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PROBLEM 23 MATHEMATICAL CALENDAR SEPTEMBER 2013

PROBLEM 23 MATHEMATICAL CALENDAR SEPTEMBER 2013. MATH COLEGIO AGUSTINIANO CIUDAD SALITRE ANDRES ALEJANDRO RAMÍREZ DIAZ-GRANADOS. G. C. D. H. F. E. A. B. MATHEMATICAL CALENDAR 2013 SEPTEMBER PROBLEM 23. INFORMATION: ABCD SQUARE. TRIANGLE EBF AND TRIANGLE FGH ARE EQUILATERALS

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PROBLEM 23 MATHEMATICAL CALENDAR SEPTEMBER 2013

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  1. PROBLEM 23 MATHEMATICAL CALENDAR SEPTEMBER 2013 MATH COLEGIO AGUSTINIANO CIUDAD SALITRE ANDRES ALEJANDRO RAMÍREZ DIAZ-GRANADOS.

  2. G C D H F E A B MATH PROBLEM 23 ANDRES RAMIREZ D.

  3. MATHEMATICAL CALENDAR 2013 SEPTEMBERPROBLEM 23 • INFORMATION: • ABCD SQUARE. • TRIANGLE EBF AND TRIANGLE FGH ARE EQUILATERALS • QUESTION: • DETERMINE THE ANGLES OF THE SHADED QUADRILATERALS MATH PROBLEM 23 ANDRES RAMIREZ D.

  4. PROCEDIMIENTO: 1.SINCE ABCD IS A SQUARE THE ANGLES C AND D COST 90 ° 2. AS THE TRIANGLES EBF AND FGH THEY ARE EQUILATERAL EACH OF THE ANGLES COST 60 °. 4. FOR THE QUADRILATERAL GCBF: THE ANGLE IN G GIVES 180 °-60 ° = 120 ° THE ANGLE IN F GIVES 180 °-60 ° = 120 ° 5. FOR THE ANGLE IN B = 360 °-(120 ° + 120 ° + 90 °) = 30 °. THE SUM OF THE INTERIOR ANGLES OF THE QUADRILATERAL ONE IS 360 ° 6.FOR THE QUADRILATERAL DAEI: THE ANGLE IN D GIVES 45 ° FOR BEING THE DIAGONAL OF THE SQUARE ABCD, THE ANGLE IN E GIVES 180 °-60 ° = 120 °, AND THE ANGLE IN I = 360 °-(90+45+120 °) = 105 °.THE SUM OF THE INTERIOR ANGLES OF THE QUADRILATERAL ONE IS 360 ° MATH PROBLEM 23 ANDRES RAMIREZ D.

  5. G C D H 90° 60° 120° 60° 45° 60° 120° F 60° 105° I 90° 60° 30° 120° 60° E A B MATH PROBLEM 23 ANDRES RAMIREZ D.

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