sa = ph + 2b - mrs. whitmore/ms. jensen's math website · 1 surface areathe amount of squares...

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1 Surface Areathe amount of squares needed to cover an object's surface. top bottom left right front back SA = ph + 2B Base_______________ height__________ Perimeter (of the base)____________ Surface Area__________ SA = ph + 2B Volume________________ Base_______________ height__________ Perimeter (of the base)____________ Surface Area__________ SA = ph + 2B Volume________________ Base_______________ height__________ Perimeter (of the base)____________ Surface Area__________ SA = ph + 2B Volume________________ 2 cm 3 cm 3 cm 3 cm 5 cm 4 cm 4 cm 3 cm 2 cm 5 cm 2 cm 1 cm

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Page 1: SA = ph + 2B - Mrs. Whitmore/Ms. Jensen's Math Website · 1 Surface Areathe amount of squares needed to cover an object's surface. top bottom left right front back SA = ph + 2B Base_____

1

Surface Area­the amount of squares needed to cover an object's surface.  

top bottomleft rightfront back

SA = ph + 2BBase_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

2 cm

3 cm

3 cm

3 cm5 cm

4 cm

4 cm3 cm

2 cm

5 cm2 cm

1 cm

Page 2: SA = ph + 2B - Mrs. Whitmore/Ms. Jensen's Math Website · 1 Surface Areathe amount of squares needed to cover an object's surface. top bottom left right front back SA = ph + 2B Base_____

2

3m 4m

8m

2 in5 in

9 in

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

4 cm

3 cm3 cm

4 m

2 m3 m

Page 3: SA = ph + 2B - Mrs. Whitmore/Ms. Jensen's Math Website · 1 Surface Areathe amount of squares needed to cover an object's surface. top bottom left right front back SA = ph + 2B Base_____

3

    4in

18 in

5 in

3 in

4 in 12 in

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Base_______________height__________Perimeter(of the base)____________

Surface Area__________ SA = ph + 2B

Volume________________

Page 4: SA = ph + 2B - Mrs. Whitmore/Ms. Jensen's Math Website · 1 Surface Areathe amount of squares needed to cover an object's surface. top bottom left right front back SA = ph + 2B Base_____

4

60

18 in

sector

circle

72

1.5 in

A = πr2 C = πd d = 2r

Area_________

Circumference_______

Area_________

Circumference_______

sector

circle

sector

circle 4524 in

sector

circle

Area_________

Circumference_______

Area_________

sector

circle

7.2 cm

       7.536 cm

sector

circle

sector

circleArea_________

Circumference_______

6 in

9.42 in5.4 cm

 A = 84.78 cm2

sector

circle

sector

circle

Circumference_________ Circumference_______

4 m

A = 20.096 m2

8 m A = 120.576 m2

sector

circle

Page 5: SA = ph + 2B - Mrs. Whitmore/Ms. Jensen's Math Website · 1 Surface Areathe amount of squares needed to cover an object's surface. top bottom left right front back SA = ph + 2B Base_____

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