Question

Coordinates (-3, 5), ( -3, 8), and (2, 5). The triangles was dilated by a scale favtor of 3. What is the area of the dilated triangle, in square units?

Answer 1

67.5 square units

Explanation:

The points (-3, 5) and ( -3, 8) form a vertical segment, and the points (-3, 5) and ( 2, 5) form a horizontal segment, so the triangle has a right angle.

The base of the triangle is the segment formed by the points (-3, 5) and ( 2, 5), and their distance is:

D1 = sqrt((-3-2)^2 + (5-5)^2) = 5

The height of the triangle is the segment formed by the points (-3, 5) and ( -3, 8), and their distance is:

D2 = sqrt((-3+3)^2 + (5-8)^2) = 3

If the triangle is dilated by a scale of 3, so the base is D1*3 = 15 and the height is D2*3 = 9, then the area is:

The area of a triangle is Base * Height / 2, so we have that:

Area = 15 * 9 / 2 = 67.5 square units

Answer 2

The area of the triangle is 7.5 unit²

Explanation:

Here we have the coordinates given as

(-3, 5) (-3, 8) and (2, 5)

Let us call the points A = (-3, 5)

B = (-3, 8) and

C = (2, 5)

Therefore the lengths of the sides of the triangle are

AB = a =
`√((-3-(-3))^2+(5-8)^2) = 3`

AC = b =
`√((-3-2)^2+(5-5)^2) = 5`

BC = c =
`√((-3-2)^2+(8-5)^2) = √(34)`

Therefore the Area can be derived from Heron's formula which is

A =
`√(s* (s-a)* (s-b)* (s-c))` where s = semi perimeter or (a + b + c)/2

Therefore, plugging the values, we get

s = (3 + 5 + √34)/2 = 6.9155≈6.92

A =
`\sqrt{6.92* (6.92-3)* (6.92-5)* (6.92-√(34) )}` = 7.499≈7.5.