Difference between revisions of "2005 AMC 8 Problems/Problem 19"
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==Solution== | ==Solution== | ||
Draw altitudes from <math>B</math> and <math>C</math> to base <math>AD</math> to create a rectangle and two right triangles. The side opposite <math>BC</math> is equal to <math>50</math>. The bases of the right triangles can be found using Pythagorean or special triangles to be <math>18</math> and <math>7</math>. Add it together to get <math>AD=18+50+7=75</math>. The perimeter is <math>75+30+50+25=\boxed{\textbf{(A)}\ 180}</math>. | Draw altitudes from <math>B</math> and <math>C</math> to base <math>AD</math> to create a rectangle and two right triangles. The side opposite <math>BC</math> is equal to <math>50</math>. The bases of the right triangles can be found using Pythagorean or special triangles to be <math>18</math> and <math>7</math>. Add it together to get <math>AD=18+50+7=75</math>. The perimeter is <math>75+30+50+25=\boxed{\textbf{(A)}\ 180}</math>. | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/h0V8qcEo0U8 Soo, DRMS, NM | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2005|num-b=18|num-a=20}} | {{AMC8 box|year=2005|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 00:30, 24 March 2022
Contents
Problem
What is the perimeter of trapezoid ?
Solution
Draw altitudes from and to base to create a rectangle and two right triangles. The side opposite is equal to . The bases of the right triangles can be found using Pythagorean or special triangles to be and . Add it together to get . The perimeter is .
Video Solution
https://youtu.be/h0V8qcEo0U8 Soo, DRMS, NM
See Also
2005 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.