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lions [1.4K]
3 years ago
15

I don’t know why i’m confused on this one

Mathematics
1 answer:
cestrela7 [59]3 years ago
8 0

dude its going to 14cm

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In Exercise 4, find the surface area of the solid<br> formed by the net.
Fittoniya [83]

Answer:

3. 150.72 in²

4. 535.2cm²

Step-by-step Explanation:

3. The solid formed by the net given in problem 3 is the net of a cylinder.

The cylinder bases are the 2 circles, while the curved surface of the cylinder is the rectangle.

The surface area = Area of the 2 circles + area of the rectangle

Take π as 3.14

radius of circle = ½ of 4 = 2 in

Area of the 2 circles = 2(πr²) = 2*3.14*2²

Area of the 2 circles = 25.12 in²

Area of the rectangle = L*W

width is given as 10 in.

Length (L) = the circumference or perimeter of the circle = πd = 3.14*4 = 12.56 in

Area of rectangle = L*W = 12.56*10 = 125.6 in²

Surface area of net = Area of the 2 circles + area of the rectangle

= 25.12 + 125.6 = 150.72 in²

4. Surface area of the net (S.A) = 2(area of triangle) + 3(area of rectangle)

= 2(0.5*b*h) + 3(l*w)

Where,

b = 8 cm

h = \sqrt{8^2 - 4^2} = \sqrt{48} = 6.9 cm} (Pythagorean theorem)w = 8 cm[tex]S.A =  2(0.5*8*6.9) + 3(20*8)

S.A =  2(27.6) + 3(160)

S.A =  55.2 + 480

S.A = 535.2 cm^2

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Step-by-step explanation:

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