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Burka [1]
3 years ago
13

Two pizzas with 8 inch and 16 inch diameters are each cut into 6 equal pieces. How does the area of each piece of the smaller pi

zza compare to the area of each piece of the larger pizza?
Mathematics
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

Step-by-step explanation:

The easiest way to think about increasing area is by taking a square, if you double the dimensions, let's say they were 1 by 1 and then was made into a 2x2 square, you can fit four of the original cube (1x1) into the 2x2. So when the dimensions are doubled the area is quadrupled.

Even easier shortcut:

Take the cube idea, when the dimensions are messed with for anything, (works for volume too) always think of the smaller piece as 1x1 or 1x1x1. Then if the dimensions are double make it 2x2 or 2x2x2. If the dimensions are tripled then do 3x3 or 3x3x3, etc... 1x1=1x1x1, they are both equal to 1. So think of that as the numerator. the 2x2 in this case is equal to 4 so the smaller piece is 1/4 of the bigger one. Shown with volume if you double the dimensions of a cube that was 1x1x1 into 2x2x2. 1x1x1 still equals 1 so that's still the numerator and 2x2x2 is equal to 8 so the 1x1x1 cube is 1/8 of the 2x2x2, in other words you can fit eight 1x1x1 cubes into the 2x2x2 cube.

Hope this helps with area and volume.

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Step-by-step explanation:

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3 years ago
Steve likes to entertain friends at parties with "wire tricks." Suppose he takes a piece of wire 60 inches long and cuts it into
Alex_Xolod [135]

Answer:

a) the length of the wire for the circle = (\frac{60\pi }{\pi+4}) in

b)the length of the wire for the square = (\frac{240}{\pi+4}) in

c) the smallest possible area = 126.02 in² into two decimal places

Step-by-step explanation:

If one piece of wire for the square is y; and another piece of wire for circle is (60-y).

Then; we can say; let the side of the square be b

so 4(b)=y

         b=\frac{y}{4}

Area of the square which is L² can now be said to be;

A_S=(\frac{y}{4})^2 = \frac{y^2}{16}

On the otherhand; let the radius (r) of the  circle be;

2πr = 60-y

r = \frac{60-y}{2\pi }

Area of the circle which is πr² can now be;

A_C= \pi (\frac{60-y}{2\pi } )^2

     =( \frac{60-y}{4\pi } )^2

Total Area (A);

A = A_S+A_C

   = \frac{y^2}{16} +(\frac{60-y}{4\pi } )^2

For the smallest possible area; \frac{dA}{dy}=0

∴ \frac{2y}{16}+\frac{2(60-y)(-1)}{4\pi}=0

If we divide through with (2) and each entity move to the opposite side; we have:

\frac{y}{18}=\frac{(60-y)}{2\pi}

By cross multiplying; we have:

2πy = 480 - 8y

collect like terms

(2π + 8) y = 480

which can be reduced to (π + 4)y = 240 by dividing through with 2

y= \frac{240}{\pi+4}

∴ since y= \frac{240}{\pi+4}, we can determine for the length of the circle ;

60-y can now be;

= 60-\frac{240}{\pi+4}

= \frac{(\pi+4)*60-240}{\pi+40}

= \frac{60\pi+240-240}{\pi+4}

= (\frac{60\pi}{\pi+4})in

also, the length of wire for the square  (y) ; y= (\frac{240}{\pi+4})in

The smallest possible area (A) = \frac{1}{16} (\frac{240}{\pi+4})^2+(\frac{60\pi}{\pi+y})^2(\frac{1}{4\pi})

= 126.0223095 in²

≅ 126.02 in² ( to two decimal places)

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