Question

Classify the polygon: Coordinates: A(1, 4), B(4, 5), C(5, 2) Explain your answers. Must show all work to support your reasoning. a. Classify ∆ABC as equilateral, isosceles, or scalene. Explain your reasoning. b. Determine if ∆ABC is a right triangle. Explain your reasoning. c. Find the perimeter and area of ∆ABC. Round decimal answers to the nearest tenth.

Answer 1

a. ΔABC is an isosceles triangle

b. ΔABC is a right triangle

c. The perimeter of ΔABC is approximately 10.8

Explanation:

The given vertices of the polygon, are;

A(1, 4), B(4, 5), C(5, 2)

By observation of the differences of the coordinates of the given points, the points are non-linear, as the rate of increase in the y-values, for a given increase in the x-value, is not constant

The length of segments of the triangle ΔABC are found using the formula for finding the distance, 'd', between points given their coordinates, (x₁, y₁), (x₂, y₂), as follows;

`d = \sqrt{\left (y_(2)-y_(1) \right )^(2)+\left (x_(2)-x_(1) \right )^(2)}`

The length of segment,
`\overline {AB}` = √((5 - 4)² + (4 - 1)²) = √10

The length of segment,
`\overline {AC}` = √((2 - 4)² + (5 - 1)²) = √20

The length of segment,
`\overline {BC}` = √((2 - 5)² + (5 - 4)² = √10

The length of segment,
`\overline {AB}` = The length of segment,
`\overline {BC}`, therefore, ΔABC is an isosceles triangle by definition of an isosceles triangle which is a triangle that have two sides of equal length

ΔABC is an isosceles triangle

b. According to Pythagoras' theorem, the sum of the squares of the two shorter sides (the legs) of a right triangle = The square of the length of the longest side

In if triangle ΔABC is a right triangle, we should have;

`\overline {AB}`² +
`\overline {BC}`² =
`\overline {AC}`²

Checking gives;

(√10)² + (√10)² = 10 + 10 = 20 = (√20)²

Whereby
`\overline {AB}`² (= (√10)²) +
`\overline {BC}`² (= (√10)²) = (√20)² =
`\overline {AC}`², ΔABC is a right triangle

ΔABC is a right triangle

c. The perimeter of ΔABC = The length of segment,
`\overline {AB}` + The length of segment,
`\overline {AC}` + The length of segment,
`\overline {BC}`

∴ The perimeter of ΔABC = √10 + √10 + √20 ≈ 10.7966912753

By rounding off to the nearest tenth, the perimeter of ΔABC ≈ 10.8.