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# Right Triangle Similarity

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### Right Triangle Similarity

1. 1. SIMILARITY IN RIGHT TRIANGLES<br />
3. 3. We draw altitude CD to the hypotenuse.<br />
4. 4. This divides the original triangle into two smaller right triangle:<br />
5. 5. This divides the original triangle into two smaller right triangle: ΔDCA<br />
6. 6. This divides the original triangle into two smaller right triangle: ΔBDC<br />
7. 7. There are a three triangles in the figure below.<br />
8. 8. Big<br />Medium<br />Small<br />
9. 9. Big<br />Medium<br />Small<br />
10. 10. We orient the three triangles to see the them clearer.<br />Big<br />Small<br />Medium<br />
11. 11. We can see that the three triangles are similar to each other. <br />~<br />~<br />Big<br />Small<br />Medium<br />
12. 12. SIMILARITY IN RIGHT TRIANGLES<br />
13. 13. Parts of a right triangle incorporated with the altitude<br />C<br />Leg adjacent to DB<br />Leg adjacent to AD<br />A<br />B<br />D<br />Segments of the hypotenuse AD and DB<br />
14. 14. Right Triangle Similarity Theorem<br />The altitude to the hypotenuse of a right triangle divides the triangle into two triangles that are similar to the original triangle and to each other.<br />C<br />A<br />B<br />D<br />ABC ~ ACD ~ CBD<br />
15. 15. Geometric Mean-Altitude Theorem 1<br />The length of the altitude to the hypotenuse is the geometric mean of the lengths of the segments of the hypotenuse.<br />C<br />A<br />B<br />D<br />𝐀𝐃𝐂𝐃=𝐂𝐃𝐃𝐁<br /> <br />𝐂𝐃=𝐀𝐃×𝐃𝐁<br /> <br />
16. 16. Geometric Mean-Altitude Theorem 2<br />The altitude to the hypotenuse to a right triangle intersects it to that the length of each leg us the geometric mean of the length of its adjacent segment of the hypotenuse and the length of the entire hypotenuse<br />BACB=CBBD<br /> <br />ABCA=CAAD<br /> <br />𝐂𝐁=𝐁𝐃×𝐁𝐀<br /> <br />𝐂𝐀=𝐀𝐃×𝐀𝐁<br /> <br />
17. 17. Summary of theGeometric Mean – Altitude Theorem<br />b<br />a<br />h<br />m<br />n<br />c<br />𝐜=𝐦+𝐧<br /> <br />𝐚=𝐦×𝐜<br /> <br />𝐡=𝐦×𝐧<br /> <br />𝐛=𝐧×𝐜<br /> <br />
18. 18. Solve for the other missing lengths given only two measurements.<br />a = 4, b = 6<br />a = 8, c = 10<br />a = 5, m = 7<br />a = 9, n = 6<br />a = 12, h = 9<br />b = 6, c = 15<br />b = 8, m = 9<br />b = 4, n = 3<br />b = 11, h = 8<br />c = 18, m = 12<br />c = 15, n = 8<br />c = 20, h = 6<br />m = 12, n = 8<br />m = 9, h = 12<br />n = 10, h = 12<br />
19. 19. Solve for the other missing lengths given only two measurements.<br />a = 4, b = 6<br />a = 8, c = 10<br />a = 5, c = 8<br />a = 9, m = 6<br />a = 12, h = 9<br />b = 6, c = 15<br />b = 12, n = 9<br />b = 4, n = 3<br />b = 10, h = 6<br />b = 11, h = 8<br />c = 18, m = 12<br />c = 15, n = 8<br />c = 20, m = 6<br />m = 12, n = 8<br />m = 9, h = 12<br />n = 10, h = 12<br />a = 9, m = 6<br />c = 18, m = 12<br />
20. 20. Relating to the Real WorldRecreation<br />At the parking lot of a State Park, the 300-m path to the snack bar and the 400-m path to the boat rental shop meet at a right angle. Marla walks straight from the parking lot to the ocean. How far is Marla from the snack bar?<br />WHICH IS STRONGER? TRIANGLE OR QUADRILATERAL?<br />