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for # 10, write out the 3 triangles made up by the one standing up. you will see some equal sides and some equal angles. play with angles adding to 180 and you should get there. sorry. equal sides since they touch. | for # 10, write out the 3 triangles made up by the one standing up. you will see some equal sides and some equal angles. play with angles adding to 180 and you should get there. sorry. equal sides since they touch. | ||
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| + | there should be no reason to add 180 or play with angles...just use thm 5a with the first given to find 2 equal sides and the fact that PQ=PQ for two more equal sides. then look at your second given!--[[User:Jrhaynie|Jrhaynie]] 16:12, 4 November 2009 (UTC) | ||
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Doing #8 and no clue! | Doing #8 and no clue! | ||
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ok. Doing 4 now. almost there. Does it just take forever? | ok. Doing 4 now. almost there. Does it just take forever? | ||
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| + | For 3 once you have the perpendicular altitudes to ABC you have to look at the altitude of A as the side of the triangle AHB with BC already being perpendicular by the def of orthocenter of ABC. Ao AH is a side, BC is perpendicular, and B is a vertex, so BC is the altitude of B in triangle ABC. Similarly, CH is the altitude of H and AC to altitude of A. | ||
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| + | To do 4, use lines QPR, Q'P'R', XQP, YR'P, and Q'RZ and triangle ABC. 7 use the same proof as the book, but try to use SAS instead of SSS. For 8 use DEG~AGB. Then intersect AD and BE with center G and prove DEG~ABG. Then do the same with CF and BE and intersect them at point H and prove similarities. You can combine these inequalities to get the answer. | ||
[[ MA460 (Fall2009Walther) Homework|Back to MA460 (Fall2009Walther) Homework]] | [[ MA460 (Fall2009Walther) Homework|Back to MA460 (Fall2009Walther) Homework]] | ||
Latest revision as of 11:13, 4 November 2009
HW 9
Does anyone have 3,4,7 and 8?
Does anyone have any hints for #8 and #10?
for # 10, write out the 3 triangles made up by the one standing up. you will see some equal sides and some equal angles. play with angles adding to 180 and you should get there. sorry. equal sides since they touch.
there should be no reason to add 180 or play with angles...just use thm 5a with the first given to find 2 equal sides and the fact that PQ=PQ for two more equal sides. then look at your second given!--Jrhaynie 16:12, 4 November 2009 (UTC)
Doing #8 and no clue!
Also would appreciate any hints for #9. Thanks.
9 isn't too bad. do all angles equal 180 degress for gha and cba. and in the middle to get 2 angles equal to 90 degrees. Makes 2 triangles equal to half way there. then same thing with 2 other sets of triangles. Think angles, not so much sides.
ok. Doing 4 now. almost there. Does it just take forever?
For 3 once you have the perpendicular altitudes to ABC you have to look at the altitude of A as the side of the triangle AHB with BC already being perpendicular by the def of orthocenter of ABC. Ao AH is a side, BC is perpendicular, and B is a vertex, so BC is the altitude of B in triangle ABC. Similarly, CH is the altitude of H and AC to altitude of A.
To do 4, use lines QPR, Q'P'R', XQP, YR'P, and Q'RZ and triangle ABC. 7 use the same proof as the book, but try to use SAS instead of SSS. For 8 use DEG~AGB. Then intersect AD and BE with center G and prove DEG~ABG. Then do the same with CF and BE and intersect them at point H and prove similarities. You can combine these inequalities to get the answer.
