Question Video: Finding the Perimeter of a Rhombus | Nagwa Question Video: Finding the Perimeter of a Rhombus | Nagwa

Question Video: Finding the Perimeter of a Rhombus

Determine the perimeter of a rhombus 𝐴𝐡𝐢𝐷, given that the points 𝐴 and 𝐡 are (βˆ’5, βˆ’1) and (βˆ’4, 4).


Video Transcript

Determine the perimeter of a rhombus 𝐴𝐡𝐢𝐷, given that the points 𝐴 and 𝐡 are negative five, negative one and negative four, four.

Let’s begin by recalling that this rhombus is a quadrilateral with all four sides congruent. The rhombus here is 𝐴𝐡𝐢𝐷, but we’re only given two of the points and we need to find out the perimeter of the entire rhombus. So let’s get some grid paper and see if we can plot these two points 𝐴 and 𝐡. So here we have our points 𝐴 at negative five, negative one and 𝐡 at negative four, four. We can even join them to create the line segment 𝐴𝐡.

So how do we go about finding the perimeter? After all, the rhombus 𝐴𝐡𝐢𝐷 could look like this, or it could look like this. But let’s think for a minute. We can remember that the perimeter of a shape is the distance around the outside edge. And we know that, in a rhombus, all four sides are congruent, so all the sides will be the same length. So we really only need to know one of the lengths of a rhombus in order to calculate the perimeter. Let’s see if we can work out the length of this line segment 𝐴𝐡.

We could do this in two ways. We could use the Pythagorean theorem, or we could use this formula to find the distance between two points, which is actually an application of the Pythagorean theorem. This formula tells us that for two points π‘₯ one, 𝑦 one and π‘₯ two, 𝑦 two, the distance is calculated as the square root of π‘₯ two minus π‘₯ one squared plus 𝑦 two minus 𝑦 one squared. So let’s use this formula, and we can show that, in actual fact, we didn’t even have to draw the grid with 𝐴 and 𝐡 on it. Let’s take the formula and designate point 𝐴 with the π‘₯ one, 𝑦 one values and point 𝐡 with the π‘₯ two, 𝑦 two values.

We would therefore have the distance equals the square root of π‘₯ two, which is negative four, subtract π‘₯ one, which is negative five, all squared plus four, that was our 𝑦 two value, subtract negative one, which is our 𝑦 one value, squared. Simplifying the first set of parentheses, negative four subtract negative five gives us negative four plus five. And that’s one, so we’d have one squared. Our second set of parentheses, the four subtract negative one becomes four plus one, which is five, and we square that two. One squared is one, and five squared is 25. Adding those would give us 26, so the distance 𝑑 is equal to the square root of 26.

We should also add in the units here. We’ve worked out the distance. So, that will be a length unit. So now we know that we have the length of 𝐴𝐡. That’s the square root of 26 length units. And we’ll keep it in this square-root form as we continue to work out the perimeter. Remember that it doesn’t matter what 𝐴𝐡𝐢𝐷 looks like. We know that the length of one side will be the square root of 26. As the rhombus has four equal length sides, then the perimeter will be four times root 26. And we can write this simply as four root 26 length units. And that’s our answer for the perimeter of rhombus 𝐴𝐡𝐢𝐷.

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