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What is the area of triangle ABC whose vertices are at -3 7 and 2 19 and 10 7 respectively on the Cartesian plane showing work?
Wiki User
∙ 7y ago
The points (-3, 7) and (10, 7) lie on the same horizontal line as the y-coordinates are the same), therefore it is best to take this as the base of the triangle; it is 10 - -3 = 10 + 3 = 13 units long.
The third point (2, 19) is 19 - 7 = 12 units above this line which is the height of the triangle.
Thus the area is ½ × base × height = ½ × 13 units × 12 units = 78 square units.
Another possible answer:-
Distance formula: square root of (x2 -x1) together with +(y2 -y1)
Using the distance formula for the given vertices it works out that it is an isosceles triangle with equal sides of 13, a base of 4 Times Square root of 13 and a height of 3 times square root of 13.
Therefore area is: 0.5*4 times sq rt 13*3 times sq rt 13 = 78 square units
Checking using Pythagoras' theorem will confirm that it is an isosceles triangle.
Wiki User
∙ 7y ago
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