Question

1. find the length of the sides of the triangle with vertices a(0,4), b(5,4) and c(-3,-2). classify the triangle by its sides (equilateral, isosceles, or scalene) and angles (right, obtuse, acute, or equiangular).

round the side lengths to the nearest hundredth (2 decimal places)
ab =
bc =
ca =
△abc is a (sides), (angles) triangle

Explanation:

Step1: Calculate length of AB

Use distance formula: $AB = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
$AB = \sqrt{(5-0)^2+(4-4)^2} = \sqrt{25+0} = 5.00$

Step2: Calculate length of BC

Apply distance formula to B(5,4) and C(-3,-2)
$BC = \sqrt{(-3-5)^2+(-2-4)^2} = \sqrt{(-8)^2+(-6)^2} = \sqrt{64+36} = \sqrt{100} = 10.00$

Step3: Calculate length of CA

Apply distance formula to C(-3,-2) and A(0,4)
$CA = \sqrt{(0-(-3))^2+(4-(-2))^2} = \sqrt{3^2+6^2} = \sqrt{9+36} = \sqrt{45} \approx 6.71$

Step4: Classify by side lengths

All sides are distinct (5.00, 10.00, 6.71), so scalene.

Step5: Classify by angles

Use Pythagorean theorem test: $AB^2 + CA^2 = 5^2 + (\sqrt{45})^2 = 25+45=70$, $BC^2=100$. Since $AB^2 + CA^2 < BC^2$, the angle opposite BC is obtuse.

Answer:

AB = 5.00
BC = 10.00
CA = 6.71
$\triangle ABC$ is a scalene (sides), obtuse (angles) triangle