Video: Finding the Width of a Parallelogram given the Perimeter and a Relation between the Parallelogram’s Dimensions

𝐴𝐡𝐢𝐷 is a parallelogram. Given that the perimeter of 𝐴𝐡𝐢𝐷 is 54, find the length of 𝐴𝐡.

04:53

Video Transcript

𝐴𝐡𝐢𝐷 is a parallelogram. Given that the perimeter of 𝐴𝐡𝐢𝐷 is 54, find the length of 𝐴𝐡.

Let’s look at the diagram carefully. The lengths of the four sides of the parallelogram have been given in terms of three variables: π‘₯, 𝑦, and 𝑧.

In order to work out the length of 𝐴𝐡, we’re going to need to know the values of these variables or at least the value of 𝑦. We’re told that the perimeter of the parallelogram 𝐴𝐡𝐢𝐷 is 54, so let’s start with that. The perimeter of a parallelogram is the sum of all four of its sides. So this means if we add together the expressions for 𝐴𝐡, 𝐡𝐢, 𝐢𝐷, and 𝐷𝐴, in terms of π‘₯, 𝑦, and 𝑧, we must get 54.

This gives the equation on screen. Now this equation can be simplified by grouping like terms together. There’s only one 𝑦-term, so that’s still seven 𝑦. The three π‘₯ and the five π‘₯ make eight π‘₯, and the seven 𝑧 and six 𝑧 make 13𝑧. The constant on the left-hand side also simplifies to negative 53. We can then add 53 to both sides of this equation, giving us seven 𝑦 plus eight π‘₯ plus 13𝑧 equals 107.

Now this is one equation with three unknown letters: π‘₯, 𝑦, and 𝑧. So there isn’t anything else we can do with this equation at this stage. Let’s look back to the diagram to see what else we can do. A key fact about parallelograms is that opposite sides are of equal length. This means that 𝐴𝐡 is equal to 𝐢𝐷, and also 𝐴𝐷 is equal to 𝐡𝐢.

Let’s substitute the expressions for each of these sides in terms of the letters π‘₯, 𝑦, and 𝑧. For the first pair of sides, we have that seven 𝑦 minus 38 is equal to six 𝑧 minus one. Adding 38 to both sides simplifies this to seven 𝑦 equals six 𝑧 plus 37. Now this is one equation with two letters so there’s nothing further we can do with this. Let’s look at the equations for 𝐴𝐷 and 𝐡𝐢. Substituting the expressions for each side gives five π‘₯ minus four is equal to three π‘₯ plus seven 𝑧 minus 10.

Subtracting three π‘₯ from both sides and also adding four simplifies this equation to two π‘₯ is equal to seven 𝑧 minus six. Now we can’t solve either these equations, but what they’ve done is give us both 𝑦 and π‘₯ in terms of 𝑧. Each of these expressions for π‘₯ and 𝑦 can be substituted into our equation for the perimeter so that instead of it involving π‘₯, 𝑦, and 𝑧, it now just involves 𝑧.

The expression six 𝑧 plus 37 can be substituted for the seven 𝑦 in the equation. If we think of eight π‘₯ as four multiplied by two π‘₯, then our expression for two π‘₯ of seven 𝑧 minus six can also be substituted into the equation for the perimeter. Making both of these substitutions gives six 𝑧 plus 37 plus four lots of seven 𝑧 minus six plus 13𝑧 is equal to 107.

Expanding brackets and grouping like terms gives 47𝑧 plus 13 equals 107. Subtracting 13 from both sides of this equation gives 47𝑧 equals 94. Finally, dividing both sides of the equation by 47 tells us that the value of 𝑧 is two. Now we could go on to use this value of 𝑧 to calculate the values of π‘₯ and 𝑦, but let’s look back at the question.

The question asked us to find the length of 𝐴𝐡. We’ve already stated that the length of 𝐴𝐡 is the same as the length of 𝐢𝐷. We have an expression for the length of 𝐢𝐷 in terms of 𝑧, and therefore we can use this to work out the length of 𝐴𝐡. So substituting the known value of 𝑧, which is two, into this expression tells us that the length of 𝐴𝐡 is six multiplied by two minus one, which is 11, and therefore we have our answer to the problem: the length of 𝐴𝐡 is 11.

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