the area of the triangle with vertices at (-4,-1),(1,2),(4,-3) is
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Answer:
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√AC2=(−4−4)2+(−1+3)2
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√AC2=(−4−4)2+(−1+3)2=(−8)2+(2)2=64+6=68
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√AC2=(−4−4)2+(−1+3)2=(−8)2+(2)2=64+6=68AC=68−−√
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√AC2=(−4−4)2+(−1+3)2=(−8)2+(2)2=64+6=68AC=68−−√AC2=AB2+BC2
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√AC2=(−4−4)2+(−1+3)2=(−8)2+(2)2=64+6=68AC=68−−√AC2=AB2+BC2area of the△=12×34−−√×34−−√
Let A be (-4, -1), B be (1, 2) and C the (4, -3) be the vertices of the triangle.AB2=(−4−1)2+(−1−2)2(−5)2+(−3)2=25+9=34AB=34−−√BC2=(4−1)2+(−3−2)2=32+(−5)2=9+25=34BC=34−−√AC2=(−4−4)2+(−1+3)2=(−8)2+(2)2=64+6=68AC=68−−√AC2=AB2+BC2area of the△=12×34−−√×34−−√=342=17 sq.units
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Answer:
Answer: if it why this quadrilateral ABCD into two Triangles ABC and ACD then the area of triangle ABC is 8 square units and the area of triangle ACD is 6 square units.
If I add the areas of this two triangles then I get the area of quadrilateral ABCD that is 8 + 6=14 square units.
Step-by-step explanation: