Trapez Problem: Geometry Explained Step-by-Step

Hey everyone, let's dive into a classic geometry problem involving a trapezoid. We'll break down each part step-by-step, making sure you understand every concept. This is a great way to brush up on your geometry skills. This problem is about a trapezoid ABCD, where AB is parallel to CD, and angles A and D are right angles. We're given that AB is 14 cm, AD and DC are both 8 cm. Point E lies on DC, and DE is three-quarters of DC. So, let’s get started and solve this geometry problem together.

Understanding the Problem and Given Information

Alright, guys, before we jump into the solution, it's super important to clearly understand the problem and what we're given. In this case, we're dealing with a trapezoid ABCD. Remember, a trapezoid is a quadrilateral with at least one pair of parallel sides. Here, AB is parallel to CD. We also know that angles A and D are right angles (90 degrees). This means the trapezoid is a right trapezoid. Think of it like a rectangle with one side slanted. We're also given some side lengths: AB = 14 cm, AD = DC = 8 cm. That is cool, because two sides are the same length. Now, the trickier part: point E lies on DC, and DE is equal to three-quarters of DC. This means DE = (3/4) * 8 cm = 6 cm. This point E splits DC into two segments.

So, to recap, we know the shape, the parallel sides, the right angles, and the lengths of AB, AD, DC, and DE. With all this information we're ready to proceed to the next parts. Always start by drawing a diagram, so you can visualize the shape and its elements. Trust me, it makes everything easier! This problem involves a trapezoid, and right angles, and calculating perimeters. We have a set of details to analyze: parallel lines, right angles, and specific side lengths. From there, we can move forward and look at solving each question. Let us move on to Part A to determine the perimeter.

Part A: Calculating the Perimeter of Triangle ADE

Now, let's tackle part (a): Show that the perimeter of triangle ADE is equal to 24 cm. Remember that the perimeter of any shape is the sum of the lengths of all its sides. In this case, we have triangle ADE. We already know AD = 8 cm, and we just calculated DE = 6 cm. But we need to find AE to calculate the perimeter.

To find AE, we can use the Pythagorean theorem because triangle ADE is a right-angled triangle (with angle D being 90 degrees). The Pythagorean theorem states that in a right-angled triangle, the square of the length of the hypotenuse (AE in this case) is equal to the sum of the squares of the lengths of the other two sides (AD and DE). So, AE² = AD² + DE².

Let’s plug in the values: AE² = 8² + 6² = 64 + 36 = 100. Taking the square root of both sides, we get AE = 10 cm. Now we have all the side lengths of triangle ADE: AD = 8 cm, DE = 6 cm, and AE = 10 cm. Now we're ready to calculate the perimeter. Perimeter of ADE = AD + DE + AE = 8 cm + 6 cm + 10 cm = 24 cm. Therefore, we have successfully shown that the perimeter of triangle ADE is 24 cm. That was not too hard, right? Always break it down into smaller steps, so it's easier to manage.

Now we proceed to the next section to show how to prove that AC ⊥ BE.

Part B: Proving AC is Perpendicular to BE

Now, for part (b): Prove that AC is perpendicular to BE. This is where things get a bit more interesting, and we'll need to use a few more geometry concepts. To prove that two lines are perpendicular, we need to show that the angle between them is 90 degrees. In this case, we need to show that angle AEB = 90 degrees.

First, let's find the length of CE. Since DE = 6 cm and DC = 8 cm, then CE = DC - DE = 8 cm - 6 cm = 2 cm. That's a key value we will use shortly. Next, let’s consider triangle ABC. We know AB = 14 cm. To find BC, we can use the Pythagorean theorem. To do so, imagine that we drop a perpendicular line from B to DC, and call the intersection point F. Then, BF is parallel to AD and has a length of 8 cm.

Now, let us find FC. Since AB = 14 cm and CD = 8 cm, we can calculate FC. Think of it like this: If we extend the sides of the trapezoid, the length of the base AB is split into the segments DE, EF, and FC. Since DE = 6 cm and DC = 8 cm, then EF would have a length equal to AB - CD, so FC = 14 cm - 8 cm = 6 cm. Now we can apply the Pythagorean theorem for the triangle BCF, and we get: BC² = BF² + FC² = 8² + 6² = 64 + 36 = 100. The BC = 10 cm. Alright, we have a lot of information now, let us see how we can prove AC is perpendicular to BE. So, we found that AE = 10 cm and BC = 10 cm. This is a very useful data for us to prove the requested question.

To prove AC is perpendicular to BE, we can utilize the slopes of the lines. If the product of the slopes of two lines is -1, then the lines are perpendicular. Now, let’s calculate the coordinates, assuming that A = (0, 8), D = (0, 0), and C = (8, 0). Then E = (6, 0). To calculate the coordinates of B, consider BCF as a right triangle, BC = 10 cm and FC = 6 cm, so BF = 8 cm. Thus, B = (8, 8). Now, we have the coordinates. The slope of AC = (0 - 8) / (8 - 0) = -1. The slope of BE = (0 - 8) / (6 - 8) = 4.

Now let's calculate the product of the slopes: (-1) * 4 = -4. Since -4 is not equal to -1, it means that this proof does not stand. The assumption about the data must be wrong. I realize now that the coordinates for point B should not be (8,8), it must be (14, 8) and CE = 2 cm, so point E must be (6, 0). The slope of AC = (0-8) / (8-0) = -1. The slope of BE = (8-0) / (14-6) = 1. Then we proceed to find the product: -1 * 1 = -1. This is correct, thus we proved that AC is perpendicular to BE. We can finally conclude that angle formed between AC and BE is 90 degrees. This confirms that AC ⊥ BE. Awesome, that was the solution for this geometry problem.

Conclusion

Great job, guys! We've successfully solved both parts of this geometry problem. We calculated the perimeter of triangle ADE and proved that AC is perpendicular to BE. Remember, the key to solving geometry problems is to break them down into smaller, manageable steps. Always start with a diagram, identify what you know, and what you need to find. Then, apply the relevant formulas and theorems. Keep practicing, and you'll become a geometry pro in no time! Also, do not forget to go back and check your work to avoid making mistakes. Practice makes perfect, so keep solving geometry problems, and you'll get better and better.