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A student solved the following problem and made an error: Triangles ABC and DEF. Angles A and F are congruent and measure 135 degrees. Coordinates for the vertices are at A 0, 2 and B 2, 4 and C 0, 0 and D 2, 0 and E 4, 4 and F 4, 2. Line 1 Segment AC equals 2. Segment FE equals 2. Segment AC is congruent to segment FE Line 2 ∠A ≅ ∠F Line 3 Length of segment AB. A (0, 2) B (2, 4) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub 1 squared, d equals square root of quantity 0 minus 2 all squared plus quantity 2 minus 4 all squared, d equals square root of negative 2 squared plus negative 2 squared, d equals square root of 4 plus 4, d equals square root of 8 segment AB = 2.83 Line 4 Length of segment DE. D (2, 0) E (4, 4) d equals square root of quantity x sub 2 minus x sub 1 squared plus quantity y sub 2 minus y sub squared, d equals square root of quantity 2 minus 4 all squared plus quantity 0 minus 4 all squared, d equals square root of negative 2 squared plus negative 4 squared, d equals square root of 4 plus 16, then d equals square root of 20 segment DE = 4.47 Line 5 segment AB is congruent with segment DE Line 6 triangle ABC is congruent with triangle FDEby SAS In which line did the student make the first mistake? Line 2 Line 5 Line 3 Line 4

Respuesta :

sqdancefan

The first false statement in the proof as it stands is in Line 5, where it is claimed that a line of length 2.83 is congruent to a line of length 4.47. This mistake cannot be corrected by adding lines to the proof.

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The first erroneous tactical move is in Line 4, where the length of DE is computed, rather than the length of FD. This mistake can be corrected by adding lines to the proof.

A correct SAS proof would use segment FD in Line 4, so it could be argued that the first mistake is there.

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