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A conical cup is made from a circular piece of paper with radius 6 cm by cutting out a sector and joining the edges as shown below. Suppose θ = 5π/3.

A conical cup is made from a circular piece of paper with radius 6 cm by cutting out a sector and joining the edges as shown below Suppose θ 5π3 class=

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The [tex]r[/tex], [tex]h[/tex], and 6 cm form a right triangle, so you find [tex]h[/tex] with the Pythagorean Theorem.

[tex]r^{2}+h^{2}=6^{2}[/tex]
We have [tex]r=5[/tex], so
[tex]5^{2}+h^{2}=6^{2}[/tex]
[tex]\rightarrow 25+h^{2}=36[/tex]
[tex]\rightarrow h^{2}=11[/tex]
[tex]\rightarrow h=\sqrt{11}[/tex]

Answer:

(a) 10π cm

(b) r = 5 cm

(c) h = √11 cm

Step-by-step explanation:

(a) The surface of the opening the cup is circular in shape and a sector is cut from circle is calculate by formula,

C = r × θ

here, r = 6 cm

and θ = [tex]\frac{5\pi}{3}[/tex]

⇒ [tex]C = 6 \times \frac{5\pi}{3} = 10\pi[/tex]

(b) For finding the radius of the cup,

As the circumference of the circle is 10π

and we know that area of cone is calculate by formula, 2πr

⇒ 2πr = 10π

⇒ r = 5 cm

(c) In cone we know slant height(l) = 6 cm

Radius = 5 cm

Thus, using Pythagoras theorem,

l² = h² + r²

⇒ h² = l² - r²

⇒ h² = 36 - 25

⇒ h = √11 cm

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