More Triangle Problems, 2

Category: Trigonometry, Plane Geometry

"Published in Vacaville, California, USA"

Solve for the values of x and y for the given figure:

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Solution:

Consider the given figure above

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There are two right triangles that are adjacent to each other. The hypotenuse of a right triangle is equal to the side of the other right triangle. Since the values of one side and two angles are given, then we can solve for x and y using trigonometric functions as follows

Area - Triangle, Given Three Vertices

Category: Analytic Geometry, Plane Geometry

"Published in Vacaville, California, USA"

Refer to triangle ABC in the figure.

(a) Show that triangle ABC is a right triangle by using the converse of the Pythagorean Theorem.
(b) Find the area of triangle ABC.

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Solution:

Consider the given figure above. To illustrate further the problem, let's label further the given figure as follows

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(a) The first thing that we have to do is to get the length of each sides of a triangle using the distance of two points formula.

The length of AB is

The length of BC is

The length of AC is

Next, apply the Pythagorean Theorem by substituting the values of the sides of the triangle. Since AC is the longest side, then it is the hypotenuse of a right triangle. 

Since both sides of the equation are equal, then ΔABC is a right triangle.

(b) If the vertices of a triangle are given, then we can get the area of a triangle as follows

Since the area of any plane figure is always in absolute value or positive value, then the area of a triangle is

Rectangular Parallelepiped Problem, 2

Category: Solid Geometry

"Published in Vacaville, California, USA"

A cardboard box has a square base, with each of the four edges of the base having length of x inches, as shown in the figure. The total length of all 12 edges of the box is 144 in.

(a) Express the volume V of the box as a function of x.
(b) Since both x and V represent positive quantities (length and volume, respectively), what is the domain of V?

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Solution:

Consider the given figure above

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The length of the square base of side x is given in the problem but the height of the rectangular box is not given. Let's assign h as the height of the rectangular box. 

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We know that the volume of a rectangular parallelepiped is

Since the base of a rectangular box is a square, then it follows that L = W = x. The height of the rectangular box is not given in the problem. If the perimeter or the total length of the sides of a rectangular box is given, then we can get the value of h which is the height as follows

(a) Therefore, the volume of a rectangular parallelepiped is

(b) The volume of a rectangular parallelepiped can be written as 

If you will examine the above equation, we can assign x = 0 up to x = 18. Obviously, we cannot have a negative value of x which is the length of the sides of a rectangular box. Also, we cannot have x > 18 because the value of V will be negative. Therefore the domain of V will be equal to