Question

# Find the area of the square whose one pair of opposite vertices are (3,4) and (5,6).

Verified
145.5k+ views
Hint: Calculate the length of diagonal of the square using distance formula between two points. Use the formula for calculating the area of square when the length of diagonal is given, which is $\dfrac{{{a}^{2}}}{2}$ where ‘a’ is the length of the diagonal. You can also calculate the length of the side of the square when a diagonal is given and then calculate the area of the square.

We know that distance between two points of the form $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ is $\sqrt{{{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( {{y}_{1}}-{{y}_{2}} \right)}^{2}}}$.
Substituting ${{x}_{1}}=3,{{y}_{1}}=4,{{x}_{2}}=5,{{y}_{2}}=6$ in the above formula, the distance between the points (3,4) and (5,6) is $\sqrt{{{\left( 3-5 \right)}^{2}}+{{\left( 4-6 \right)}^{2}}}=\sqrt{{{2}^{2}}+{{2}^{2}}}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$ units.
We know that the area of a square when the length of the edge is ‘a’ is ${{a}^{2}}$. So, the length of the diagonal is $\sqrt{2}a$.
If we substitute $x=\sqrt{2}a$, the area of the square will be $\dfrac{{{x}^{2}}}{2}$.
Substituting $x=2\sqrt{2}$, the area of square $=\dfrac{{{a}^{2}}}{2}=\dfrac{{{\left( 2\sqrt{2} \right)}^{2}}}{2}=\dfrac{8}{2}=4$ sq. units.