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Find the area of the triangle, whose vertices are (a,c+a), (a,c) and (-a,c-a). |
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Answer» Area of triangle `=1/2[x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))]` `=1/2[a(c-c+a)+a(c-a-a)-a(c+a-c)]` `=1/2[a^(2)-2a^(2)-a^(2)]=-a^(2)` But area of triangle connot be negative `therefore` Area of triangle `=a^(2)` square units |
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