We read your photo as:

Given the vertices A(2;-1), B(-3;4), and C(4;5), sketch the triangle and calculate the lengths of its sides and its perimeter.

Mathematics

ScanSolve AI Answer

Answers

3 parts
ii

AB=52,BC=52,AC=210AB = 5\sqrt{2}, BC = 5\sqrt{2}, AC = 2\sqrt{10}

iii

P=102+210≈20.464P = 10\sqrt{2} + 2\sqrt{10} \approx 20.464

iv

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

Step-by-step solution

3 steps

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

Part i

Sketch the Δ\Delta using the co-ordinates

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

Part ii

AB=52,BC=52,AC=210AB = 5\sqrt{2}, BC = 5\sqrt{2}, AC = 2\sqrt{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,−1)(2,-1) and B(−3,4)(-3,4).

    AB=(−3−2)2+(4−(−1))2AB = \sqrt{(-3 - 2)^2 + (4 - (-1))^2} AB=(−5)2+(4+1)2AB = \sqrt{(-5)^2 + (4 + 1)^2} AB=25+(5)2AB = \sqrt{25 + (5)^2} AB=25+25AB = \sqrt{25 + 25} AB=50AB = \sqrt{50} AB=25×2AB = \sqrt{25 \times 2} AB=52AB = 5\sqrt{2}

  2. Calculate the length of side BC.

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

    BC=(4−(−3))2+(5−4)2BC = \sqrt{(4 - (-3))^2 + (5 - 4)^2} BC=(4+3)2+(1)2BC = \sqrt{(4 + 3)^2 + (1)^2} BC=(7)2+1BC = \sqrt{(7)^2 + 1} BC=49+1BC = \sqrt{49 + 1} BC=50BC = \sqrt{50} BC=25×2BC = \sqrt{25 \times 2} BC=52BC = 5\sqrt{2}

  3. Calculate the length of side AC.

    Given A(2,−1)(2,-1) and C(4,5)(4,5).

    AC=(4−2)2+(5−(−1))2AC = \sqrt{(4 - 2)^2 + (5 - (-1))^2} AC=(2)2+(5+1)2AC = \sqrt{(2)^2 + (5 + 1)^2} AC=4+(6)2AC = \sqrt{4 + (6)^2} AC=4+36AC = \sqrt{4 + 36} AC=40AC = \sqrt{40} AC=4×10AC = \sqrt{4 \times 10} AC=210AC = 2\sqrt{10}

    The lengths of the sides are:

    AB=52,BC=52,AC=210\boxed{AB = 5\sqrt{2}, BC = 5\sqrt{2}, AC = 2\sqrt{10}}

Part iii

P=102+210≈20.464P = 10\sqrt{2} + 2\sqrt{10} \approx 20.464

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=52+52+210P = 5\sqrt{2} + 5\sqrt{2} + 2\sqrt{10} P=102+210P = 10\sqrt{2} + 2\sqrt{10}

To get a numerical value, we approximate 2≈1.414\sqrt{2} \approx 1.414 and 10≈3.162\sqrt{10} \approx 3.162.

P≈10(1.414)+2(3.162)P \approx 10(1.414) + 2(3.162) P≈14.14+6.324P \approx 14.14 + 6.324 P≈20.464P \approx 20.464

The perimeter of the triangle is:

P=102+210≈20.464\boxed{P = 10\sqrt{2} + 2\sqrt{10} \approx 20.464}

Part iv

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

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=52AB = 5\sqrt{2}
BC=52BC = 5\sqrt{2}
AC=210AC = 2\sqrt{10}

Since AB=BCAB = BC, two sides of the triangle are equal.
Therefore, the triangle is an isosceles triangle.

The triangle is isosceles.\boxed{\text{The triangle is isosceles.}}
Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Mathematics Questions

QGiven the vertices of a triangle ABC as A(−2,−3), B(−4,1), and C(4,5), sketch the triangle…QGiven triangle ABC with vertices A(1, 4), B(2, 1), and C(5, 2).QGiven vertices A(1, 1), B(−3, 1), and C(−3, 3) of a rectangle ABCD, find the coordinates of vertex D…QShow that the points A(5,−6), B(−3,0) and C(−1,2) form an isosceles triangle. 2) Show that the…QPlot the points A(1,2), B(1,6), and C(5,5) on a Cartesian plane and draw triangle ABC.
Browse by topic:Geometry
See all Mathematics questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.
Handwritten step-by-step solution preview

Handwritten Step-by-Step Solution

Get this as a handwritten image on WhatsApp — like a tutor wrote it for you.

Try on WhatsApp