Question

$\color{red} \sf \: Find \: what \: length \: of \: canvas \: 2 \: m \: in \: width \: is \: required \: to \: make \: a \: conical \: tent \: 20 \: m \: in \\ \color{red} \sf\: diameter \: and \: 42 \: m \: in \: slant \: height \: allowing \: 10\% \: for \: folds \: and \: the \: stitching .\: Also \: \\ \color{red} \sf \: find \: the \: cost \: of \: the \: canvas \: at \: the \: rate \: of \: ₹\:80 \: per \: metre. \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$
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Answer 1

Answer:

35

Step-by-step explanation:

take out ratio of respective triangles to find area then subtract from total area

Answer 2

Answer:

$\color{darkcyan} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

Step-by-step explanation:

Diameter of the base of the conical tent =20 m

$\color{olive} \pmb{\[ \begin{array}{l} \tt \therefore \text { Radius }(r)=\dfrac{20}{2}=10 m \\ \\ \tt \text { and slant height }(h)=42 m \\ \\ \tt \therefore \text { Area of canvas used }=\pi r l \\ \\ \tt =\dfrac{22}{7} \times 10 \times 42 m ^{2}=1320 m ^{2} \\\\ \tt\text { Width of cloth }=2 m \\ \\ \tt\therefore \text { Length of canvas cloth }=\dfrac{1320}{2}=660 m \\ \\ \tt\text { Cloth used in stiching and folding } \\\\ \tt =10 \% \text { of } 660 m \\\\ \tt =\dfrac{660 \times 16}{100}=66 m \\\\ \tt \therefore \text { Total length of cloth required }=660 m +66 m \\ \\ \tt=726 m \\\\ \tt\text { Given rate of cloth }=\text { ₹ } 80 \text { per metre } \\ \\ \tt\therefore \text { Total cost of canvas }=\text { ₹ } 726 \times 80=\text { ₹ } 58080 \end{array} \]}$