Question

if the altitude of an isosceles right triangle has a length of x units, what is the length of one leg of the large right triangle in terms of x?

x units

x√2 units

x√3 units

2x units

Explanation:

Step1: Recall properties of isosceles right triangle

In an isosceles right triangle, the altitude to the hypotenuse (or in this case, the altitude within the triangle structure) creates two smaller isosceles right triangles. Let the leg of the large isosceles right triangle be \( l \). The altitude to the hypotenuse (or the relevant altitude) in an isosceles right triangle relates to the leg. For an isosceles right triangle, if we consider the altitude \( x \), the relationship between the altitude and the leg: In an isosceles right triangle, the altitude to the hypotenuse is equal to half the hypotenuse, and also, using the properties of 45 - 45 - 90 triangles (isosceles right triangles), the legs are equal, and the hypotenuse is \( l\sqrt{2} \). But when we draw an altitude from the right angle to the hypotenuse (wait, in the given diagram, maybe the altitude is from the right angle? Wait, no, the diagram shows a triangle with a right angle, and an altitude inside, creating two smaller triangles. Wait, actually, in an isosceles right triangle, if we have a leg of length \( l \), and we draw an altitude (maybe from the right angle to the hypotenuse, but no, in the diagram, the triangle has two equal sides (marked with ticks), so it's isosceles right, so the two legs are equal, and the right angle is between them. Wait, maybe the altitude here is the leg of the smaller triangle? Wait, no, let's think again.

Wait, in an isosceles right triangle, the altitude from the right angle to the hypotenuse: Let the legs be \( l \), hypotenuse \( l\sqrt{2} \). The altitude \( h \) to the hypotenuse is given by \( h=\frac{l\sqrt{2}}{2}=\frac{l}{\sqrt{2}} \). But in the problem, the altitude is \( x \), so \( x = \frac{l}{\sqrt{2}} \), then \( l = x\sqrt{2} \)? Wait, no, maybe I got it reversed. Wait, no, let's consider the smaller triangle. Wait, the diagram shows a triangle with a right angle, and an altitude (marked \( x \)) creating two smaller triangles. Since the large triangle is isosceles right, the two smaller triangles are also isosceles right. So the altitude \( x \) is a leg of the smaller isosceles right triangle, and the leg of the large triangle is the hypotenuse of the smaller isosceles right triangle. In a 45 - 45 - 90 triangle, the hypotenuse is leg \( \times\sqrt{2} \). So if the leg of the smaller triangle is \( x \), then the hypotenuse (which is the leg of the large triangle) is \( x\sqrt{2} \)? Wait, no, wait. Wait, maybe the altitude is the leg of the large triangle? No, the problem says "the altitude of an isosceles right triangle has a length of \( x \) units, what is the length of one leg of the large right triangle". Wait, maybe the altitude is from the right angle to the hypotenuse, but in an isosceles right triangle, the altitude from the right angle to the hypotenuse is equal to half the hypotenuse, and also, the legs are equal. Let me denote the leg of the large triangle as \( l \). Then hypotenuse is \( l\sqrt{2} \). The altitude to the hypotenuse is \( \frac{l\sqrt{2}}{2}=\frac{l}{\sqrt{2}} \). But the problem states that the altitude is \( x \), so \( x=\frac{l}{\sqrt{2}} \), so solving for \( l \), we get \( l = x\sqrt{2} \). Wait, but that would mean the leg is \( x\sqrt{2} \). But wait, another approach: In an isosceles right triangle, the two legs are equal, and the angles are 45 - 45 - 90. If we have a smaller isosceles right triangle with leg \( x \), then the hypotenuse of the smaller triangle (which is the leg of the larger triangle) would be \( x\sqrt{2} \). Wait, no, maybe the altitude is the leg of the larger triangle? No, the problem says "the altitude of an isosceles right triangle has a length of \( x \) units", so the altitude is \( x \), and we need to find the leg \( l \). Wait, maybe I made a mistake. Wait, let's consider the 45 - 45 - 90 triangle ratios. In a 45 - 45 - 90 triangle, the sides are in the ratio \( 1:1:\sqrt{2} \) (leg:leg:hypotenuse). Now, if the altitude (which is a leg of the smaller triangle) is \( x \), then the hypotenuse of the smaller triangle (which is the leg of the larger triangle) would be \( x\sqrt{2} \)? Wait, no, maybe the altitude is the leg of the larger triangle? No, the problem says "the altitude of an isosceles right triangle has a length of \( x \) units", so altitude is \( x \), leg is \( l \). Wait, maybe the altitude is equal to the leg? No, that can't be. Wait, let's look at the options. The options are \( x \), \( x\sqrt{2} \), \( x\sqrt{3} \), \( 2x \). Let's think of the altitude in an isosceles right triangle. If the triangle is isosceles right, with legs \( l \), then the altitude from the right angle to the hypotenuse is \( \frac{l}{\sqrt{2}} \). But if the altitude is \( x \), then \( x = \frac{l}{\sqrt{2}} \), so \( l = x\sqrt{2} \). Wait, but that would mean the leg is \( x\sqrt{2} \). But wait, maybe the altitude is one of the legs? No, because the altitude from the right angle to the hypotenuse is shorter than the leg. Wait, maybe the diagram is different. The diagram shows a triangle with two equal sides (ticks) and a right angle, and an altitude (marked \( x \)) inside, creating two smaller triangles. So the large triangle is isosceles right, so legs are equal, right angle between them. The altitude \( x \) is the leg of the smaller isosceles right triangle, and the leg of the large triangle is the hypotenuse of the smaller triangle. So in the smaller triangle, legs are \( x \), so hypotenuse (leg of large triangle) is \( x\sqrt{2} \). Wait, no, that would be if the smaller triangle is isosceles right. Wait, yes, because the large triangle is isosceles right, so the two smaller triangles formed by the altitude are also isosceles right. So if the altitude (leg of smaller triangle) is \( x \), then the hypotenuse of the smaller triangle (which is the leg of the large triangle) is \( x\sqrt{2} \). Wait, but let's check with an example. Let \( x = 1 \), then the leg of the large triangle would be \( \sqrt{2} \), which is \( 1\times\sqrt{2} \), so that matches. Alternatively, if the leg of the large triangle is \( 2x \), then \( x \) would be half, but that doesn't fit. Wait, maybe I got the altitude wrong. Wait, another approach: In an isosceles right triangle, the legs are equal, let's call them \( l \). The area is \( \frac{1}{2}l^2 \). Also, the area can be expressed as \( \frac{1}{2}\times\text{hypotenuse}\times\text{altitude} \). The hypotenuse is \( l\sqrt{2} \), so area is \( \frac{1}{2}\times l\sqrt{2}\times x \). Setting equal to \( \frac{1}{2}l^2 \), we get \( \frac{1}{2}l\sqrt{2}x=\frac{1}{2}l^2 \), so \( l\sqrt{2}x = l^2 \), so \( l = x\sqrt{2} \). Yes, that makes sense. So the leg length \( l = x\sqrt{2} \). Wait, but the options include \( x\sqrt{2} \) as an option. Wait, but let's check again. Wait, maybe the altitude is the leg, not the altitude to the hypotenuse. If the altitude is a leg, then the leg would be \( x \), but that's not correct because the altitude to the hypotenuse is shorter. Wait, the problem says "the altitude of an isosceles right triangle has a length of \( x \) units". Maybe the altitude is one of the legs? But in an isosceles right triangle, the legs are altitudes to each other. So if the altitude (from one leg to the other, but no, the altitude from a vertex to the opposite side. So the altitude from the right angle to the hypotenuse is different. Wait, I think I confused the altitude. Let's clarify: In a right triangle, the altitude from the right angle to the hypotenuse is equal to the product of the legs divided by the hypotenuse. In an isosceles right triangle, legs are \( l \), hypotenuse \( l\sqrt{2} \), so altitude \( h=\frac{l\times l}{l\sqrt{2}}=\frac{l}{\sqrt{2}} \). So \( h = \frac{l}{\sqrt{2}} \), so \( l = h\sqrt{2} \). So if \( h = x \), then \( l = x\sqrt{2} \). So the leg length is \( x\sqrt{2} \) units. Wait, but let's check the options. The options are:

- \( x \) units

- \( x\sqrt{2} \) units

- \( x\sqrt{3} \) units

- \( 2x \) units

So according to this, the correct answer should be \( x\sqrt{2} \) units. Wait, but wait, maybe I made a mistake. Let's take \( l = 2 \), then hypotenuse is \( 2\sqrt{2} \), altitude to hypotenuse is \( \frac{2}{\sqrt{2}}=\sqrt{2} \), so if altitude \( x = \sqrt{2} \), then leg \( l = 2 = \sqrt{2}\times\sqrt{2} \), which is \( x\sqrt{2} \) (since \( x = \sqrt{2} \), \( x\sqrt{2} = 2 \)). Yes, that works. So the leg length is \( x\sqrt{2} \) units.

Step2: Verify with the 45-45-90 triangle ratio

In a 45-45-90 triangle, the sides are in the ratio \( 1:1:\sqrt{2} \) (leg:leg:hypotenuse). If the altitude (which is a leg of the smaller 45-45-90 triangle) is \( x \), then the hypotenuse of the smaller triangle (which is a leg of the larger triangle) follows the 45-45-90 ratio, so hypotenuse = leg \( \times\sqrt{2} \), so leg of larger triangle = \( x\sqrt{2} \).

Answer:

\( x\sqrt{2} \) units (the option corresponding to \( x\sqrt{2} \) units, e.g., if the options are labeled as:

• \( x \) units

• \( x\sqrt{2} \) units

• \( x\sqrt{3} \) units

• \( 2x \) units

Then the correct option is the one with \( x\sqrt{2} \) units, e.g., "B. \( x\sqrt{2} \) units" (assuming the second option is \( x\sqrt{2} \) units))