\(\displaystyle SA=2lw+2lh+2hw=2*4*6+2*4*5+2*5*6=48+40+60=148\)

Find the surface area of the box by summing the area of all six faces: two 8 by 10, two 8 by 5, two 10 by 5.

\(\displaystyle 2(80)+2(40)+2(50)=340\ in^{2}\)

Since the price is $.10 for each square inch,

\(\displaystyle Total\ cost=340(.10)=\$34\)

Let \(\displaystyle h\) be the height of the prism. Then the width is \(\displaystyle 3h\), and the length is \(\displaystyle 3h+ 16\). Since the sum of the three dimensions is one meter, or 100 centimeters, we solve for \(\displaystyle h\) in this equation:

\(\displaystyle h + 3h + (3h+16) = 100\)

\(\displaystyle 7h+16 = 100\)

\(\displaystyle 7h+16-16 = 100-16\)

\(\displaystyle 7h = 84\)

\(\displaystyle 7h \div 7 = 84 \div 7\)

\(\displaystyle h = 12\)

The height of the prism is 12 cm; the width is three times this, or 36 cm; the length is sixteen centimeters greater than the width, which is 52 cm.

Set  \(\displaystyle l= 52,w=36,h=12\) in the formula for the surface area of a rectangular prism:

\(\displaystyle A = 2lw + 2wh + 2lh\)

\(\displaystyle = 2 \cdot 52 \cdot 36 + 2 \cdot 36 \cdot 12 + 2 \cdot 52 \cdot 12\)

\(\displaystyle = 3,744 + 864 + 1,248\)

\(\displaystyle = 5,856\) square centimeters

We can set the lengths of three sides to be \(\displaystyle x\), \(\displaystyle y\), \(\displaystyle z\), respectively. The volume is 120 means that \(\displaystyle xyz=120\). Also we know the areas of two sides, so we can use \(\displaystyle xy\) to represent the area of 20 and \(\displaystyle xz\) to represent the area of 30. Now the question is to figure out \(\displaystyle yz\). Then we can use

\(\displaystyle xy * xz * yz = x^2 * y^2 * z^2 =(xyz)^2\) to solve \(\displaystyle yz\).

\(\displaystyle 20 * 30 * yz = 120^2\)

\(\displaystyle yz=24\)

The area of a right triangle is equal to half the product of its legs, so each base has area

\(\displaystyle \frac{1}{2} \cdot 12 \cdot 16 = 96\)

The measure of the hypotenuse of each base is determined using the Pythagorean Theorem:

\(\displaystyle c = \sqrt{12^{2}+16^{2}}= \sqrt{144+256} = \sqrt{400}= 20\)

Therefore, the perimeter of each base is 

\(\displaystyle P=12+16+20 = 48\),

and the height of the prism is half this, or \(\displaystyle h = \frac{1}{2} \cdot 48 = 24\).

The lateral area of the prism is the product of its height and the perimeter of a base; this is

\(\displaystyle L= Ph= 24 \cdot 48 = 1,152\)

The surface area is the sum of the lateral area and the two base areas, or

\(\displaystyle S= L+B+B =1,152+96+96 =1,344\).

By the 45-45-90 Theorem, dividing the length of the hypotenuse of an isosceles right triangle by \(\displaystyle \sqrt{2}\) yields the length of one leg; therefore, the length of one leg of each base is

\(\displaystyle \frac{10}{\sqrt{2}} = \frac{10 \cdot \sqrt{2}}{\sqrt{2} \cdot \sqrt{2}} = \frac{10 \sqrt{2}}{2} = 5\sqrt{2}\).

The area of a right triangle is half the product of its legs, so the area of each base is 

\(\displaystyle B = \frac{1}{2} \cdot 5 \sqrt{2} \cdot 5 \sqrt{2} = \frac{1}{2} \cdot 25 \cdot 2 = 25\)

The perimeter of each base is the sum of its sides, which here is

\(\displaystyle P = 5 \sqrt{2}+ 5 \sqrt{2}+ 10 = 10 + 10 \sqrt{2}\)

The height of the prism is the length of one leg of a base, which is \(\displaystyle h = 5\sqrt{2}\).

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

\(\displaystyle L = Ph\)

\(\displaystyle = \left (10+10 \sqrt{2} \right ) \cdot 5 \sqrt{2}\)

\(\displaystyle = 10 \cdot 5 \sqrt{2} +10 \sqrt{2} \cdot 5 \sqrt{2}\)

\(\displaystyle = 50 \sqrt{2} +10\cdot 5 \cdot 2\)

\(\displaystyle = 100+ 50 \sqrt{2}\)

The surface area is the sum of the lateral area and the areas of the bases:

\(\displaystyle S=L+B+B= 100+ 50 \sqrt{2}+25+25 = 150+ 50 \sqrt{2}\)

The area of each equilateral triangle base can be determined by setting \(\displaystyle s = 6\) in the formula

\(\displaystyle B= \frac{s^{2}\sqrt{3}}{4} = \frac{6^{2}\sqrt{3}}{4} = \frac{36\sqrt{3}}{4} = 9\sqrt{3}\)

The perimeter of each base is \(\displaystyle P=6 \cdot 3 = 18\), and the height of the prism is three times this, or \(\displaystyle h = 18 \cdot 3 = 54\). The lateral area of a prism is equal to the perimeter of a base multiplied by the height, so

\(\displaystyle L =Ph= 18 \cdot 54 = 972\)

Add this to the areas of two bases - the surface area is

\(\displaystyle S= L+ B + B= 972 + 9\sqrt{3}+ 9\sqrt{3}= 972 + 18 \sqrt{3}\).

By the 45-45-90 Theorem, multiplying the length of a leg of an isosceles right triangle by \(\displaystyle \sqrt{2}\) yields the length of its hypotenuse; therefore, the length of the hypotenuse of each base is \(\displaystyle 16 \sqrt{2}\).

The area of a right triangle is half the product of its legs, so the area of each base is 

\(\displaystyle B = \frac{1}{2} \cdot 16 \cdot 16 =128\)

The perimeter of each base is the sum of its sides, which here is

\(\displaystyle P =16+16+16 \sqrt{2} = 32 + 16 \sqrt{2}\)

The height of the prism is the length of the hypotenuse of the base, which is \(\displaystyle h = 16\sqrt{2}\).

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

\(\displaystyle L = Ph\)

\(\displaystyle = \left ( 32 + 16 \sqrt{2}\right ) \cdot 16 \sqrt{2}\)

\(\displaystyle = 32 \cdot16 \sqrt{2} +16 \sqrt{2} \cdot 16 \sqrt{2}\)

\(\displaystyle = 32 \cdot16 \sqrt{2} +16 \cdot 16 \cdot \sqrt{2} \cdot \sqrt{2}\)

\(\displaystyle = 512 \sqrt{2} +16 \cdot 16 \cdot 2\)

\(\displaystyle = 512+ 512 \sqrt{2}\)

The surface area is the sum of the lateral area and the areas of the bases:

\(\displaystyle S=L+B+B= 512+ 512 \sqrt{2} + 128 +128 = 768 + 512 \sqrt{2}\)

A right triangle with one leg half the length of the hypotenuse is a 30-60-90 triangle. Its other leg has measure \(\displaystyle \sqrt{3}\) times the length of the first leg, which in the case of each base is \(\displaystyle 10 \sqrt{3}\). 

The area of each base is half the product of the legs, or 

\(\displaystyle B = \frac{1}{2} \cdot 10 \cdot 10 \sqrt{3} = 50 \sqrt{3}\).

The perimeter of each base is 

\(\displaystyle P = 20+10+10 \sqrt {3} = 30 + 10 \sqrt {3}\)

and the height of the prism is equal to the length of the longer leg, or

\(\displaystyle h= 10 \sqrt{3}\).

The lateral area of the prism is equal to the product of the height of the prism and the perimeter of a base, so

\(\displaystyle L = Ph\)

\(\displaystyle =\left ( 30 + 10 \sqrt {3} \right ) \cdot 10 \sqrt {3}\)

\(\displaystyle = 30 \cdot 10 \sqrt {3}+ 10 \sqrt {3}\cdot 10 \sqrt {3}\)

\(\displaystyle = 30 \cdot 10 \sqrt {3}+ 10 \cdot 10\cdot \sqrt {3}\cdot \sqrt {3}\)

\(\displaystyle = 30 \cdot 10 \sqrt {3}+ 10 \cdot 10\cdot 3\)

\(\displaystyle = 300+ 30 0 \sqrt {3}\)

The surface area is the sum of the lateral area and the areas of the bases:

\(\displaystyle S=L+B+B= 300+300 \sqrt{3} + 50 \sqrt{3} + 50 \sqrt{3}= 300+400 \sqrt{3}\)

The volume of a rectangular solid is the product of its length, height, and width. Since \(\displaystyle L = 18\) , \(\displaystyle W = 20\), and \(\displaystyle H = 5,400\),

\(\displaystyle LWH = V\)

\(\displaystyle 18 \cdot 20 \cdot H =5,400\)

\(\displaystyle 360H = 5,400\)

\(\displaystyle H = 15\)

The surface area of a rectangular solid is 

\(\displaystyle S = 2LH + 2LW + 2WH\)

\(\displaystyle = 2 \cdot 18 \cdot 15 + 2 \cdot 18 \cdot 20 + 2 \cdot 20\cdot 15\)

\(\displaystyle = 540+720 + 600\)

\(\displaystyle = 1,860\)