Make a flow chart showing that the triangles below are similar. NO LINKS OR CRAZY LANGUAGES.!!!!!!
1 answer:
Answer: See the attached image
You have the correct idea for the boxes you've filled out. For the first three boxes in column 1, I would be specific which segments you are dividing. So for instance, in the first box, it would be EG/EB = 55/11 = 5. Then the second box would be EF/EC = 35/7 = 5, and so on. The order of the boxes doesn't matter. The three boxes then combine together to help show that the triangles are similar. Specifically . The order of the letters is important to help show how the angles pair up and how the sides pair up. We use the SSS similarity theorem here.
The second problem is the same idea, but we use one pair of congruent angles. So we'll use the SAS similarity theorem this time.
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2 ways: make a little divide tree based on how many times you can divide it. Like you have sqrt88 you know it can divide by 2. So its 2 and 44, and 44 can be broken to 2 and 22, then 2 and 11. 11 can’t be divided anymore.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.
Now choose numbers that are the same and pair them together. These go outside the radicand. You had three 2’s and one 11. So a 2 would go on the outside and the remaining 2 and 11 would be multiplied leaved 2sqrt22.
Another way is to divide by the highest squared number. Taking sqrt88 again. 4 would be the highest squared number. So 88 would be 4 and 22, then 2 and 11. Since you have a 4 which is a perfect square you put that squareroot outside. The inside would remain the same because it wouldn’t break down into perfect squares or pairs anymore. So that would be 2sqrt22.