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Objective: Reasoning With Equations and InequalitiesSolve equations and inequalities in one variable. (Linear inequalities; literal that are linear in the variables being solved for; quadratics with real solutions.)17.) Solve linear equations and inequalities in one variable, including equations with coefficients represented by letters. [A­REI3]

Purpose:I know I've got it when I can plug information into a given formula and solve for the missing variable.

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1. 22cm

2. 22 in.

3. 24 cm2

4. 56 in2

5. 216 cm3

6. 48 cm3

7. 352 cm3, 351.858cm3

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Problem 1: A rectangular prism of volume 3200 mm3 has arectangular base of length 10 mm and width 8 mm. Find the height h ofthe prism.

Solution to Problem 1:• Volume is given by by

volume = length * width * height = 10 mm * 8 mm * h = 3200mm3

Solve for h

h = 3200 mm3 / 80 mm2 = 40 mm

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Problem 2: The area of one square face of a cube is equal to 64 cm2.Find the volume of the cube.

Solution to Problem 2:• The area of one square face is given by

s * s = 64 cm2

• Solve for s

s = SQRT(64 cm2) = 8 cm The volume V of the given cube is given by

V = s3 = 83 = 512 cm3

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Problem 3: The triangular base of a prism is a right triangle of sides aand b = 2a. The height h of the prism is equal to 10 mm and its volumeis equal to 40 mm3, find the lengths of the sides a and b of the triangle.

Solution to Problem 3:• The volume V of the prism is given by

V = (1/2) a * b * h = 40 mm3

• Substitute b by 2a and h by its value

40 mm3 = a2 * 10 mm Solve for a and calculate b

a = 2 mm b = 2a = 4 mm

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Problem 4: Find the volume of the given L­shaped rectangularstructure.

Solution to Problem 4:We can think of the given shape as a larger rectangular prism ofdimensions 60, 80 and 10 mm from which a smaller prism ofdimensions 40, 60 and 10 mm has been cut. Hence the volume V ofthe given 3D shape

V = 60 * 80 * 10 mm3 ­ 40 * 60 * 10 mm3 = 24000 mm3

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Problem 5: Find the thickness x of the hollow cylinder of height 100 cmif the volume between the inner and outer cylinders is equal to 11000 Pimm3 and the outer diameter is 12 mm.

Solution to Problem 5:• If R and r are the outer and inner radii of the hollow cylinder thevolume V between the inner and outer cylinders is given by

V = h*(Pi R2 ­ Pi r2) = 11000 Pi • Also R = 6 and h = 100 cm = 1000 mm, hence

1000 * (36 Pi ­ Pi r2) = 11000 Pi • Solve for r

r = 5 mm Find x

x = R ­ r = 1 mm

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Problem 6: Find x so that the volume of the U­shaped rectangularstructure is equal to 165 cm3.

Solution to Problem 6:• We can think of the given shape as a larger rectangular prism ofdimensons 8, 3 and 10 cm from which a smaller prism ofdimensions x, x and 3 cm has been cut. Hence the volume Vof the given 3D shape is given by

V = 8 * 3 * 10 mm3 ­ x * x * 3 mm3 = 165 cm3

Solve for x

x = 5 cm

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Problem 7: Find the volume of the hexagonal prism whose base is aregular hexagon of side x = 10 cm.

Solution to Problem 7:• The hexagon is made up of 6 equilateral triangles, hence thearea A of the base

A = 6 (x2 SQRT(3) / 4) Hence the volume V of the prism

V = 24 * 6 (102 SQRT(3) / 4) cm 2 = 6235.4 cm3 (rounded to 1 decimalplace)