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An isosceles right triangle has legs that are the same length and acute angles each measuring \(45^{\circ} .\) (GRAPH NOT COPY) a. Write an expression in terms of \(a\) that represents the length of the hypotenuse. b. Use your result from part (a) to write a sentence that describes the length of the hypotenuse of an isosceles right triangle in terms of the length of a leg.

Short Answer

Expert verified
The length of the hypotenuse of an isosceles right triangle with legs of length \(a\) is \(a\sqrt{2}\). In other words, the hypotenuse is \(\sqrt{2}\) times the length of a leg.

Step by step solution

01

Apply Pythagorean Theorem

Since the triangle is a right triangle, the Pythagorean theorem applies. The theorem is: \(a^{2} + b^{2} = c^{2}\) where \(a\) and \(b\) are the lengths of the two legs and \(c\) is the length of the hypotenuse. Since this is an isosceles right triangle, we know that \(a = b\). This simplifies the theorem to: \(a^{2} + a^{2} = c^{2}\). Combine like terms to get \(2a^{2} = c^{2}\).
02

Solve for c

Now we need to express hypotenuse length (\(c\)) in terms of \(a\). For this, we need to solve the equation from step 1 for \(c\), which gives us: \(c = \sqrt{2a^{2}}\). This is the length of the hypotenuse in terms of \(a\).
03

Simplify the expression

We can further simplify this expression by factoring out \(a^{2}\) under the square root to get: \(c = a\sqrt{2}\). So the length of the hypotenuse is the length of one leg times \(\sqrt{2}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pythagorean Theorem
The Pythagorean Theorem is a cornerstone of geometry, especially useful when dealing with right triangles. It helps us find the length of the sides in right-angled triangles. The basic formula is: \(a^{2} + b^{2} = c^{2}\), where \(a\) and \(b\) are the two perpendicular sides, and \(c\) is the hypotenuse.

In this exercise featuring an isosceles right triangle, the theorem's beauty shines. Since both legs of an isosceles right triangle are equal, imagine both being \(a\). By substituting \(a = b\) into the theorem, it becomes \(a^{2} + a^{2} = c^{2}\). This reduces to \(2a^{2} = c^{2}\), a simplified scenario that only happens due to the symmetrical nature of the triangle.
  • The Pythagorean Theorem is powerful for right triangles.
  • In isosceles variants, it simplifies calculations.
  • It reveals relationships between triangle sides.
Hypotenuse
In any right triangle, the hypotenuse is the longest side. It is the side opposite the right angle. This hypotenuse holds a special place, especially in the Pythagorean Theorem.

For our isosceles right triangle, knowing the length of the hypotenuse involves realizing it is the sum square roots of the other two sides. Once the equal legs are known as \(a\), applying the theorem gives us \(c = \sqrt{2a^{2}}\). After simplifying, we find that the hypotenuse length is actually \(c = a\sqrt{2}\).

This relationship is significant because it shows a consistent pattern in isosceles right triangles: the hypotenuse length is always the leg length times \(\sqrt{2}\). Consider these points about the hypotenuse:
  • It completes the Pythagorean theorem's relation.
  • In isosceles right triangles, it's predictable and consistent.
  • Unique for measuring isosceles right triangles.
Square Root Simplification
Simplifying square roots is an essential skill in mathematics, especially in geometry when dealing with lengths and measures.

In our exercise, we started with the expression \(c = \sqrt{2a^{2}}\) for the hypotenuse. This can be seen a bit complex at first. Breaking it down, you decompose the square root by recognizing \(\sqrt{2a^{2}} = \sqrt{2} \times \sqrt{a^{2}}\).

Since \(\sqrt{a^{2}} = a\), the expression simplifies nicely to \(c = a\sqrt{2}\). This process of breaking down and simplifying is crucial because it turns an intricate-looking expression into something more accessible and understandable. Here's a summary of square root simplification in this context:
  • It turns complex terms into simpler forms.
  • Helps extract straightforward relationships in geometry.
  • Makes the solution process smoother and more intuitive.

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