Question

find d.
right triangle with angles 45°, 45°, 90°, hypotenuse √6, side d opposite one 45° angle
write your answer in simplest radical form.
blank units

Explanation:

Step1: Identify triangle type

The triangle is a 45-45-90 right triangle, so legs are equal, hypotenuse \( h = leg \times \sqrt{2} \). Here, hypotenuse is \( \sqrt{6} \)? Wait, no—wait, in 45-45-90, legs are equal, hypotenuse is leg\( \times \sqrt{2} \). Wait, the triangle has two 45° angles, so it's isosceles right triangle. So the two legs are equal, and hypotenuse is leg\( \times \sqrt{2} \). Wait, but here, the side labeled \( \sqrt{6} \) is the hypotenuse? Wait, no, the right angle is between the two legs, and the other two angles are 45°, so the legs are the two sides with 45° angles adjacent, and hypotenuse is opposite right angle. Wait, no—wait, the triangle: right angle, two 45° angles, so legs (the two sides forming right angle) are equal, hypotenuse is the side opposite right angle. Wait, in the diagram, the side labeled \( d \) and the vertical leg (with 45°) are legs, and the side \( \sqrt{6} \) is hypotenuse? Wait, no, wait: in a 45-45-90 triangle, legs are equal, hypotenuse \( h = l\sqrt{2} \), where \( l \) is leg length. Wait, maybe I got it reversed. Let's check: if the two legs are equal (since two angles are 45°), then hypotenuse is leg\( \times \sqrt{2} \). So if hypotenuse is \( \sqrt{6} \), then leg \( d = \frac{\text{hypotenuse}}{\sqrt{2}} \). Wait, no: hypotenuse \( h = l\sqrt{2} \), so \( l = \frac{h}{\sqrt{2}} \). So here, hypotenuse is \( \sqrt{6} \), so leg \( d = \frac{\sqrt{6}}{\sqrt{2}} \). Simplify \( \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{\frac{6}{2}} = \sqrt{3} \)? Wait, no, wait: \( \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{6/2} = \sqrt{3} \)? Wait, no, wait, maybe the side \( \sqrt{6} \) is a leg? Wait, no, the right angle is between the two legs, so the two legs are \( d \) and the vertical leg (with 45°), and the hypotenuse is \( \sqrt{6} \). Wait, no, that can't be, because in 45-45-90, legs are equal, so \( d \) and the vertical leg are equal, and hypotenuse is \( d\sqrt{2} \). So if hypotenuse is \( \sqrt{6} \), then \( d\sqrt{2} = \sqrt{6} \), so \( d = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \)? Wait, no, wait: \( \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{6/2} = \sqrt{3} \). Wait, but maybe I mixed up leg and hypotenuse. Wait, let's re-express: in a 45-45-90 triangle, the legs are the two sides with 45° angles adjacent, and hypotenuse is opposite right angle. So if the triangle has right angle, and two 45° angles, then the two legs (the sides forming the right angle) are equal, and hypotenuse is leg\( \times \sqrt{2} \). So if one leg is \( d \), the other leg is also \( d \), and hypotenuse is \( d\sqrt{2} \). But in the diagram, the side labeled \( \sqrt{6} \) is the hypotenuse? Wait, no, looking at the diagram: the right angle is at the top left, one leg is \( d \) (horizontal), one leg is vertical (with 45° at bottom), and the hypotenuse is the side with \( \sqrt{6} \) (connecting the two 45° angles). So hypotenuse is \( \sqrt{6} \), so \( d\sqrt{2} = \sqrt{6} \), so \( d = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \)? Wait, no, wait: \( \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{6/2} = \sqrt{3} \). Wait, but that seems small. Wait, maybe the side \( \sqrt{6} \) is a leg? Wait, if \( \sqrt{6} \) is a leg, then hypotenuse would be \( \sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3} \), but that's not the case. Wait, no, let's check again. Wait, the triangle: right angle, two 45° angles, so it's isosceles right triangle. So legs are equal, hypotenuse is leg\( \times \sqrt{2} \). So if the hypotenuse is \( \sqrt{6} \), then leg \( d = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \). Wait, but maybe I had it backwards. Let's do the calculation: \( \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{\frac{6}{2}} = \sqrt{3} \). So \( d = \sqrt{3} \)? Wait, no, wait: wait, maybe the side \( \sqrt{6} \) is a leg, and \( d \) is a leg, so they should be equal? But that can't be, because \( \sqrt{6} \) is longer than \( \sqrt{3} \). Wait, no, I think I messed up the hypotenuse. Wait, in the diagram, the right angle is at the top left, so the two legs are the top horizontal (d) and the left vertical (with 45°), and the hypotenuse is the right side (labeled \( \sqrt{6} \)). So hypotenuse is \( \sqrt{6} \), so legs are equal, so \( d = \frac{\text{hypotenuse}}{\sqrt{2}} = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \). Wait, but let's rationalize: \( \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{6} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \). Yes, that's correct. So \( d = \sqrt{3} \)? Wait, no, wait, maybe I made a mistake. Wait, let's check: in a 45-45-90 triangle, legs are equal, hypotenuse is leg\( \times \sqrt{2} \). So if leg is \( d \), hypotenuse is \( d\sqrt{2} \). So if hypotenuse is \( \sqrt{6} \), then \( d\sqrt{2} = \sqrt{6} \implies d = \frac{\sqrt{6}}{\sqrt{2}} = \sqrt{3} \). Yes, that's correct. So the length of \( d \) is \( \sqrt{3} \)? Wait, no, wait, maybe the side \( \sqrt{6} \) is a leg, and \( d \) is the other leg, so they should be equal? But that would mean \( d = \sqrt{6} \), but then hypotenuse would be \( \sqrt{6} \times \sqrt{2} = \sqrt{12} = 2\sqrt{3} \), which is not labeled. So the diagram must have the hypotenuse as \( \sqrt{6} \), so legs are \( \sqrt{3} \). So \( d = \sqrt{3} \). Wait, but let's confirm with trigonometry. In a right triangle, \( \sin(45^\circ) = \frac{\text{opposite}}{\text{hypotenuse}} \). So \( \sin(45^\circ) = \frac{d}{\sqrt{6}} \). Since \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), then \( \frac{\sqrt{2}}{2} = \frac{d}{\sqrt{6}} \implies d = \sqrt{6} \times \frac{\sqrt{2}}{2} = \frac{\sqrt{12}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3} \). Yes, that's correct. So \( d = \sqrt{3} \).

Wait, no, wait, maybe I had the angle wrong. Wait, the 45° angle is at the top, so the side \( d \) is adjacent to the 45° angle, and the leg \( d \) is adjacent, and the other leg (vertical) is also adjacent? No, in a right triangle, for an acute angle, adjacent and opposite sides. So for the 45° angle at the top, the adjacent side is \( d \), the opposite side is the vertical leg (let's call it \( l \)), and hypotenuse is \( \sqrt{6} \). Since it's 45°, \( \tan(45^\circ) = \frac{l}{d} = 1 \), so \( l = d \). Then by Pythagoras, \( d^2 + l^2 = (\sqrt{6})^2 \implies d^2 + d^2 = 6 \implies 2d^2 = 6 \implies d^2 = 3 \implies d = \sqrt{3} \). Ah, there we go. So using Pythagoras: two legs \( d \) and \( d \), hypotenuse \( \sqrt{6} \). So \( d^2 + d^2 = (\sqrt{6})^2 \implies 2d^2 = 6 \implies d^2 = 3 \implies d = \sqrt{3} \). Yes, that's correct. So the length of \( d \) is \( \sqrt{3} \).

Step1: Recognize triangle type

It's a 45-45-90 right triangle (isosceles right triangle), so legs are equal, hypotenuse \( h = l\sqrt{2} \) ( \( l \) = leg length).

Step2: Apply Pythagorean theorem

Let legs be \( d \) (both legs are equal, since two angles are 45°). Hypotenuse \( = \sqrt{6} \).
By Pythagoras: \( d^2 + d^2 = (\sqrt{6})^2 \)
Simplify: \( 2d^2 = 6 \)

Step3: Solve for \( d \)

Divide both sides by 2: \( d^2 = 3 \)
Take square root: \( d = \sqrt{3} \) (since length is positive)

Answer:

\( \sqrt{3} \)