Question

solve for ( x ) to the nearest tenth.
answer attempt 1 out of 3
( x approx )
submit answer

Explanation:

Step1: Find the hypotenuse of the small right triangle.

Using the Pythagorean theorem \( c = \sqrt{a^2 + b^2} \), where \( a = 2 \) and \( b = 4 \).
\( c = \sqrt{2^2 + 4^2} = \sqrt{4 + 16} = \sqrt{20} \)

Step2: Find \( x \) using the geometric mean theorem (or altitude-on-hypotenuse theorem) in the large right triangle.

The geometric mean theorem states that in a right triangle, the length of a leg is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. Here, the vertical side of the large triangle is 10, and the segment adjacent to \( x \) (from the small triangle's hypotenuse) is \( \sqrt{20} \)? Wait, no, actually, let's re - examine. The two right triangles: the small one with legs 2 and 4, and the large one with one leg \( x \) and the other leg related to 10. Wait, another approach: The hypotenuse of the small right triangle is \( \sqrt{2^2 + 4^2}=\sqrt{4 + 16}=\sqrt{20} \). Now, in the large right triangle, we can consider the two right triangles (the small one and the one with leg \( x \)) are similar? Wait, no, actually, the large triangle is a right triangle, and we can use the Pythagorean theorem in a different way. Wait, the vertical side of the large figure is 10, and the horizontal side is 4. Wait, no, the key is that the triangle with leg \( x \) and the other leg (let's say the segment from the right angle of the small triangle to the bottom vertex) and the hypotenuse 10? Wait, no, let's use the geometric mean. The length of the altitude to the hypotenuse of a right triangle is the geometric mean of the lengths of the two segments. Wait, maybe a better way: Let's call the hypotenuse of the small triangle \( h \), so \( h=\sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20}=2\sqrt{5}\approx4.472 \). Now, in the large right triangle, we have a right triangle where one leg is \( x \), another leg is related to the vertical side of length 10, and the hypotenuse is 10? No, wait, the large triangle has a vertical side of length 10, and the horizontal side is 4. Wait, actually, the triangle with legs 2 and 4, and the triangle with leg \( x \) and the other leg (let's say the segment from the top right vertex to the bottom vertex) are similar? Wait, no, let's use the Pythagorean theorem in the large right triangle. Wait, the correct approach: The two right triangles (the small one with legs 2 and 4, and the one with leg \( x \)) are similar to the large right triangle? Wait, no, let's consider the following: The length of the segment along the vertical side: the total vertical side is 10, and the upper segment is 2, so the lower segment is \( 10 - 2=8 \)? No, that's not correct. Wait, no, the figure is composed of two right triangles: one with legs 2 and 4 (let's call this triangle A), and another right triangle with leg \( x \) and the other leg being the same as the hypotenuse of triangle A? Wait, no, let's use the geometric mean formula. In a right triangle, if we draw an altitude to the hypotenuse, then \( leg_1^2=segment_1\times hypotenuse \) and \( leg_2^2 = segment_2\times hypotenuse \). But in our case, the triangle with leg \( x \) and the triangle with legs 2 and 4: Let's assume that the triangle with legs 2 and 4 is similar to the triangle with leg \( x \) and the other leg (let's say length \( y \)), and the hypotenuse of the large triangle is 10. Wait, maybe a better way: Let's use the Pythagorean theorem in the large right triangle. The hypotenuse of the large right triangle is 10, one segment of the hypotenuse (from the top right vertex to the right angle of the small triangle) is \( \sqrt{2^{2}+4^{2}}=\sqrt{20} \), and the other segment is \( x \)? No, that's not right. Wait, I think I made a mistake. Let's start over.

The small right triangle has legs \( a = 2 \) and \( b = 4 \). The hypotenuse of this small right triangle is \( c=\sqrt{a^{2}+b^{2}}=\sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20} \). Now, in the large right triangle (the one with the vertical side of length 10), we can consider that the length of \( x \) is the geometric mean between the length of the vertical side (10) and the length of the segment corresponding to the leg of the small triangle (2)? No, wait, the correct geometric mean relationship here is that \( x^{2}=2\times10 \)? No, that's not. Wait, no, the two right triangles: the small one with legs 2 and 4, and the large one with leg \( x \) and the other leg (let's say the horizontal segment 4) and hypotenuse related to 10? I'm getting confused. Wait, another approach: The triangle with leg \( x \) is a right triangle, and we can use the Pythagorean theorem if we know the other two sides. Wait, the vertical side of the entire figure is 10, and the horizontal side is 4. The triangle with legs 2 and 4, and the triangle with leg \( x \) and the other leg (the segment from the right angle of the small triangle to the bottom vertex) form a larger right triangle? Wait, no, let's use the following formula: In a right triangle, if you have a segment from the right angle to the hypotenuse, then the square of the leg is equal to the product of the hypotenuse and the adjacent segment. So, for the small triangle, the leg of length 2: \( 2^{2}=segment_1\times hypotenuse_{large} \), and the leg of length 4: \( 4^{2}=segment_2\times hypotenuse_{large} \), and \( segment_1 + segment_2=hypotenuse_{large} \). But we know that the vertical side of the large triangle is 10. Wait, I think I messed up the figure. Let's look at the figure again: There is a small right triangle with legs 2 and 4, and a larger right triangle with one leg \( x \) and the other leg (the vertical side of length 10) and the hypotenuse? No, the correct way is: The triangle with legs 2 and 4, and the triangle with leg \( x \) are similar. Wait, the small triangle has legs 2 and 4, and the large triangle has legs \( x \) and 4, and hypotenuse 10? No, that's not. Wait, let's calculate the length of the hypotenuse of the small triangle: \( \sqrt{2^{2}+4^{2}}=\sqrt{20} \). Now, in the large right triangle, we have a right triangle where one leg is \( x \), another leg is \( \sqrt{20} \), and the hypotenuse is 10? No, that can't be, because \( \sqrt{20}^{2}+x^{2}=10^{2} \), so \( 20+x^{2}=100 \), then \( x^{2}=80 \), so \( x = \sqrt{80}\approx8.9 \). Wait, that makes sense! Because the two right triangles (the small one with legs 2 and 4, and the large one with leg \( x \) and the hypotenuse of the small triangle) are part of a larger right triangle with hypotenuse 10. So, using the Pythagorean theorem in the large right triangle: \( (\sqrt{2^{2}+4^{2}})^{2}+x^{2}=10^{2} \).

So, Step 1: Calculate the hypotenuse of the small right triangle.
Let \( h \) be the hypotenuse of the small right triangle with legs \( a = 2 \) and \( b = 4 \). By the Pythagorean theorem, \( h=\sqrt{a^{2}+b^{2}}=\sqrt{2^{2}+4^{2}}=\sqrt{4 + 16}=\sqrt{20} \)

Step 2: Apply the Pythagorean theorem to the large right triangle.
In the large right triangle, the two legs are \( h=\sqrt{20} \) and \( x \), and the hypotenuse is 10. So, by the Pythagorean theorem \( h^{2}+x^{2}=10^{2} \)
Substitute \( h^{2}=20 \) into the equation: \( 20+x^{2}=100 \)
Subtract 20 from both sides: \( x^{2}=100 - 20=80 \)
Take the square root of both sides: \( x=\sqrt{80}\approx8.944 \)

Step 3: Round to the nearest tenth.
Rounding \( 8.944 \) to the nearest tenth gives \( 8.9 \) (wait, no, \( 8.944 \) to the nearest tenth: the hundredth digit is 4, which is less than 5, so we round down? Wait, no, \( 8.944 \) is approximately \( 8.9 \)? Wait, no, \( \sqrt{80}\approx8.94427191 \), so to the nearest tenth, we look at the hundredth place, which is 4. Since \( 4<5 \), we round down, so \( x\approx8.9 \)? Wait, no, wait, \( 8.944 \) to the nearest tenth: the tenths place is 9, hundredths is 4, so it's \( 8.9 \)? Wait, no, \( 8.944 \) is closer to \( 8.9 \) than \( 9.0 \)? Wait, no, \( 8.944 \) is \( 8.9 + 0.044 \), so when rounding to the nearest tenth, we have \( 8.9 \) (because the digit in the hundredths place is 4, which is less than 5). Wait, but let's check the calculation again. Wait, maybe I made a mistake in the triangle. Wait, the large triangle: is the hypotenuse 10? Let's look at the figure again. The vertical side is 10, the horizontal side is 4. The triangle with legs 2 and 4, and the triangle with leg \( x \): maybe the large triangle has a vertical side of length 10, and the horizontal side of length 4, and the hypotenuse is the side of length 10? No, that can't be. Wait, another way: The two right triangles (the small one with legs 2 and 4, and the one with leg \( x \)) are similar to the large right triangle. Wait, the small triangle has legs 2 and 4, so the ratio of legs is \( 2:4 = 1:2 \). The large triangle has a vertical side of length 10, so if the ratio is 1:2, then the horizontal side would be 5, but that's not 4. So my previous approach was wrong.

Wait, let's start over. Let's consider the entire figure: There is a right angle at the top left, with a vertical leg of length 2 and horizontal leg of length 4. Then there is another right angle below it, and a leg \( x \) going down to the bottom vertex, and a vertical leg of length 10 on the right. So, the segment from the top right vertex to the bottom vertex is 10. The triangle with legs 2 and 4, and the triangle with leg \( x \) and the other leg (the segment from the top right vertex to the right angle of the small triangle) are similar. Wait, the length of the segment from the top right vertex to the right angle of the small triangle is the hypotenuse of the small triangle, which is \( \sqrt{2^{2}+4^{2}}=\sqrt{20} \). Then, in the right triangle with leg \( x \), hypotenuse 10, and the other leg \( \sqrt{20} \), we can use the Pythagorean theorem: \( x^{2}+(\sqrt{20})^{2}=10^{2} \), which is what I did before. So \( x^{2}=100 - 20 = 80 \), so \( x=\sqrt{80}\approx8.944 \), which rounds to \( 8.9 \) when rounded to the nearest tenth. Wait, but let's check with another method. Let's use the geometric mean. The length of a leg of a right triangle is the geometric mean of the length of the hypotenuse and the length of the adjacent segment. So, for the leg of length \( x \), the adjacent segment on the hypotenuse (of the large triangle) is the length of the hypotenuse of the small triangle, which is \( \sqrt{20} \), and the hypotenuse of the large triangle is 10. Wait, no, the geometric mean formula is \( leg^{2}=segment\times hypotenuse \). So if the hypotenuse of the large triangle is \( L \), and the segment is \( s \), then \( x^{2}=s\times L \). But we know that the other leg of the large triangle (the one with length 2) satisfies \( 2^{2}=(L - s)\times L \). Wait, let's set \( L = 10 \) (the hypotenuse of the large triangle), then \( 2^{2}=(10 - s)\times10 \), so \( 4 = 100-10s \), then \( 10s=96 \), \( s = 9.6 \). Then \( x^{2}=s\times L=9.6\times10 = 96 \), so \( x=\sqrt{96}\approx9.8 \). Wait, now I'm really confused. There must be a mistake in identifying the triangle. Let's look at the figure again: The figure has a right angle at the top left (legs 2 and 4), then another right angle below it, and a leg \( x \) going down to the bottom, and a vertical side of length 10 on the right. So the vertical side is composed of two parts: the top part (length 2) and the bottom part (length \( y \)), so \( 2 + y=10 \), so \( y = 8 \). Now, the two right triangles: the small one with legs 2 and 4, and the large one with legs \( x \) and 4, and hypotenuse related to \( y = 8 \)? No, the horizontal side is 4 for both triangles. Wait, now I see! The two right triangles are: one with legs 2 and 4 (top), and one with legs \( x \) and 4 (bottom), and the vertical side of the bottom triangle is \( 10 - 2=8 \). Wait, no, the bottom triangle has a vertical side of length 8 and horizontal side of length 4, so its hypotenuse \( x \) is \( \sqrt{8^{2}+4^{2}}=\sqrt{64 + 16}=\sqrt{80}\approx8.944 \), which is what I calculated before. And the top triangle has hypotenuse \( \sqrt{2^{2}+4^{2}}=\sqrt{20}\approx4.472 \). Then the total vertical side is \( 2 + 8 = 10 \), which matches the figure. So the bottom triangle has legs 8 and 4, so \( x=\sqrt{8^{2}+4^{2}}=\sqrt{64 + 16}=\sqrt{80}\approx8.944 \), which rounds to \( 8.9 \) when rounded to the nearest tenth. Wait, but \( \sqrt{80}\approx8.944 \), so to the nearest tenth, it's \( 8.9 \) (since the hundredth digit is 4, we round down). Wait, but let's check with the geometric mean. The altitude to the hypotenuse of a right triangle is the geometric mean of the two segments. Wait, in the large right triangle (with legs 10 and 4? No, the large triangle would have legs 10 and 4, but that's not right. Wait, no, the correct triangles are: the small triangle (legs 2, 4, hypotenuse \( \sqrt{20} \)) and the large triangle (legs \( x \), 4, hypotenuse \( \sqrt{8^{2}+4^{2}} \))? No, I think the key is that the bottom triangle has legs 8 (10 - 2) and 4, so \( x=\sqrt{8^{2}+4^{2}}=\sqrt{64 + 16}=\sqrt{80}\approx8.9 \) when rounded to the nearest tenth.

Answer:

\( x\approx8.9 \)