Triangle Cde F120B6

1. **State the problem:** We need to find the perimeter and area of triangle $CDE$ with vertices $C(4, -1)$, $D(4, -5)$, and $E(2, -3)$. 2. **Find the lengths of the sides for the perimeter:** Use the distance formula between points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$: $$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$ - Length $CD$: $$CD = \sqrt{(4 - 4)^2 + (-5 - (-1))^2} = \sqrt{0^2 + (-4)^2} = \sqrt{16} = 4$$ - Length $DE$: $$DE = \sqrt{(2 - 4)^2 + (-3 - (-5))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$ - Length $EC$: $$EC = \sqrt{(4 - 2)^2 + (-1 - (-3))^2} = \sqrt{2^2 + 2^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \approx 2.83$$ 3. **Calculate the perimeter:** $$\text{Perimeter} = CD + DE + EC = 4 + 2.83 + 2.83 = 9.66$$ 4. **Find the area using the coordinate formula for the area of a triangle:** $$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$$ Using points $C(4, -1)$, $D(4, -5)$, and $E(2, -3)$: $$\text{Area} = \frac{1}{2} |4(-5 + 3) + 4(-3 + 1) + 2(-1 + 5)|$$ $$= \frac{1}{2} |4(-2) + 4(-2) + 2(4)| = \frac{1}{2} |-8 -8 + 8| = \frac{1}{2} |-8| = 4$$ 5. **Final answers:** - Perimeter is about $9.66$ units. - Area is $4$ square units.