Triangle ABC is an isosceles triangle in which ∠C = 90°. If AC = 8 cm, find AB.
8√2 cm
Understanding the Isosceles Right Triangle Problem
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.
- Given that triangle ABC is isosceles.
- Given that ∠C = 90°.
- Given that AC = 8 cm.
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.
Applying the Pythagorean Theorem
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:
\(AB = \sqrt{128 \text{ cm}^2}\)
\(AB = \sqrt{128} \text{ cm}\)
Simplifying the Square Root of 128
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.
Summary of Calculation Steps
| 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.
Revision Table: Isosceles Right Triangle Key Properties
| 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 \(x^2 + x^2 = (\text{hypotenuse})^2 \implies 2x^2 = (\text{hypotenuse})^2 \implies \text{hypotenuse} = \sqrt{2x^2} = x\sqrt{2}\). |
Additional Information: The Pythagorean Theorem and Triangle Types
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:
- By Angles: Acute (all angles < 90°), Right (one angle = 90°), Obtuse (one angle > 90°).
- By Sides: Scalene (all sides different), Isosceles (two sides equal), Equilateral (all three sides equal).
An isosceles right triangle, as in this problem, combines characteristics of both classifications. It has a right angle and two equal sides.
Important Questions from Plane Figures
If length of a rectangle is increased to its three times and breadth is decreased to its half, then the ratio of the area of given rectangle to the area of new rectangle is:
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The perimeter of a circular lawn is 1232 m. There is 7 m wide path around the lawn. The area (in m 2) of the path is:
Take \(\left(\pi=\frac{22}{7}\right)\)
