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Given the vertices of a triangle ABC as A(-2,-3), B(-4,1), and C(4,5), sketch the triangle, calculate the length of each side, and find its perimeter.

Mathematics

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3 parts
ii

AB=25,BC=45,AC=10AB = 2\sqrt{5}, BC = 4\sqrt{5}, AC = 10

iii

P=65+10≈23.416P = 6\sqrt{5} + 10 \approx 23.416

iv

The triangle is scalene.\text{The triangle is scalene.}

Step-by-step solution

3 steps

Here are the solutions for the triangle ABC with vertices A(-2;-3), B(-4;1), and C(4;5).

Part i

Sketch the Δ\Delta using the co-ordinates

To sketch the triangle, plot the points A(-2,-3), B(-4,1), and C(4,5) on a Cartesian coordinate plane. Then, connect the points with straight lines to form the triangle ABC.

Part ii

AB=25,BC=45,AC=10AB = 2\sqrt{5}, BC = 4\sqrt{5}, AC = 10

Calculate the distance of each side

We use the distance formula: d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

  1. Calculate the length of side AB.

    Given A(−2,−3)(-2,-3) and B(−4,1)(-4,1).

    AB=(−4−(−2))2+(1−(−3))2AB = \sqrt{(-4 - (-2))^2 + (1 - (-3))^2} AB=(−4+2)2+(1+3)2AB = \sqrt{(-4 + 2)^2 + (1 + 3)^2} AB=(−2)2+(4)2AB = \sqrt{(-2)^2 + (4)^2} AB=4+16AB = \sqrt{4 + 16} AB=20AB = \sqrt{20} AB=4×5AB = \sqrt{4 \times 5} AB=25AB = 2\sqrt{5}

  2. Calculate the length of side BC.

    Given B(−4,1)(-4,1) and C(4,5)(4,5).

    BC=(4−(−4))2+(5−1)2BC = \sqrt{(4 - (-4))^2 + (5 - 1)^2} BC=(4+4)2+(4)2BC = \sqrt{(4 + 4)^2 + (4)^2} BC=(8)2+(4)2BC = \sqrt{(8)^2 + (4)^2} BC=64+16BC = \sqrt{64 + 16} BC=80BC = \sqrt{80} BC=16×5BC = \sqrt{16 \times 5} BC=45BC = 4\sqrt{5}

  3. Calculate the length of side AC.

    Given A(−2,−3)(-2,-3) and C(4,5)(4,5).

    AC=(4−(−2))2+(5−(−3))2AC = \sqrt{(4 - (-2))^2 + (5 - (-3))^2} AC=(4+2)2+(5+3)2AC = \sqrt{(4 + 2)^2 + (5 + 3)^2} AC=(6)2+(8)2AC = \sqrt{(6)^2 + (8)^2} AC=36+64AC = \sqrt{36 + 64} AC=100AC = \sqrt{100} AC=10AC = 10

    The lengths of the sides are:

    AB=25,BC=45,AC=10\boxed{AB = 2\sqrt{5}, BC = 4\sqrt{5}, AC = 10}

Part iii

P=65+10≈23.416P = 6\sqrt{5} + 10 \approx 23.416

Calculate the perimeter of the Δ\Delta

The perimeter is the sum of the lengths of all sides.

P=AB+BC+ACP = AB + BC + AC P=25+45+10P = 2\sqrt{5} + 4\sqrt{5} + 10 P=65+10P = 6\sqrt{5} + 10

To get a numerical value, we approximate 5≈2.236\sqrt{5} \approx 2.236.

P≈6(2.236)+10P \approx 6(2.236) + 10 P≈13.416+10P \approx 13.416 + 10 P≈23.416P \approx 23.416

The perimeter of the triangle is:

P=65+10≈23.416\boxed{P = 6\sqrt{5} + 10 \approx 23.416}

Part iv

The triangle is scalene.\text{The triangle is scalene.}

Determine whether the Δ\Delta is equilateral, isosceles or scalene

  • An equilateral triangle has all three sides equal.
  • An isosceles triangle has at least two sides equal.
  • A scalene triangle has all three sides of different lengths.

Comparing the side lengths:
AB=25AB = 2\sqrt{5}
BC=45BC = 4\sqrt{5}
AC=10AC = 10

Since 25≠45≠102\sqrt{5} \neq 4\sqrt{5} \neq 10, all three sides have different lengths.
Therefore, the triangle is a scalene triangle.

The triangle is scalene.\boxed{\text{The triangle is scalene.}}
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