In this video, we will identify interesting properties of and practice solving problems involving a rhombus. First off, we will take a close look at this rhombus. We can tell it is a rhombus since all four sides of the quadrilateral are congruent, as indicated by the hash marks… As a result of all four sides being congruent, the shape becomes a parallelogram in which opposite angles are congruent, adjacent angles are supplementary, and diagonals bisect one another – that means they meet at the midpoint of each. An interesting feature special to rhombi is that the diagonals also form perpendicular lines. By looking at the same rhombus with measurements shown, we can see a collection of congruent pieces. Please pause the video and take a look at all of the interesting measurements… Now we take a look at a dynamic example of a similar rhombus. By dragging points around to change the shape, we see the relationships remain constant even though the dimensions change… We could even arrange a rhombus to look like a square…

Looking back at our worksheet, we have a few problems to solve. A few of which require us to examine smaller parts of the rhombus such as the four, congruent, right triangles that fit inside the rhombus. By the way, a proof can be made to show these triangles are congruent by the SAS Congruence Postulate. Let us look at the first problem – What is the length of diagonal BD? I would suggest looking at the triangle AOB first. It is a right triangle since the diagonals meet at right angles… it has a hypotenuse length of 7 cm since all four sides of a rhombus are congruent… and it has one leg length of 5.6 cm from the given information. Pause the video and use the Pythagorean equation to solve for the missing side length BO. Resume playback in a moment to check your work… The length of BO is 4.2 cm. Since the diagonals of a rhombus are bisected at the intersection, we know the full length of the diagonal is twice that of this piece, or 8.4 cm…

When asked to find the area of the rhombus, you might be tempted to multiply 7 by 7 since that is the formula to find the area of a square, which is a special type of rhombus. But actually that will not work since finding the area of a rhombus is more like finding the area of a parallelogram, which requires the base length times the vertical height… This vertical height is difficult to obtain from the diagram, so I'll share an easier method. As we discovered earlier, the rhombus is composed of four congruent triangles. If we find the area of one, we could quadruple that area to obtain the total area of the rhombus. Triangle AOB has a base of 5.6 cm and a height of 4.2 cm according to our prior work… And the area of a triangle is one-half of the base times height, or one-half times 5.6 times 4.2, or 11.76 square centimeters. When quadrupled, we see the total area of the rhombus is approximately 47.04 square centimeters. A little bit less than the 49 square centimeters we would have gotten if we multiplied 7 by 7.

Finding angle d₁ is easy. It is congruent to the adjacent angle in this diagram. So it must also be 38 degrees. Notice a₁ and a₂ are congruent to each other and c₁ and c₂, while b₁ and b₂ are congruent to each other and d₁.

Finally, recalling the definition of a Tangent ratio, we could find the tangent ratio of angle c₂. The tangent ratio of an angle within a right triangle is the opposite side length divided by the adjacent side length. So in this case, side DO of triangle COD is opposite of angle c₂, and side CO is adjacent to c₂. Side CO is congruent to AO and must have a length of 5.6 cm. Side DO is congruent to BO and must have a length of 4.2 cm as discovered earlier… Therefore the tangent of angle c₂ is equal to 4.2 divided by 5.6, or 0.75.

I hope these examples help you to understand some of the impressive features of rhombi. Please make sure to watch this video again, and review any of the problems that may have been difficult for you to follow the first time through. Good luck!