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8. K th 4. Use the formula S.A. = 2 X (IT X ) + (it xdxh) to calculate the surface area of each object. Give each answer to the nearest hundredth of d or to Example 1 on pages 59-61. a square unit. nets for this cylinder. 9. How could you simplify S.A. = 2 x (nt Xr2) + (st x d x h)? a) d= 2.5 cm rea to the nearest tenth 10 cm r= 10 cm b) d=5 cm b) 1 22 cm 7 cm ect to the

Sagot :

We are asked to calculate the surface area of the objects using the following formula

[tex]SA=2\times(\pi\times r^2)+(\pi\times d\times h)[/tex]

Where r is the radius, d is the diameter, and h is the height of the circular cylinder.

Part a)

The diameter (d) is 2.5 cm

The height (h) is 10 cm

The radius is half of the diameter

[tex]r=\frac{d}{2}=\frac{2.5}{2}=1.25\; cm[/tex]

Let us substitute the given values into the formula

[tex]\begin{gathered} SA=2\times(\pi\times r^2)+(\pi\times d\times h) \\ SA=2\times(\pi\times1.25^2)+(\pi\times2.5\times10) \\ SA=2\times1.5625\pi+25\pi \\ SA=3.125\pi+25\pi \\ SA=28.125\pi \\ SA=88.36\; cm^2 \end{gathered}[/tex]

The surface area of the object is 88.36 cm^2

Part b)

The diameter (d) is 5 cm

The height (h) is 7 cm

The radius is half of the diameter

[tex]r=\frac{d}{2}=\frac{5}{2}=2.5\; cm[/tex]

Let us substitute the given values into the formula

[tex]\begin{gathered} SA=2\times(\pi\times r^2)+(\pi\times d\times h) \\ SA=2\times(\pi\times2.5^2)+(\pi\times5\times7) \\ SA=2\times(6.25\pi)+(35\pi) \\ SA=12.5\pi+35\pi \\ SA=47.5\pi \\ SA=149.23\; cm^2 \end{gathered}[/tex]

The surface area of the object is 149.23 cm^2