Question

please help me
CLASS 9
CHAPTER 13
EXERCISE 13.3
QUESTION 5
please don't give irrelevant answers ​

Answer 1

$\sf\large{\underline{\underline{\bold{\star Solution:-}}}}$

$\sf\large{Tarpulin\:will\:be\:on\:the\:Curved\:Surface}$$\sf\large{area\:of\:tent.}$

$\sf\large{Lets\:find\:curved\:surface\:area\:first}$

$\sf\large{Curved\:Surface\:area\:of\:Tent= pie\: r l}$

$\sf\large{\underline{\underline{Given:-}}}$

$\sf\large{\implies Radius\:(r) = 6\:Cm}$

$\sf\large{\implies Height\:(h) = 8\:m}$

$\sf\large{Let\:slant\:height\:be = l}$

$\sf\large{\underline{\underline{We\:know\:that,}}}$

$\sf\large{\implies {l}^{2} = {h}^{2} + {r}^{2}}$

$\sf\large{Putting\:the\:values:}$

$\sf\large{\implies {l}^{2} = {(8)}^{2} + {(6)}^{2}}$

$\sf\large{\implies {l}^{2} = 64 + 36}$

$\sf\large{\implies {l}^{2} = 100}$

$\sf\large{\implies l = \sqrt{100}}$

$\sf\large{\implies l = \sqrt{({10}^{2})}}$

$\sf\large{\implies l = 10m}$

$\sf\large{Length=10m}$

$\sf\large{\star Curved\:Surface\:Area\:of\:tent = pie\: rl}$

$\sf\large{ \: \: \: \: \: \: \: \: \: = (3.14 × 6 × 10)\: {m}^{2}}$

$\sf\large{ \: \: \: \: \: \: \: \: \: = 188.4\: {m}^{2}}$

$\sf\large{Now,}$

$\sf\large{Area\:of\:tarpulin\:material = Area\:of\:Tent}$

$\sf\large{\implies (Length) × (Breadth) = Area \:of \:Tent}$

$\sf\large{\implies (Length) × 3 = Area\:of\:Tent}$

$\sf\large{\implies Length = \dfrac{1}{3}(Area\:of\:Tent)}$

$\sf\large{\implies Length = \dfrac{188.4}{3}}$

$\sf\large{\implies Length = 62.8\:m}$

$\sf\large{Now,}$

$\sf\large{Given\:that\:margin\:is\:20\:cm}$

$\sf\large{\implies Total\:Length = Length\:Calculated + Margin}$

$\sf\large{\implies Total\:Length = 62.8\:m + 20\:cm}$

$\sf\large{\implies Total\:Length = 62.8 + 20 × \dfrac{1}{100}\:m}$

$\sf\large{\implies Total\:Length = 62.8\:m + 0.2\:m}$

$\sf\large{\implies Total\:Length = 63\:m}$

$\sf\large{\underline{\underline{Therefore,}}}$

$\sf\large{\therefore{ Length \:of \:Tarpulin \:Required= 63\:m}}$

Answer 2

QUESTION :-

What length of tarpaulin 3 m wide will be required to make conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that will be required for stitching margins and wastage in cutting is approximately 20 cm.

ANSWER :-

63 m

STEP BY STEP

EXPLANATION :-

➠ LET

• Height of the conical tent → h

• Radius of the conical tent → r

so ,

• h = 8
• r = 6

Slant height of the tent

$→ (l) = \sqrt{ {r}^{2} + {h}^{2} }$

$→ \sqrt{ {(6)}^{2} + {(8)}^{2} }$

$→ \sqrt{36 + 64}$

$→ \sqrt{100}$

→ 10 m

Area of tarpaulin = Curved surface area of tent →

→ πrl

→ 3.14 x 6 x 10

→ 188.4 m²

Width of tarpaulin = 3 m

Let

• Length of tarpaulin → L

★ Area of tarpaulin = Length × Breadth

→ L x 3 = 3L

Now According to question,

3L = 188.4

L = 188.4/3 = 62.8 m

The extra length

→20 cm

→ 0.2 m

So the total length of tarpaulin bought is

→ (62.8 + 0.2) m

→ 63 m