Answer:

1. a. 26 m

b. 1,37,155 inr

2. 550 cm²

Step-by-step explanation:

1. a. slant height: root(height² + radius²)

slant height = root(10² + 24²)

= root(100 + 576)

= root(676)

= 26 m

b. curved surface area = pie r l

= 3.14×24×26

= 1959.36

costing = 70 × 1959.36 = 1,37,155.2 inr

2. slant height = root(height² + radius²)

= root( 24² + 7²)

= root (576 + 49)

= root (625)

= 25

csa = pie r l

= 22/7 × 7 × 25

= 22×25

= 550 cm²

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