Question

the Length breadth and height of the cuboidal box is 2m 10cm,1m and 80 cm respectively . find the area of canvas required to cover this box

Answer 1

Length=2m10cm In cm=210cm B=1m Incm=100cm Height =80cm As we know that the tsa of cuboid= the area of the canvas.  Tsa of cuboid=2*(lb+lh+bh)    =2*(210*100+210*80+100*80)cm2   =2*(21000+16800+8000)cm2   =2*45800cm2   =91600 cm2 In m2=91600/10000            =9.16m2

Answer 2

$\huge{\underline{\underline{\mathrm{GIVEN:-}}}}$

• Length of cuboidal box = 2m 10 cm
• Breadth of Cuboidal box = 1m
• Height of cuboidal box = 80 cm

$\huge{\underline{\underline{\mathrm{TO\:FIND:-}}}}$

Find the area of canvas required to cover this box = ?

$\huge{\underline{\underline{\mathrm{FORMULA\: USED:-}}}}$

$\large{\underline{\overline{\mathbf{\red{T.S.A. \:of \: cuboid = 2(lb+bh+ hl)}}}}}$

$\huge{\underline{\underline{\mathrm{SOLUTION:-}}}}$

Convert m into cm

1 m = 100 cm

Length = 2m 10cm = 210 cm

Breadth = 1m = 100 cm

Height = 80 cm

Now, we have to find the T.S.A. of the cuboid

Putting the given values in the formula, we get

$\large\rm{ T.S.A. = 2(lb+bh+hl)}$

$\implies$$\rm{T.S.A. = 2(210\times 100 + 100 \times 80 + 80 \times 210)}$

$\implies$$\rm{T.S.A. = 2(21000 + 8000 + 16800)}$

$\implies$$\rm{T.S.A. = 2 \times 45800}$

$\implies$$\large\bf\red{T.S.A. = 91600\: {cm}^{2}}$

Convert cm into m

1 cm = $\bf{\frac{1}{100}\:m}$

91600 cm = $\bf{\frac{91600}{100}\:m}$

$\large{\underline{\overline{\mathbf{\red{T.S.A. = 916\: {cm}^{2}}}}}}$

Thus, the area of canvas required to cover this box is 916 cm²

$\huge{\underline{\underline{\mathrm{FINAL\:ANSWER:-}}}}$

$\huge{\boxed{\boxed{\bf{\red{T.S.A. = 196\: {cm}^{2}}}}}}$