Understanding the Triangle Problem

We are given a triangle ABC where AB = AC, meaning it is an isosceles triangle. We know the perimeter of the triangle and a relationship between the length of BC and the length of AB. Our goal is to find the area of triangle ABC.

Let's break down the given information:

• Triangle ABC is isosceles with AB = AC.
• Perimeter of ΔABC is \(8(2 + \sqrt{2})\) cm.
• The length of BC is \(\sqrt{2}\) times the length of AB.

Setting up the Side Lengths

Let's denote the length of AB as \(x\) cm.

Since AB = AC, the length of AC is also \(x\) cm.

The problem states that the length of BC is \(\sqrt{2}\) times the length of AB. So, BC = \(\sqrt{2} \times AB = \sqrt{2}x\) cm.

Now, the perimeter of ΔABC is the sum of the lengths of its sides:

Perimeter = AB + AC + BC

Perimeter = \(x + x + \sqrt{2}x\)

Perimeter = \(2x + \sqrt{2}x\)

Perimeter = \(x(2 + \sqrt{2})\) cm

Using the Given Perimeter to Find Side Lengths

We are given that the perimeter of ΔABC is \(8(2 + \sqrt{2})\) cm.

We can set our expression for the perimeter equal to the given value:

\(x(2 + \sqrt{2}) = 8(2 + \sqrt{2})\)

To find the value of \(x\), we can divide both sides of the equation by \((2 + \sqrt{2})\):

\(x = \frac{8(2 + \sqrt{2})}{2 + \sqrt{2}}\)

\(x = 8\)

So, the length of AB is 8 cm. Since AB = AC, the length of AC is also 8 cm.

The length of BC is \(\sqrt{2}\) times the length of AB:

BC = \(\sqrt{2} \times 8 = 8\sqrt{2}\) cm.

The side lengths of ΔABC are AB = 8 cm, AC = 8 cm, and BC = \(8\sqrt{2}\) cm.

Calculating the Area of the Isosceles Triangle

To find the area of a triangle, we can use the formula: Area = \(\frac{1}{2} \times \text{base} \times \text{height}\). In an isosceles triangle, the altitude from the vertex angle (the angle between the equal sides) to the base bisects the base and is perpendicular to the base. Let's draw the altitude from A to BC and call the point where it meets BC, D.

AD is the height of the triangle with respect to the base BC. D is the midpoint of BC.

Length of BD = Length of DC = \(\frac{BC}{2} = \frac{8\sqrt{2}}{2} = 4\sqrt{2}\) cm.

Now consider the right-angled triangle ADB. We know the hypotenuse AB = 8 cm and the base BD = \(4\sqrt{2}\) cm. We can find the height AD using the Pythagorean theorem:

\(AB^2 = AD^2 + BD^2\)

\(8^2 = AD^2 + (4\sqrt{2})^2\)

\(64 = AD^2 + (16 \times 2)\)

\(64 = AD^2 + 32\)

Now, solve for \(AD^2\):

\(AD^2 = 64 - 32\)

\(AD^2 = 32\)

Taking the square root of both sides to find AD:

\(AD = \sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4\sqrt{2}\) cm.

The height of the triangle AD is \(4\sqrt{2}\) cm.

Now we can calculate the area of ΔABC using the base BC and height AD:

Area = \(\frac{1}{2} \times BC \times AD\)

Area = \(\frac{1}{2} \times 8\sqrt{2} \times 4\sqrt{2}\)

Area = \(\frac{1}{2} \times (8 \times 4) \times (\sqrt{2} \times \sqrt{2})\)

Area = \(\frac{1}{2} \times 32 \times 2\)

Area = \(\frac{1}{2} \times 64\)

Area = 32 cm\(^2\).

Final Area Calculation

The calculated area of triangle ABC is 32 cm\(^2\).

Revision Table: Key Concepts and Formulas

Additional Information on Isosceles Triangles and Area

An isosceles triangle has special properties that can simplify calculations. Since two sides are equal (AB = AC), the angles opposite these sides (\(\angle ABC\) and \(\angle ACB\)) are also equal. The altitude from the vertex angle A to the base BC acts as a median and an angle bisector as well.

In our case, since AB = 8, AC = 8, and BC = \(8\sqrt{2}\), we might also notice a special relationship between the sides. Consider the angles. If we drop the altitude AD, triangle ADB is a right triangle with BD = \(4\sqrt{2}\) and AB = 8. We found AD = \(4\sqrt{2}\). Since BD = AD = \(4\sqrt{2}\), triangle ADB is a right-angled isosceles triangle. This means \(\angle BAD = \angle ABD = 45^\circ\). Similarly, \(\angle CAD = \angle ACD = 45^\circ\). Therefore, \(\angle BAC = \angle BAD + \angle CAD = 45^\circ + 45^\circ = 90^\circ\). This triangle is actually a right-angled isosceles triangle!

If we recognised this earlier, we could calculate the area simply using the two perpendicular sides (AB and AC) as base and height:

Area = \(\frac{1}{2} \times \text{product of perpendicular sides}\)

Area = \(\frac{1}{2} \times AB \times AC\)

Area = \(\frac{1}{2} \times 8 \times 8\)

Area = \(\frac{1}{2} \times 64\)

Area = 32 cm\(^2\).

Both methods lead to the same result, confirming our calculation.