Question

find the value of x using the figure to the right.
x = \square (simplify your answer.)

Explanation:

Step1: Recall the geometric mean theorem (altitude-on-hypotenuse theorem)

In a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse segment adjacent to that leg and the length of the entire hypotenuse. For the given right triangle, if we consider the leg of length \( x \), the adjacent segment to \( x \) on the hypotenuse is \( x \) itself? Wait, no, let's re - examine. Wait, the two segments of the hypotenuse are \( x \) and \( 21 \), and the leg adjacent to \( x \) is \( 10 \)? Wait, no, the geometric mean theorem states that in a right triangle, when an altitude is drawn to the hypotenuse, each leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. Let's denote the hypotenuse as \( c=x + 21 \), one leg as \( a = 10 \), and the segments of the hypotenuse as \( m=x \) and \( n = 21 \). Then by the geometric mean theorem, \( a^{2}=m\times c\)? Wait, no, the correct formula is that if we have a right triangle with hypotenuse divided into segments of length \( m \) and \( n \) by the altitude, and the legs are \( l_1 \) and \( l_2 \), then \( l_1^{2}=m\times(m + n)\) and \( l_2^{2}=n\times(m + n) \), and also \( l_1^{2}=m\times(m + n)\), \( l_2^{2}=n\times(m + n) \), and the altitude \( h^{2}=m\times n \). Wait, in our case, the leg with length \( 10 \) is adjacent to the segment \( x \) of the hypotenuse? Wait, no, looking at the figure, the right triangle has a leg of length \( 10 \), and the hypotenuse is split into two parts: one part is \( x \) and the other is \( 21 \). The leg of length \( 10 \) is opposite to the segment \( 21 \)? No, the correct application is that for the leg of length \( x \), the adjacent segment on the hypotenuse is \( x \), and the other segment is \( 21 \), and the leg of length \( 10 \) is related to the segment \( x \) and \( 21 \). Wait, actually, the formula is that if we have a right triangle, and we draw an altitude to the hypotenuse, then \( x^{2}=10\times10\)? No, wait, no. Wait, the correct formula is that in a right triangle, the square of a leg is equal to the product of the hypotenuse and the adjacent segment. Wait, let's denote the hypotenuse as \( H=x + 21 \), the segment adjacent to \( x \) is \( x \), and the segment adjacent to \( 21 \) is \( 21 \). The leg of length \( 10 \) is adjacent to the segment \( x \)? No, I think I made a mistake. Let's start over.

Let the right triangle be \( \triangle ABC \) with right angle at \( C \), and altitude \( CD \) drawn to hypotenuse \( AB \), where \( D \) is on \( AB \). Let \( AD=x \), \( DB = 21 \), and \( CD = 10 \)? Wait, no, the leg is \( 10 \), not the altitude. Wait, the figure shows a right triangle with a leg of length \( 10 \), and the hypotenuse is split into two parts: \( x \) and \( 21 \), and there is an altitude drawn to the hypotenuse, forming two smaller right triangles. So, the leg of length \( 10 \) is one of the legs of the original right triangle, and the two segments of the hypotenuse are \( x \) and \( 21 \). Then, by the geometric mean theorem, \( 10^{2}=x\times21 \)? No, that's not right. Wait, no, the correct formula is that if the leg is \( l \), and the hypotenuse is divided into segments \( m \) and \( n \), then \( l^{2}=m\times(m + n) \) if \( l \) is adjacent to \( m \), or \( l^{2}=n\times(m + n) \) if adjacent to \( n \). Wait, in our case, the leg of length \( 10 \) is adjacent to the segment \( x \), so \( 10^{2}=x\times(x + 21) \)? No, that can't be. Wait, no, I think the other leg (the one with length \( x \)) is adjacent to the segment \( x \), and the leg of length \( 10 \) is adjacent to the segment \( 21 \). Wait, the correct formula is \( x^{2}=10\times10\)? No, I'm confused. Wait, the geometric mean theorem: In a right triangle, the square of a leg is equal to the product of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So, if we have a right triangle with hypotenuse \( c=x + 21 \), and one leg \( a \) adjacent to segment \( m=x \), then \( a^{2}=m\times c \), and the other leg \( b \) adjacent to segment \( n = 21 \), then \( b^{2}=n\times c \). But in our case, the leg with length \( 10 \) is adjacent to segment \( x \)? No, looking at the figure, the leg of length \( 10 \) is opposite to the segment \( 21 \). Wait, maybe the correct formula is that the leg of length \( x \) and the leg of length \( 10 \) are related to the segments of the hypotenuse. Wait, no, the correct application is that \( x^{2}=10\times10\)? No, that's not. Wait, let's use the formula for the geometric mean in right triangles: If a right triangle has a hypotenuse divided into two segments of lengths \( m \) and \( n \) by the altitude, then the length of the leg adjacent to \( m \) is \( \sqrt{m(m + n)} \), and the leg adjacent to \( n \) is \( \sqrt{n(m + n)} \), and the altitude is \( \sqrt{mn} \). In our case, the leg of length \( 10 \) is adjacent to the segment \( x \)? No, the leg of length \( 10 \) is adjacent to the segment \( 21 \)? Wait, no, the problem is to find \( x \), and we have a leg of length \( 10 \), and the hypotenuse segments are \( x \) and \( 21 \). So, by the geometric mean theorem, \( x^{2}=10\times10\)? No, that's not. Wait, I think I messed up the formula. Let's recall: In a right triangle, when an altitude is drawn to the hypotenuse, then:

1. \( (\text{leg}_1)^{2}=\text{hypotenuse}\times\text{adjacent segment}_1 \)
2. \( (\text{leg}_2)^{2}=\text{hypotenuse}\times\text{adjacent segment}_2 \)
3. \( (\text{altitude})^{2}=\text{segment}_1\times\text{segment}_2 \)

In our figure, the leg of length \( 10 \) is one leg, and the other leg is related to \( x \). Wait, the hypotenuse is \( x + 21 \), the segment adjacent to \( x \) is \( x \), and the segment adjacent to \( 21 \) is \( 21 \). The leg of length \( 10 \) is adjacent to the segment \( x \)? No, the leg of length \( 10 \) is adjacent to the segment \( 21 \)? Wait, no, let's assume that the leg with length \( x \) is adjacent to the segment \( x \) of the hypotenuse, and the leg with length \( 10 \) is adjacent to the segment \( 21 \) of the hypotenuse. Then, by the geometric mean theorem:

\( x^{2}=x\times(x + 21) \)? No, that's not. Wait, no, the correct formula is that the square of the leg is equal to the product of the hypotenuse and the adjacent segment. So, if the leg is \( x \), and the adjacent segment on the hypotenuse is \( x \), and the hypotenuse is \( x + 21 \), then \( x^{2}=x\times(x + 21) \) is wrong. Wait, I think the correct formula is that the leg of length \( 10 \) is such that \( 10^{2}=x\times21 \)? No, that would be if \( 10 \) is the altitude. Wait, no, the altitude is the perpendicular segment from the right angle to the hypotenuse. In the figure, the right angle is at the bottom, and the altitude is drawn to the hypotenuse, creating two smaller right triangles. So, the altitude is not \( 10 \), \( 10 \) is a leg of the original right triangle. So, the original right triangle has legs \( 10 \) and \( x \), and hypotenuse \( x + 21 \)? No, that can't be, because by Pythagoras, \( 10^{2}+x^{2}=(x + 21)^{2} \). Let's try that.

Step2: Apply the Pythagorean theorem

If we consider the right triangle with legs \( 10 \) and \( x \), and hypotenuse \( x + 21 \), then by the Pythagorean theorem:

\( 10^{2}+x^{2}=(x + 21)^{2} \)

Expand the right - hand side: \( (x + 21)^{2}=x^{2}+42x + 441 \)

So, the equation becomes:

\( 100+x^{2}=x^{2}+42x + 441 \)

Subtract \( x^{2} \) from both sides:

\( 100=42x + 441 \)

Subtract \( 441 \) from both sides:

\( 100-441 = 42x \)

\( - 341=42x \), which is impossible since length can't be negative. So, my assumption about the legs is wrong.

Wait, maybe the leg of length \( 10 \) is adjacent to the segment \( x \) of the hypotenuse, and the other segment is \( 21 \), and the leg of length \( x \) is adjacent to the segment \( 21 \) of the hypotenuse. Then, by the geometric mean theorem:

\( 10^{2}=x\times(x + 21) \)? No, that's not. Wait, the correct formula is that the square of the leg is equal to the product of the hypotenuse and the adjacent segment. So, if the leg is \( 10 \), and the adjacent segment on the hypotenuse is \( x \), and the hypotenuse is \( x + 21 \), then \( 10^{2}=x\times(x + 21) \) is wrong. Wait, I think the correct application is that the leg of length \( x \) and the leg of length \( 10 \) are related to the segments of the hypotenuse such that \( x^{2}=10\times21 \)? No, that's the altitude formula. Wait, no, the altitude formula is \( h^{2}=m\times n \), where \( h \) is the altitude, \( m \) and \( n \) are the segments of the hypotenuse. In our case, if \( 10 \) is the altitude, then \( 10^{2}=x\times21 \), so \( x=\frac{100}{21}\), but that doesn't make sense. Wait, no, the figure shows a right triangle with a leg of length \( 10 \), and the hypotenuse is split into \( x \) and \( 21 \), and there is an altitude drawn to the hypotenuse, so the two smaller right triangles are similar to the original triangle and to each other.

Let the original right triangle be \( \triangle ABC \) with right angle at \( C \), altitude \( CD \) to hypotenuse \( AB \), where \( AD=x \), \( DB = 21 \), and \( AC = 10 \)? No, \( BC = 10 \). Then, \( \triangle BCD\sim\triangle BAC \), so \( \frac{BC}{BA}=\frac{BD}{BC} \), so \( BC^{2}=BD\times BA \). Here, \( BC = 10 \), \( BD = 21 \), \( BA=x + 21 \). Wait, no, that would be \( 10^{2}=21\times(x + 21) \), which gives \( 100 = 21x+441 \), \( 21x=100 - 441=- 341 \), still negative.

Wait, maybe \( \triangle ACD\sim\triangle ABC \), so \( \frac{AC}{AB}=\frac{AD}{AC} \), so \( AC^{2}=AD\times AB \). Let \( AC = 10 \), \( AD=x \), \( AB=x + 21 \). Then \( 10^{2}=x\times(x + 21) \), \( 100=x^{2}+21x \), \( x^{2}+21x - 100 = 0 \). Using the quadratic formula \( x=\frac{-21\pm\sqrt{21^{2}+400}}{2}=\frac{-21\pm\sqrt{441 + 400}}{2}=\frac{-21\pm\sqrt{841}}{2}=\frac{-21\pm29}{2} \). We take the positive root: \( x=\frac{-21 + 29}{2}=\frac{8}{2}=4 \). Wait, that works. So, the correct application is that the leg \( AC = 10 \) is related to the segment \( AD=x \) and the hypotenuse \( AB=x + 21 \) by the geometric mean theorem (similar triangles), so \( AC^{2}=AD\times AB \).

So, step by step:

1. Identify the similar triangles: \( \triangle ACD\sim\triangle ABC \) (by AA similarity, since \( \angle A \) is common and both are right triangles).
2. Set up the proportion from similar triangles: \( \frac{AC}{AB}=\frac{AD}{AC} \)
3. Substitute the known values: \( AC = 10 \), \( AD=x \), \( AB=x + 21 \), so \( 10^{2}=x(x + 21) \)
4. Expand the equation: \( 100=x^{2}+21x \)
5. Rearrange into quadratic form: \( x^{2}+21x - 100 = 0 \)
6. Use the quadratic formula \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \), where \( a = 1 \), \( b = 21 \), \( c=- 100 \)
- Calculate the discriminant: \( \Delta=b^{2}-4ac=21^{2}-4\times1\times(-100)=441 + 400 = 841 \)
- \( \sqrt{\Delta}=\sqrt{841}=29 \)
- \( x=\frac{-21\pm29}{2} \)
- We take the positive solution: \( x=\frac{-21 + 29}{2}=\frac{8}{2}=4 \) (we discard the negative solution \( x=\frac{-21 - 29}{2}=\frac{-50}{2}=-25 \) since length can't be negative)

Answer:

\( 4 \)