InterviewSolution
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InterviewSolution
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If the extremities of the base of an isosceles triangle are the points `(2a ,0)`and (0, a), and the equation of one of the side is `x=2a ,`then the area of the triangle is`5a^2s qdotu n i t s`(b) `(5a^2)/2s qdotu n i t s``(25 a^2)/2s qdotu n i t s`(d) none oftheseA. `5a^(2) "sq. units"`B. `5a^(2)//2 "sq. units"`C. `25a^(2)//2 "sq. units"`D. none of these |
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Answer» Correct Answer - B Given vertices are A(2a,0) and B(0,a) Let the coordinates of the third vertex C be (2a,t). Now, AC = BC. Hence, `t = sqrt(4a^(2) + (a-t)^(2)) " or " t =(5a)/(2)` So, the coordinates of the third vertex C are (2a,5a/2). Therefore, area of the triangle is `(1)/(2)||{:(2a, 5a//2, 1),(2a, 0, 1),(0, a, 1):}|| = (5a^(2))/(2)` sq. units |
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