USING POLYNOMIAL TO FIND VOLUME OF 3D SHAPES

Find the volume of the 3D shapes given below.

Problem 1 :

Solution :

Length = 2x, width = x and height = 3x

Volume of cuboid = Base area ⋅ height

Base area = 2x (x) ==> 2x²

Volume of cuboid = 2x²(3x)

V = 6x³

Problem 2 :

Solution :

length = c, Breadth = a, Height = b

Base is in the shape of triangle. 

Area of triangle = 1/2 × base x height

= (1/2) ⋅ a ⋅ b

Volume of triangular prism = Base area × height

= 1/2 × a × b × c

V = (1/2)abc

Problem 3 :

Solution :

Height = 4r, Diameter = 2r, Radius r = 2r/2 = r

Base area = πr²

Volume of cylinder = Base area x height

= π × r² × 4r

V = 4πr³

Problem 4 :

Solution :

Hight = 3x, Diameter = 2x, Radius = 2x/2 = x

Volume of cone = 1/3 ×Base area x height

Base area = πr²

= πx²

= (1/3) × π x² × 3x

V = πx³

Problem 5 :

Solution :

Base = a, Height = h

Volume of pyramid = 1/3 × (base area) × height

Base is in the shape of square. Area of base square = a²

= 1/3 × a² × h

V = a²h/3

Problem 6 : 

Solution :

Height = 4x, Diameter = 2x, Radius = 2x/2 = x

Base is in the shape of semi circle.

Area of semi circle = (1/2)πr² ==> (1/2)πx²

Volume of semi circular = 1/2 × πr²h

= 1/2 × π × x² × 4x

V = 2πx³

Problem 7 : 

The diagram shows a solid triangular prism. All the measurements are in centimeters. The volume of the prism is V cm³. Find a formula for V in terms of x. Give your answer in simplified form

Solution :

Volume of triangular prism = base area ⋅ height

base of triangle = 2x + 1

Height of triangle = x

Height of prism = x + 8

= (1/2) ⋅ (2x + 1) ⋅ x ⋅ (x + 8)

= (1/2) x (2x² + 16x + x + 8)

= (1/2) x (2x² + 17x + 8)

= (1/2) (2x³ + 17x² + 8x)

Problem 8 : 

A toy company sells two different toy chests. The toy chests have different dimensions, but the same volume. What is the width w of Toy Chest 2?

Solution :

Volume of toy chest 1 = volume of toy chest 2

30 ⋅ 16 ⋅ 12 = 24 ⋅ w ⋅ 16

w = (30 ⋅ 16 ⋅ 12) / (24 ⋅ 16)

= 15 inches

So, the required width of toy chest 2 is 15 inches.

Problem 9 : 

The prism has a volume of 150 cubic feet. Find the length of the prism.

Solution :

Volume of rectangular prism = length ⋅ width ⋅ height

Length = l, width = 5 ft and height = 4 ft

Volume of the rectangular prism = 150 cubic feet

l ⋅ 5 ⋅ 4 = 150

20l = 150

l = 150/20

l = 7.5

So, the required length of the rectangular prism is 7.5 ft.

Problem 10 : 

The volume V of the rectangular prism is given by

V = 2x³ + 17x² + 46x + 40

Find an expression for the missing dimension

Solution :

Volume of rectangular prism = length ⋅ (x + 4) ⋅ (x + 2)

2x³ + 17x² + 46x + 40 = length ⋅ (x + 4) ⋅ (x + 2) 

-4 is the solution, by factoring 2x² + 9x + 10

= 2x² + 4x + 5x + 10

= 2x(x + 2) + 5(x + 2)

= (2x + 5) (x + 2)

So, the factors are (x + 4)(x + 2) and (2x + 5). By observing these factors, it is clear that two of the factors are already there. Since the missing factor is 2x + 5, the required length of the rectangular prism is 2x + 5.

Problem 11 : 

The volume (in cubic inches) of a shipping box is modeled by

V = 2x³ − 19x² + 39x

where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.

Solution :

By solving the polynomial V = 2x³ − 19x² + 39x, we will get three factors and solutions.

V = 2x³ − 19x² + 39x

= x(2x² − 19x + 39)

= x(2x² − 13x - 6x + 39)

= x [x (x - 13) - 3 (2x  - 13)] 

Volume = x (x - 3)(2x - 13)

Length = x, width = x - 3 and height = 2x - 13

The factors are x (x - 3) and (2x - 13). By solving these factors, we get

x = 0, x - 3 = 0 and 2x - 13 = 0

x = 0, x = 3 and x = 6.5

So, the solutions are 0, 3 and 6.5

Problem 12 : 

The volume (in cubic inches) of a rectangular birdcage can be modeled by

V = 3x³ − 17x² + 29x − 15

where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.

Solution :

x = 3 is the solution.

Solving 3x² - 8x + 5 = 0

3x² - 5x - 3x + 5 = 0

x(3x - 5) - 1(3x - 5) = 0

(x - 1)(3x - 5) = 0

x = 1 and x = 5/3

So, the possible solutions for x are 1, 3 and 5/3.