Q1) A conical tent is 10 m high and the radius of its base is 24 m. Find
(i) slant height of the tent.
(ii) cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70.
Q2) A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
Answers & Comments
Sure, let's solve these problems step by step.
Q1)
(i) The slant height (l) of the tent can be found using the Pythagorean theorem in a right-angled triangle, where the radius (r) and the height (h) are the two perpendicular sides, and the slant height is the hypotenuse. So, l = sqrt(r² + h²) = sqrt(24² + 10²) = 26 m.
(ii) The area of the canvas required to make the tent is the curved surface area of the cone, which is given by πrl. So, the area = π * 24 * 26 = 1974 m². Therefore, the cost of the canvas required = area * cost per m² = 1974 * 70 = Rs 138180.
Q2)
The area of the sheet required to make one cap is the curved surface area of the cone, which is given by πrl. So, the area for one cap = π * 7 * 25 = 550 cm². Therefore, the area of the sheet required to make 10 such caps = 10 * area for one cap = 10 * 550 = 5500 cm².
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Verified answer
Answer:
1) (i) Slant height of the tent is 26 m
(ii) Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70 is Rs 137280
2) Area of sheet required to make 10 such caps is 5500 cm²
Step-by-step explanation:
Given that, a conical tent is 10 m high and the radius of its base is 24.
So, we have
Height of conical tent, h = 10 m
Radius of conical tent, r = 24 m
We know, Slant height (l), height (h) and radius (r) is connected by a relationship
[tex] \sf \: {l}^{2} = {h}^{2} + {r}^{2} \\ [/tex]
[tex] \sf \: {l}^{2} = {(10)}^{2} + {(24)}^{2} \\ [/tex]
[tex] \sf \: {l}^{2} =100 + 576 \\ [/tex]
[tex] \sf \: {l}^{2} =676 \\ [/tex]
[tex] \sf \: l = \sqrt{676} = \sqrt{ {(26)}^{2} } \\ [/tex]
[tex]\implies\sf\:l = 26 \: m \\ [/tex]
Thus, Slant height of a conical tent is 26 m
Now, Canvas required to make a conical tent is equals to Curved Surface Area of conical tent.
Thus,
[tex] \sf \: Amount\:of\:canvas \: required = \pi \: r \: l \\ [/tex]
[tex] \sf \: Amount\:of\:canvas \: required = \dfrac{22}{7} \times 24 \times 26\\ [/tex]
[tex] \implies\sf\: Amount\:of\:canvas \: required = \dfrac{13728}{7} \: {m}^{2} \\ [/tex]
Now, Further given that
[tex] \sf \: Cost\:of\: {1 \:m }^{2} \: of \: canvas = Rs \: 70 \\ [/tex]
So,
[tex] \sf \: Cost\:of\: \dfrac{13728}{7} \: {m }^{2} \: of \: canvas = \dfrac{13728}{7} \times 70 \\ [/tex]
[tex]\implies\sf\: Cost\:of\: \dfrac{13728}{7} \: {m }^{2} \: of \: canvas = Rs \: 137280 \\ [/tex]
Hence,
(i) Slant height of the tent is 26 m
(ii) Cost of the canvas required to make the tent, if the cost of 1 m² canvas is Rs 70 is Rs 137280
[tex]\rule{190pt}{2pt}[/tex]
Given that, A joker's cap is in the form of a right circular cone of base radius 7 cm and height 24 cm.
So, we have
Height of conical tent, h = 24 cm
Radius of conical tent, r = 7 cm
We know, Slant height (l), height (h) and radius (r) is connected by a relationship
[tex] \sf \: {l}^{2} = {h}^{2} + {r}^{2} \\ [/tex]
[tex] \sf \: {l}^{2} = {(24)}^{2} + {(7)}^{2} \\ [/tex]
[tex] \sf \: {l}^{2} =576 + 49 \\ [/tex]
[tex] \sf \: {l}^{2} = 625 \\ [/tex]
[tex] \sf \: l = \sqrt{625} = \sqrt{ {(25)}^{2} } \\ [/tex]
[tex]\implies\sf\:l = 25 \: cm \\ [/tex]
Now, Area of sheet required to make such 10 conical caps is equals to 10 times the Curved Surface Area of cone.
So, we have
[tex] \sf \: Area\:of\:sheet\:required = 10 \: \pi \: r \: l \\ [/tex]
[tex] \sf \: Area\:of\:sheet\:required = 10 \times \dfrac{22}{7} \times 7 \times 25 \\ [/tex]
[tex] \sf \: Area\:of\:sheet\:required = 10 \times 22 \times 25 \\ [/tex]
[tex] \sf \: Area\:of\:sheet\:required = 220 \times 25 \\ [/tex]
[tex]\implies\sf\:Area\:of\:sheet\:required = 5500 \: {cm}^{2} \\[/tex]
Hence, area of sheet required to make 10 such caps is 5500 cm²