid)} \: = \: 30 \times 55$

$\rm\implies \:\boxed{ \rm{ \:\rm \: CSA_{(Cuboid)} \: = \: 1650 \: {cm}^{2} \: }}$

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Additional Information :-

$\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}$