1. Tardigrade
  2. Question
  3. Mathematics
  4. In the square ABCD with side AB =2, two points M and N are on the adjacent sides of the square such that MN is parallel to the diagonal BD. If x is the distance of MN from the vertex A and f ( x )= Area ( triangle AMN ), then range of f ( x ) is :

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Q. In the square $ABCD$ with side $AB =2$, two points $M$ and $N$ are on the adjacent sides of the square such that $MN$ is parallel to the diagonal $BD$. If $x$ is the distance of $MN$ from the vertex $A$ and $f ( x )=$ Area $(\triangle AMN )$, then range of $f ( x )$ is :

Relations and Functions - Part 2

Solution:

$AP = x ; MN = y ; BD =2 \sqrt{2}$
hence, $\frac{ y }{2 \sqrt{2}}=\frac{2 \sqrt{2}- x }{\sqrt{2}} \Rightarrow \Delta^{\prime}$ 's CNM and CDB are similar $y =2(2 \sqrt{2}- x )$
$f(x)=\frac{x y}{2}=x(2 \sqrt{2}-x)=2-(x-\sqrt{2})^2$
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