Video: Using Vectors to Find the Area of a Right Trapezoid

Trapezoid 𝐴𝐡𝐢𝐷 has vertices 𝐴(4, 14), 𝐡(4, βˆ’4), 𝐢(βˆ’12, βˆ’4), and 𝐷(βˆ’12, 9). Given that 𝐀𝐁 βˆ₯ 𝐃𝐂 and 𝐀𝐁 βŠ₯ 𝐂𝐁, find the area of that trapezoid.

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Video Transcript

Trapezoid 𝐴𝐡𝐢𝐷 has vertices 𝐴: four, 14; 𝐡: four, negative four; 𝐢: negative 12, negative four; and 𝐷: negative 12, nine. Given that vector 𝐀𝐁 is parallel to vector 𝐃𝐂 and vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁, find the area of that trapezoid.

Let’s consider what we are told about the trapezoid. We know that vector 𝐀𝐁 is parallel to vector 𝐃𝐂. We also know that vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁. This means that they meet at right angles. It therefore follows that vector 𝐃𝐂 also meets vector 𝐂𝐁 at right angles. We know that the area of a trapezoid can be found using the formula π‘Ž plus 𝑏 over two multiplied by β„Ž, where π‘Ž and 𝑏 are the parallel sides and β„Ž is the perpendicular height. We need to find the length of the sides 𝐴𝐡 and 𝐷𝐢 and the perpendicular height 𝐢𝐡.

In order to calculate the length of the sides, we need to work out the magnitude of the vectors. We will begin by calculating the magnitude of 𝐀𝐁. This is equal to the square root of four minus four squared plus negative four minus 14 squared. Four minus four is equal to zero. And negative four minus 14 is negative 18. Squaring this gives us 324, and then square rooting the answer gives us 18. The magnitude of vector 𝐀𝐁 is 18. We can repeat this process to calculate the magnitude of vector 𝐃𝐂. This is equal to the square root of negative 12 minus negative 12 squared plus negative four minus nine squared. This gives us an answer of 13.

As the magnitude of 𝐀𝐁 is greater than the magnitude of 𝐃𝐂, we can see that our sketch has not been drawn to scale. It would therefore make more sense to relabel it as shown. Vector 𝐀𝐁 is still parallel to vector 𝐃𝐂, and vector 𝐀𝐁 is perpendicular to vector 𝐂𝐁. We can now add the lengths onto our diagram. We now need to calculate the magnitude of vector 𝐁𝐂. Using the same method, we see that this is equal to 16. We now have the lengths of the parallel sides as well as the length of the perpendicular height of the trapezoid. The area is therefore equal to 13 plus 18 divided by two multiplied by 16. 13 plus 18 is equal to 31. Multiplying 31 over two by 16 gives us 248. The area of the trapezoid is therefore equal to 248 square units.

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