Question

Sameera wants to celebrate the fifth birthday of her daughter with a party. She
bought thick paper to make the conical party caps. Each cap is to have a base
diameter of 10 cm and height 12 cm. A sheet of the paper is 25 cm by 40 cm and
approximately 82% of the sheet can be effectively used for making the caps after
cutting. What is the minimum number of sheets of paper that Sameera would need to
buy, if there are to be 15 children at the party? (Use = 3.14)​

Answer 1

$\underline { \blue { Dimensions \: a \: sheet \:paper :}}$

Length (l) = 40 cm,

Breadth (b) = 25 cm ,

$Area \: of \:the \: sheet (A) = l \times b \\= 40 \:cm \times 25 \: cm \\= 1000 \:cm^{2}$

$Area \: of \: Used \: sheets = 82 \% \:of \: A\\= \frac{82}{100}\times 1000\\= 820 \:cm^{2}$

$\underline { \blue { Dimensions \: a \: cap :}}$

Diameter (d) = 10 cm,

Radius (r) = 5 cm ,

Height (h) = 12 cm

$Slant \: height(l) = \sqrt{r^{2} + h^{2}}\\= \sqrt{5^{2} + 12^{2}}\\= \sqrt{25+144}\\= \sqrt{169}\\= \sqrt{13^{2}} \\= 13 \:cm$

$Area \: of \: sheet \: required \: to \:make \:1\:cap(A_{1})\\= Curved \:surface \:area \:of \:cone \\= \pi rl\\= 3.14 \times 5 \times 13 \\= 204.1 \:cm^{2}$

$Let \\number \: caps \:made \: by \: 1 \:sheet= n$

$n = \frac{A}{A_{1}} \\= \frac{820}{204.1} \\= 4.01\\≈ 4 \: caps$

$Number \: sheets \: required \:to \:make \\15 \:caps = \frac{15}{4} \\= 3.75 sheets \\= 3\frac{3}{4} \: sheets$

Therefore.,

$Number \: sheets \: required \:to \:make \\15 \:caps = \green { 3\frac{3}{4} \: sheets }$

•••♪

Answer 2

• Diameter of cone = 10 cm=> radius = 5 cm
• height = 12 cm

by pythagoras therome

radius² + height² = slant length²

5² + 12² = L²

25 + 144 =L²

L² = 169

L = 13 cm

Now,

Curved Surface Area of cone = π r l

 = 3.14 × 5 × 13

 = 204.1 cm²

No of caps = 15 (given)

∴ total area required = 204.1 ×15 = 3061.5 cm²

• Area of sheet = 25 ×40 = 1000 cm²

 82 % can be effectively used.(given)

∴ Area that can be used = 820 cm²

No of sheets required =3061.5/820

∴ No of sheets required = 4