Areas Of Similar Triangles: A Step-by-Step Solution

by TextBrain Team 52 views

Hey guys! Let's dive into a classic geometry problem involving similar triangles. We've got triangles ABC and A1B1C1 that are similar, meaning they have the same shape but potentially different sizes. The ratio of their corresponding sides is 3:4, and the area of triangle ABC is 70 square meters less than the area of triangle A1B1C1. Our mission, should we choose to accept it, is to find the areas of both triangles. Buckle up, because we're about to break it down!

Understanding Similar Triangles and Area Ratios

Before we jump into calculations, let's quickly recap the key concepts about similar triangles. When two triangles are similar, their corresponding angles are equal, and their corresponding sides are in proportion. This proportionality extends to their areas, but in a squared manner. This is a crucial point to remember: the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

Think of it this way: if you double the sides of a triangle, you're not just doubling the area; you're scaling it up by a factor of four (2 squared). This principle is fundamental to solving our problem. To really drive this home, imagine you have a small triangle drawn on a piece of paper. Now, imagine you photocopy it and enlarge it so that every side is twice as long. The new, larger triangle will have an area four times bigger than the original! This happens because area is a two-dimensional measurement, so scaling the sides affects both the length and the width, effectively squaring the scaling factor. This relationship is not just a handy trick for solving problems; it reflects a deep connection between the geometry of shapes and the way area scales with size. Understanding this concept will not only help you solve this particular problem but also equip you with a powerful tool for tackling various geometric challenges in the future. We'll see how this works practically in the next section.

Setting up the Equations

Now, let's translate the problem's information into mathematical equations. This is where the magic happens, guys! We'll use the area ratio concept we just discussed. First, let's define our variables:

  • Let the area of triangle ABC be denoted as S. This is the area of the smaller triangle, which we'll use as our baseline.
  • Consequently, the area of triangle A1B1C1 is S + 70, reflecting the given difference of 70 square meters.

Remember that the ratio of corresponding sides is 3:4. This means the ratio of the areas is (3/4)^2, which simplifies to 9/16. This 9/16 ratio is the key to connecting the areas of our two triangles. We can now express the relationship between their areas using this ratio. The area of the smaller triangle (ABC) divided by the area of the larger triangle (A1B1C1) equals 9/16. Mathematically, this is represented as:

S / (S + 70) = 9 / 16

This equation is the cornerstone of our solution. It succinctly captures the relationship between the two areas and incorporates the information about both the side ratio and the area difference. From here, it's all about the algebra. By solving this equation for S, we'll uncover the area of triangle ABC. Once we have that, finding the area of triangle A1B1C1 is a simple matter of adding 70. So, with our variables defined and our core equation in place, we're well-equipped to solve for the unknowns and crack this geometric puzzle!

Solving for the Areas

Alright, let's get our hands dirty with some algebra! We have the equation:

S / (S + 70) = 9 / 16

To solve for S, we can cross-multiply. This involves multiplying the numerator of the first fraction by the denominator of the second, and vice-versa. This step helps us eliminate the fractions and get everything on a single line, making the equation easier to manipulate. So, let's do it:

16 * S = 9 * (S + 70)

Now, we need to distribute the 9 on the right side of the equation. Remember the distributive property? It says that a number multiplied by a sum is the same as that number multiplied by each term in the sum individually. So, we multiply 9 by both S and 70:

16S = 9S + 630

Our next step is to isolate S on one side of the equation. To do this, we can subtract 9S from both sides. This keeps the equation balanced while moving all the S terms to one side:

16S - 9S = 630

This simplifies to:

7S = 630

Finally, to solve for S, we divide both sides of the equation by 7:

S = 630 / 7

S = 90

So, we've found that the area of triangle ABC (S) is 90 square meters. Now, to find the area of triangle A1B1C1, we simply add 70 square meters (as the problem stated the difference in areas):

Area of A1B1C1 = S + 70 = 90 + 70 = 160 square meters.

Therefore, the area of triangle ABC is 90 square meters, and the area of triangle A1B1C1 is 160 square meters. Woohoo! We solved it!

Final Answer and Key Takeaways

So, there you have it, guys! We've successfully navigated the world of similar triangles and found the areas of our two triangles. To recap:

  • The area of triangle ABC is 90 m^2.
  • The area of triangle A1B1C1 is 160 m^2.

But beyond the specific answer, let's highlight some key takeaways from this problem. These are the concepts and strategies that will serve you well in future geometric adventures:

  1. The Area Ratio of Similar Triangles: Remember, the ratio of the areas of similar triangles is equal to the square of the ratio of their corresponding sides. This is a fundamental relationship that pops up frequently in geometry problems.
  2. Translating Word Problems into Equations: The ability to convert a written problem into mathematical equations is a crucial skill in mathematics. Identifying the unknowns, assigning variables, and expressing the relationships as equations is half the battle.
  3. Algebraic Manipulation: Once you have your equations, proficiency in algebraic techniques like cross-multiplication, distribution, and isolating variables is essential for solving them accurately.
  4. Checking Your Work: It's always a good idea to plug your answers back into the original equations or conditions to make sure they make sense. In this case, you could check if the area ratio (90/160) simplifies to 9/16 and if the area difference is indeed 70.

This problem is a great example of how different mathematical concepts – geometry, ratios, and algebra – come together to solve a single problem. By mastering these concepts and practicing problem-solving strategies, you'll be well-equipped to tackle any geometric challenge that comes your way. Keep practicing, keep exploring, and most importantly, keep enjoying the beauty of math!