What percentage greater is the area of a 12 in. plate compared with a 9 in plate?

Question

The NBC Nightly News on June 21, 2010, featured a story about food portions, stating that dinner plates were 9 in. in diameter in the 1960s but now they are 12 in. in diameter "making room for one-third more food." What percentage greater is the area of a 12 in. plate compared with a 9 in. plate?

Answer 1

 

A₁ = πd₁²/4

 

A₂ = πd₂²/4

 

A₂/A₁ = (d₂/d₁)²

 

A₂ = (d₂/d₁)² A₁

 

(d₂/d₁)² = (12/9)² ≅ 1.778

 

A₂ ≅ 1.778 A₁

 

A₂ is approximately 77.8% larger than A₁.

Answer 2

It is very instructive to ask, “If the radius of a circle is x% longer, how much bigger is the area?”

For this problem, a 12-inch plate (radius=6) has a radius that is 33.3% (that is, 1/3) longer than a 9-inch plate (radius=4.5). So, how has the area increased?

The area of a circle is pi*radius^2.

 

The % increase is (difference)/(beginning). That is,
   Percent increase = ((pi*r₂² - pi*r₁²)/pi*r₁²)*100%

                          = ((r₂² - r₁²)/r₁²)*100%              [cancel the pi]

                           = (((1.333r₁)² – r₁²)/r₁²)*100%             [solve for x=33.3% larger]

                          = 77.8%

The 12-inch plate increases the area by more than 75% (that is, more than 3/4ths more food rather that one-third more food).

Answer 3

Area 12 inch:  144π

Area 9 inch:  81π

 

81π = (x/100)(144π)

 

x = 81π(100)/144π

x = 81(100)/144

x = 56.25

 

The are of 9 inch plates are 56.25% 

of the 12 inch plates.  The increase is

100-56.25 = 43.75%

Answer 4

(new area- old area)/old area= Percent larger

 

Because all of the Pi cancel, they can be omitted.

 

((12²/4)-(9²/4))/(9²/4)= answer

Answer 5

(12 - 9) / 9 = ?