Answer:
Therefore,
![AB=6\ unit\\BC=6\ unit\\CD=6\ unit\\DA=6\ unit\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/e9s205vw39v1u17iibo3g11bo5a601rr58.png)
Since , ABCD has four Right angles and four Congruent sides, it is a Square
Explanation:
The four points for the Figure are
point A( x₁ , y₁) ≡ ( 0 , 6)
point B( x₂ , y₂) ≡ (6 , 6)
point C(x₃ , y₃ ) ≡ (6 , 0)
point D(x₄ , y₄ ) ≡ (0 , 0)
∠A = ∠B = ∠C = ∠D = 90°
To Prove:
ABCD is a Square
Proof:
∠A = ∠B = ∠C = ∠D = 90° .........Given:
Now By Distance Formula we have
![l(AB) = \sqrt{((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2) )}](https://img.qammunity.org/2021/formulas/mathematics/high-school/2vq37tt8z78j2dqe55jrmva1qn6d4rf2bh.png)
Substituting we get
![l(AB) = \sqrt{(6-0)^(2)+(6-6)^(2))}=\sqrt{6^(2)}=6\ unit](https://img.qammunity.org/2021/formulas/mathematics/high-school/f1mbtic3fv08hgtjqjrcngsvyj6q3ihu8e.png)
Similarly for BC ,CD ,DA we have
![l(BC) = \sqrt{(6-6)^(2)+(0-6)^(2))}=\sqrt{(-6)^(2)}=6\ unit](https://img.qammunity.org/2021/formulas/mathematics/high-school/2bcwfd2az3y37prb7vbl6pw4no0cx969mp.png)
![l(CD) = \sqrt{(0-6)^(2)+(6-6)^(2))}=\sqrt{(-6)^(2)}=6\ unit](https://img.qammunity.org/2021/formulas/mathematics/high-school/pgnng82kzk39movmxzhr2q30u5g6q9mdep.png)
![l(DA) = \sqrt{(0-0)^(2)+(6-0)^(2))}=\sqrt{6^(2)}=6\ unit](https://img.qammunity.org/2021/formulas/mathematics/high-school/d6x41rx0r2rxyq8kdg0cjjabeeqqj0vizw.png)
Therefore,
![AB=6\ unit\\BC=6\ unit\\CD=6\ unit\\DA=6\ unit\\](https://img.qammunity.org/2021/formulas/mathematics/high-school/e9s205vw39v1u17iibo3g11bo5a601rr58.png)
Since , ABCD has four Right angles and four Congruent sides, it is a Square