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Two mutually perpendicular straight lines through the origin from an isosceles triangle with the line `2x + y = 5`. Then the area of the triangle is |
Answer» `AD=|(0+0-5)/sqrt(2^2+1^2)|=5/sqrt5=sqrt5` in `/_ABD` `tan 45^o=(AD)/(BD)=sqrt5/(BD)=1` BD=`sqrt5` so, DC=`sqrt5` BC=DC+`sqrt5` BC=`sqrt5+sqrt5` BC=`2sqrt5` area of `/_ABC = 1/2*B*H` =`1/2*BC*AD` =`1/2*2sqrt5*sqrt5` =`5units^2` |
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