In this video, we will investigate different ways to use right triangles to help solve problems involving rectangles.  We do this because it is often necessary to take preliminary knowledge of a subject and apply that knowledge in more complex situations.  In these examples, you will see that we can use our right triangle knowledge because of the right triangles that are present in each rectangle due to each vertex measuring ninety degrees.  To be clear, in this rectangle ABCD, the diagonals are the segments BD and AC.  These diagonals intersect at point E.  Additionally, we are told that the sine ratio for angle CAD is five divided by fifteen.

In problem one, we are asked to find the length of segment BE.  For this it would be helpful to remember that point E is the midpoint of a diagonal since this a rectangle… And since E is the midpoint, then the length of BE is half of the length of BD.  According to our given information, the sine ratio for angle CAD is five divided by fifteen, so we could identify the length of BD as 15 since it is the hypotenuse of the right triangle BCD.  The sine ratio of an angle measure is equal to the opposite side length divided by the length of the hypotenuse, five divided by fifteen…  Therefore the length of BE is half of fifteen, or seven and a half units long…

In problem two, we must find the length of AD.  Our big hint here is that we should use the Pythagorean Theorem.  Of course this would be helpful because, again, we have a right triangle inside this rectangle.  We are provided one leg length and the length of the hypotenuse because of the given sine ratio, therefore, we can find the other leg length by using the equation associated with the Pythagorean Theorem: … Let’s say “a” is five, and “c” is fifteen.  Remember, the leg lengths can be written in place of “a” or “b”, but the hypotenuse measure MUST be written in place of “c”.  Now, pause the video and solve for “b”.  Resume playback in a moment to check your work… “b” is approximately fourteen point one four units in length.

In problem three, we will investigate the areas of two right triangles that combine to form the rectangle ABCD.  Triangle ADC has a height of five units and a width of fourteen point one four units.  Using the area formula, , we can find the area of this triangle… This triangle has an area of thirty-five point three five square units.  Now, on your own, find the area of triangle ABC and the area of the rectangle ABCD.  Look for any correlations between the triangles areas and the area of the rectangle.  Resume playback in a moment to check your work… The area of triangle ABC is also thirty-five point three five units and the area of rectangle ABCD is seventy point seven square units.  Each triangle contains exactly half the area of the rectangle.  In fact, you may have noticed that the area of a rectangle can be found using the formula, , and the area formula for a triangle utilizes the same base and height, but we modify it by multiplying it by one half.

In problem four, we again rely on previous knowledge to solve a more complex problem.  We learned that another formula that can be used to calculate the area of a triangle.  As long as we have two side lengths and the included angle, .  In triangle AED, we know one side length is fourteen point one four, the other side length is seven point five, and the sine of the included angle is five divided by fifteen.  So the area of this triangle is one half, times fourteen point one four, times seven point five, times five divided by fifteen, seventeen point six eight square units.  By the way, you might have assumed the height of this triangle to be two point five units because it must be half the height of the rectangle.  That would be a correct assumption, and one half times fourteen point one four, times two point five is also seventeen point six eight square units.

Finally, we turn to problem five.  We must find the cosine ratio and the tangent ratio of angle CAD.  This is easy now that we know all three side lengths of the right triangle.  The cosine ratio of angle CAD is equal to the adjacent side divided by the hypotenuse, or fourteen point one four divided by fifteen.  The tangent ratio of angle CAD is equal to the opposite side length divided by the adjacent side, or five divided by fourteen point one four.
I hope you found these examples helpful, and that you were able to follow along with the procedures in each of the problems.  Be sure to review these techniques and practice them until you are comfortable.  Then you may move forward in the course.  Good Luck!