Example Question - leg lengths
Here are examples of questions we've helped users solve.
Finding the Hypotenuse of a Right Triangle with Given Leg Lengths
<p>The question asks to find the hypotenuse \( c \) of a right triangle with legs \( a = 3 \) and \( b = 4 \).</p> <p>We use the Pythagorean theorem: \( a^2 + b^2 = c^2 \).</p> <p>Substitute the given values: \( 3^2 + 4^2 = c^2 \).</p> <p>Calculate the squares: \( 9 + 16 = c^2 \).</p> <p>Add the results: \( 25 = c^2 \).</p> <p>Take the square root of both sides: \( \sqrt{25} = \sqrt{c^2} \).</p> <p>Thus, \( c = 5 \).</p>
Solving for the Lengths of Legs in an Isosceles Right-Angled Triangle
This image shows a right-angled triangle with one of the angles being \(45^\circ\), which makes it an isosceles right-angled triangle (since the other non-right angle must also be \(45^\circ\)). The hypotenuse of this triangle is given as 5 units. In an isosceles right-angled triangle, the lengths of the two legs (sides opposite the \(45^\circ\) angles) are equal. If we let one of the legs be \(x\), we can use the Pythagorean theorem to solve for \(x\): \[x^2 + x^2 = 5^2\] \[2x^2 = 25\] \[x^2 = \frac{25}{2}\] \[x = \sqrt{\frac{25}{2}}\] \[x = \frac{5}{\sqrt{2}}\] Multiplying the numerator and denominator by \(\sqrt{2}\) to rationalize the denominator, we get: \[x = \frac{5\sqrt{2}}{2}\] So the lengths of the two legs are each \(\frac{5\sqrt{2}}{2}\) units.
Solving for the Length of Legs in a 45-45-90 Triangle
The image shows a right-angle triangle with a 45-degree angle. Since the angles in a triangle add up to 180 degrees and we already have a right angle (90 degrees) and one 45-degree angle, the other angle must also be 45 degrees. This makes the triangle a 45-45-90 triangle, which is a form of an isosceles right triangle. In a 45-45-90 triangle, the lengths of the legs (the two shorter sides) are equal, and the length of the hypotenuse (the longest side, opposite the right angle) is √2 times the length of a leg. The figure gives the hypotenuse as length 3. Therefore, to find the length of each leg (let's call it 'L'), we can use the proportion that L is to 3 (the hypotenuse) as 1 is to √2: L / 3 = 1 / √2 Multiplying both sides by 3 to solve for L: L = 3 / √2 To rationalize the denominator: L = (3 / √2) * (√2 / √2) L = (3√2) / 2 So, the length of each leg of the triangle is (3√2) / 2 units.
Solving for Leg Lengths in a 45-45-90 Triangle
The image shows a right triangle with one angle of 45 degrees and the hypotenuse having a length of 3 units. Given that the triangle is a right triangle with a 45-degree angle, we can infer that this is a 45-45-90 triangle, which is an isosceles right triangle. In such triangles, the lengths of the legs are equal, and the length of the hypotenuse is √2 times the length of each leg. Let's use the properties of a 45-45-90 triangle to find the lengths of the two legs. If \( a \) is the length of one leg, we have: \( a \cdot \sqrt{2} = 3 \) Now we solve for \( a \): \( a = \frac{3}{\sqrt{2}} \) To rationalize the denominator, we multiply the numerator and the denominator by \( \sqrt{2} \): \( a = \frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} \) \( a = \frac{3 \cdot \sqrt{2}}{2} \) So both legs of the triangle are \( \frac{3 \cdot \sqrt{2}}{2} \) units long.
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