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Answer :
the complete question in the attached figure
Part a) The side lengths of the green pedestal are one-half the side
lengths of the red pedestal. How much paint do you
need to paint the green pedestal?
[surface area to paint of red pedestal]=[perimeter of the base]*height+[area of the bottom of the base]
[surface area to paint of red pedestal]=[16*4]*24+[16*16]*2---> 1536+512
[surface area to paint of red pedestal]=2048 in²
0.5 pint of paint is needed to paint 2048 in²
step 1
the side lengths of the green pedestal are
16*(1/2)-------> 8 in
16*(1/2)-------> 8 in
24*(1/2)-------> 12 in
8x8x12 in
[surface area to paint of green pedestal]=[8*4]*12+[8*8]*2---> 384+128
[surface area to paint of green pedestal]= 512 in²
if for 2048 in²------------------> is needed 0.5 pint of paint
for 512 in²-------------------> X
X=512*0.5/2048---------> x=0.125 pint of paint
the answer Part A) is 0.125 pint of paint
Part b)The side lengths of the blue pedestal are triple
the side lengths of the green pedestal. How
much paint do you need to paint the
blue pedestal?
step 1
the side lengths of the blue pedestal are
8*(3)-------> 24 in
8*(3)-------> 24 in
12*(3)-------> 36 in
24x24x36 in
[surface area to paint of blue pedestal]=[24*4]*36+[24*24]*2---> 3456+1152
[surface area to paint of blue pedestal]=4608 in²
if for 2048 in²------------------> is needed 0.5 pint of paint
for 4608 in²-------------------> X
X=4608*0.5/2048---------> x= 1.125 pint of paint
the answer Part b) is 1.125 pint of paint
Part c) Compare the ratio of paint amounts
to the ratio of side lengths for the
green and red pedestals. Repeat for
the green and blue pedestals. What
do you notice?
Paint amounts: [green pedestal]/[red pedestal]=0.125 pint/0.5 pint=1/4
Sides lengths: [green pedestal]/[red pedestal]=8 in/16 in=1/2
(1/2)²=1/4
Paint amounts: [green pedestal]/[blue pedestal]=0.125 pint/1.125 pint=1/9
Sides lengths: [green pedestal]/[blue pedestal]=8 in/24 in=1/3
(1/3)²=1/9
the answer Part c) is
The ratio of paint amounts (or surface areas) is equal
to the square of the ratio of side lengths.
Part a) The side lengths of the green pedestal are one-half the side
lengths of the red pedestal. How much paint do you
need to paint the green pedestal?
[surface area to paint of red pedestal]=[perimeter of the base]*height+[area of the bottom of the base]
[surface area to paint of red pedestal]=[16*4]*24+[16*16]*2---> 1536+512
[surface area to paint of red pedestal]=2048 in²
0.5 pint of paint is needed to paint 2048 in²
step 1
the side lengths of the green pedestal are
16*(1/2)-------> 8 in
16*(1/2)-------> 8 in
24*(1/2)-------> 12 in
8x8x12 in
[surface area to paint of green pedestal]=[8*4]*12+[8*8]*2---> 384+128
[surface area to paint of green pedestal]= 512 in²
if for 2048 in²------------------> is needed 0.5 pint of paint
for 512 in²-------------------> X
X=512*0.5/2048---------> x=0.125 pint of paint
the answer Part A) is 0.125 pint of paint
Part b)The side lengths of the blue pedestal are triple
the side lengths of the green pedestal. How
much paint do you need to paint the
blue pedestal?
step 1
the side lengths of the blue pedestal are
8*(3)-------> 24 in
8*(3)-------> 24 in
12*(3)-------> 36 in
24x24x36 in
[surface area to paint of blue pedestal]=[24*4]*36+[24*24]*2---> 3456+1152
[surface area to paint of blue pedestal]=4608 in²
if for 2048 in²------------------> is needed 0.5 pint of paint
for 4608 in²-------------------> X
X=4608*0.5/2048---------> x= 1.125 pint of paint
the answer Part b) is 1.125 pint of paint
Part c) Compare the ratio of paint amounts
to the ratio of side lengths for the
green and red pedestals. Repeat for
the green and blue pedestals. What
do you notice?
Paint amounts: [green pedestal]/[red pedestal]=0.125 pint/0.5 pint=1/4
Sides lengths: [green pedestal]/[red pedestal]=8 in/16 in=1/2
(1/2)²=1/4
Paint amounts: [green pedestal]/[blue pedestal]=0.125 pint/1.125 pint=1/9
Sides lengths: [green pedestal]/[blue pedestal]=8 in/24 in=1/3
(1/3)²=1/9
the answer Part c) is
The ratio of paint amounts (or surface areas) is equal
to the square of the ratio of side lengths.
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