Triangle ABC is an isosceles triangle in which ∠C = 90°. If AC = 8 cm, find AB.
8√2 cm
The question describes a triangle ABC which is an isosceles triangle and also a right-angled triangle, with the right angle at vertex C (∠C = 90°).
In an isosceles triangle, two sides are equal in length. Since the angle at C is 90°, C is the vertex where the two legs meet. The hypotenuse (the side opposite the right angle, which is AB) is always the longest side in a right-angled triangle. Therefore, the two equal sides in this isosceles right triangle must be the sides that form the right angle at C, which are AC and BC.
Since AC and BC are the legs forming the right angle and the triangle is isosceles, we know that the lengths of AC and BC must be equal.
Therefore, BC = AC = 8 cm.
In any right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides (legs). This is known as the Pythagorean theorem.
For triangle ABC, where ∠C = 90°, the Pythagorean theorem states:
\(AB^2 = AC^2 + BC^2\)
We know AC = 8 cm and BC = 8 cm. We can substitute these values into the equation:
\(AB^2 = (8 \text{ cm})^2 + (8 \text{ cm})^2\)
\(AB^2 = 8^2 \text{ cm}^2 + 8^2 \text{ cm}^2\)
\(AB^2 = 64 \text{ cm}^2 + 64 \text{ cm}^2\)
\(AB^2 = 128 \text{ cm}^2\)
To find the length of AB, we need to take the square root of 128:
\(AB = \sqrt{128 \text{ cm}^2}\)
\(AB = \sqrt{128} \text{ cm}\)
To simplify \(\sqrt{128}\), we look for the largest perfect square factor of 128. We know that \(64 \times 2 = 128\), and 64 is a perfect square (\(8^2 = 64\)).
\(\sqrt{128} = \sqrt{64 \times 2}\)
Using the property \(\sqrt{a \times b} = \sqrt{a} \times \sqrt{b}\), we get:
\(\sqrt{128} = \sqrt{64} \times \sqrt{2}\)
Since \(\sqrt{64} = 8\):
\(\sqrt{128} = 8 \times \sqrt{2}\)
\(\sqrt{128} = 8\sqrt{2}\)
Therefore, the length of AB is \(8\sqrt{2}\) cm.
| Step | Description | Calculation |
|---|---|---|
| 1 | Identify equal sides in isosceles right triangle | AC = BC = 8 cm |
| 2 | Apply Pythagorean theorem | \(AB^2 = AC^2 + BC^2\) |
| 3 | Substitute values | \(AB^2 = 8^2 + 8^2\) |
| 4 | Calculate squares and sum | \(AB^2 = 64 + 64 = 128\) |
| 5 | Find AB by taking square root | \(AB = \sqrt{128}\) |
| 6 | Simplify the square root | \(AB = \sqrt{64 \times 2} = 8\sqrt{2}\) |
The length of AB is \(8\sqrt{2}\) cm.
| Property | Description |
|---|---|
| Angles | One angle is 90°, the other two angles are equal (\(45°\) each). |
| Sides | Two sides (legs) are equal in length. The hypotenuse is \(\sqrt{2}\) times the length of a leg. |
| Pythagorean Theorem | \((\text{leg})^2 + (\text{leg})^2 = (\text{hypotenuse})^2\). If leg length is \(x\), then \(x^2 + x^2 = (\text{hypotenuse})^2 \implies 2x^2 = (\text{hypotenuse})^2 \implies \text{hypotenuse} = \sqrt{2x^2} = x\sqrt{2}\). |
The Pythagorean theorem is a fundamental concept in geometry that applies specifically to right-angled triangles. It establishes a relationship between the lengths of the sides of a right triangle. Let the lengths of the two legs be \(a\) and \(b\), and the length of the hypotenuse be \(c\). The theorem is expressed as:
\(a^2 + b^2 = c^2\)
This theorem is crucial for solving problems involving right triangles, such as finding the length of an unknown side when the other two sides are known.
Triangles can be classified by their angles and sides:
An isosceles right triangle, as in this problem, combines characteristics of both classifications. It has a right angle and two equal sides.
If the area of a square is 625 cm 2, then what is the perimeter of the square?
The area and the perimeter of a sheet of paper are 240 cm 2and 68 cm, respectively. What would be its length and breadth?
One side of rectangular field is 15 meters and one of its diagonals is 17 meters. Then find the area of the field.
The bisector of ∠B in ΔABC meets AC at D. If AB = 12 cm, BC = 18 cm and AC = 15 cm, then the length of AD (in cm) is:
The perimeter and the length of one of the diagonals of a rhombus is 26 cm and 5 cm respectively. Find the length of its other diagonal (in cm).