If two figures are similar then:
Surface area of similar figures :
If two solids are similar, then the ratio of their surface areas is equal to the square of the ratio of their corresponding linear measures.
Volume of similar figures :
If two solids are similar, then the ratio
of their volumes is equal to the cube
of the ratio of their corresponding
linear measures.
The scale factor between two similar figures is given. The surface area and volume of the smaller figure are given. Find the surface area and volume of the larger figure.
Problem 1 :
Scale factor = 1 : 2
SA = 90 yd^{2}
V = 216 yd^{3}
Solution :
Scale factor (k) = 1 : 2 = 1/2
i) Surface area of smaller figure / surface area of larger
= (1/2)^{2}
Surface area of smaller figure = 90 yd^{2}
90/Surface area of larger figure = (1/4)
Surface area of larger figure = 90(4)
= 360 yd^{2}
So, surface area of larger figure is 360 yd^{2}.
ii) Volume of smaller figure / Volume of larger = (1/2)^{3}
Volume of smaller figure = 216 yd^{3}
216/volume of larger figure = (1/8)
Volume of larger figure = 216(8)
= 1728 yd^{3}
So, volume of larger figure is 1728 yd^{3}.
Problem 2 :
Scale factor = 4 : 9
SA = 256 km^{2}
V = 1536 km^{3}
Solution :
i) Surface area of smaller figure / surface area of larger
= (4/9)^{2}
Surface area of smaller figure = 256 km^{2}
256/Surface area of larger figure = (4/9)^{2}
Surface area of larger figure = 256(81/16)
= 1296 yd^{2}
So, surface area of larger figure is 1296 yd^{2}.
ii) Volume of smaller figure / Volume of larger = (4/9)^{3}
Volume of smaller figure = 1536 yd^{3}
1536/volume of larger figure = (64/729)
Volume of larger figure = 1536(729/64)
= 17496 km^{3}
So, volume of larger figure is 17496 km^{3}.
Problem 3 :
The dimensions of the touch tank at an aquarium are doubled. What is the volume of the new touch tank?
A) 150 ft^{3} B) 4000 ft^{3 }C) 8000 ft^{3} D) 16,000 ft^{3}
Solution :
The dimensions are doubled, so the ratio of the dimensions in the original tank to the dimensions in the new tank is 1 : 2.
(Original volume/new volume) = (original dimension/new dimension)^{3}
(2000/new volume)= (1/2)^{3}
(2000/new volume)= (1/8)
New volume = 2000(8)
= 16000 ft^{3}
The solids are similar. Find the volume of the red solid. Round your answer to the nearest tenth.
Problem 4 :
Solution :
Scale factor (k) = 5/12
Volume of red solid / volume of blue solid = k^{3}
Volume of red solid / 288 = (5/12)^{3}
Volume of red solid / 288 = 125/1728
Volume of red solid = (125/1728)⋅288
= 20.8 cm^{3}
Problem 5 :
Solution :
Scale factor (k) = 3/4
Volume of blue solid / volume of red solid = k^{3}
9 / Volume of red solid = (3/4)^{3}
9 / Volume of red solid = (27/64)
Volume of red solid = 9(64/27)
= 21.3 in^{3}
May 21, 24 08:51 PM
May 21, 24 08:51 AM
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