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Case Study - 2
A field is in the form of a rectangle. The coordinates of the rectangular field $ABCD$ are $A(10, 10), B(40, 10), C(40, 50)$ and $D(x, y)$. Anil and Anita, two friends decided to have a race. Anita started from point $A$ and moved to point $E$ along the diagonal $AC$, where $E$ is the point of intersection of both the diagonals of $ABCD$. From point $E$, she moved to point $B$ along the other diagonal $DB$ and then moved back to point $A$ along $BA$. While Anil started from point $C$ and ran to point $A$ via $D$ along the boundary of the field.
Based on the above information, answer the following questions :
(i) Find the coordinates of point $E$.
(ii) Find the distance between the points $B$ and $C$.
(iii) (a) Find the coordinates of point $D$ and the distance $BD$.
OR
(b) Find the total distance travelled by Anita.
A field is in the form of a rectangle. The coordinates of the rectangular field $ABCD$ are $A(10, 10), B(40, 10), C(40, 50)$ and $D(x, y)$. Anil and Anita, two friends decided to have a race. Anita started from point $A$ and moved to point $E$ along the diagonal $AC$, where $E$ is the point of intersection of both the diagonals of $ABCD$. From point $E$, she moved to point $B$ along the other diagonal $DB$ and then moved back to point $A$ along $BA$. While Anil started from point $C$ and ran to point $A$ via $D$ along the boundary of the field.
Based on the above information, answer the following questions :
(i) Find the coordinates of point $E$.
(ii) Find the distance between the points $B$ and $C$.
(iii) (a) Find the coordinates of point $D$ and the distance $BD$.
OR
(b) Find the total distance travelled by Anita.
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Solution: (i) Coordinates of $E$ are $(25, 30)$
(ii) Distance $BC = \sqrt{(40 - 40)^2 + (50 - 10)^2} = 40$
(iii) (a) Co-ordinates of $D$ are $(\frac{x + 40}{2}, \frac{y + 10}{2}) = (25, 30) \Rightarrow x = 10, y = 50$. Co-ordinates of $D$ are $(10, 50)$. $BD = \sqrt{(30)^2 + (-40)^2} = 50$
OR
(iii) (b) Distance travelled by Anita $= AE + EB + BA = \sqrt{(15)^2 + (20)^2} + \sqrt{(-15)^2 + (20)^2} + \sqrt{(30)^2 + (0)^2} = 25 + 25 + 30 = 80$
(ii) Distance $BC = \sqrt{(40 - 40)^2 + (50 - 10)^2} = 40$
(iii) (a) Co-ordinates of $D$ are $(\frac{x + 40}{2}, \frac{y + 10}{2}) = (25, 30) \Rightarrow x = 10, y = 50$. Co-ordinates of $D$ are $(10, 50)$. $BD = \sqrt{(30)^2 + (-40)^2} = 50$
OR
(iii) (b) Distance travelled by Anita $= AE + EB + BA = \sqrt{(15)^2 + (20)^2} + \sqrt{(-15)^2 + (20)^2} + \sqrt{(30)^2 + (0)^2} = 25 + 25 + 30 = 80$