Question

#Maths Aryabhattas only:
Open challenge. [Co-ordinate geometry]
The area of square with one of its vertices as (3,4) and point of intersection of diagonals as (5,3) is ?
[Ans. 10]​

Answer 1

Answer:

refer to the attachment for the explanation

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Answer 2

One of the vertices of the square is (3, 4) and the point of intersection of diagonals is (5, 3).

The distance between (3, 4) and (5, 3) is, by distance formula,

$\longrightarrow x=\sqrt{(5-3)^2+(3-4)^2}$

$\longrightarrow x=\sqrt{2^2+(-1)^2}$

$\longrightarrow x=\sqrt{4+1}$

$\longrightarrow x=\sqrt{5}$

This distance is half the length of diagonal.

Because the point of intersection of the diagonals (5, 3) is the midpoint of the diagonals.

So the length of the diagonal is,

$\longrightarrow d=2x$

$\longrightarrow d=2\sqrt5$

Then, area of the square,

$\longrightarrow A=\dfrac{d^2}{2}$

$\longrightarrow A=\dfrac{(2\sqrt5)^2}{2}$

$\longrightarrow A=\dfrac{20}{2}$

$\longrightarrow\underline{\underline{A=10}}$

Hence 10 is the answer.