모양을 판별하여 넓이 구하기

$\displaystyle \overline{\mathrm{AB}} ^{2} = \left( 2 + 2 \right) ^{2} + \left( - 1 - 1 \right) ^{2} = 20$ $\displaystyle \overline{\mathrm{BC}} ^{2} = \left( 4 - 2 \right) ^{2} + \left( 3 + 1 \right) ^{2} = 20$ $\displaystyle \overline{\mathrm{AC}} ^{2} = \left( 4 + 2 \right) ^{2} + \left( 3 - 1 \right) ^{2} = 40$ $\displaystyle \overline{\mathrm{AB}} ^{2} + \overline{\mathrm{BC}} ^{2} = \overline{\mathrm{AC}} ^{2}$이므로 삼각형 $\mathrm{ABC}$는 $\mathrm{B}$가 직각인 직각삼각형이고 $\displaystyle \overline{\mathrm{AB}} = \overline{\mathrm{BC}} = 2 \sqrt{5}$ 따라서 넓이는 $\displaystyle \frac{1}{2} \times 2 \sqrt{5} \times 2 \sqrt{5} = 10$ ③