Question

geometry - honors - 9(ac)10(a) - 2025-2028
7-4: mathxl for school: practice & problem-solving copy 1
due jan 21 - 11:45 pm late
the diagram shows the parts of a right triangle with an altitude to the hypotenuse. for the two given measures, find the other four.
h = 18, s₁ = 9

( s_2 = square )
( a = square )
( u_1 = square )
( u_2 = square )
(simplify your answers. type exact answers, using radicals as needed.)

Explanation:

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

In a right triangle, the altitude to the hypotenuse is the geometric mean of the segments into which it divides the hypotenuse. Also, each leg is the geometric mean of the hypotenuse and the adjacent segment.

Given \( h = 18 \) (altitude), \( s_1 = 9 \) (one segment of hypotenuse). Let \( s_2 \) be the other segment. By geometric mean theorem: \( h^2 = s_1 \times s_2 \)
So, \( 18^2 = 9 \times s_2 \)
\( 324 = 9s_2 \)

Step2: Solve for \( s_2 \)

Divide both sides by 9: \( s_2=\frac{324}{9}= 36 \)

Step3: Find \( a \) (leg, geometric mean of \( s_1 \) and hypotenuse part? Wait, no: \( a^2 = s_1 \times (s_1 + s_2) \)? No, wait: \( a^2 = s_1 \times (s_1 + s_2) \)? Wait, no, the leg \( a \) (the one adjacent to \( s_1 \)): \( a^2 = s_1 \times (s_1 + s_2) \)? Wait, no, correct formula: \( a^2 = s_1 \times (s_1 + s_2) \)? Wait, no, actually, \( a^2 = s_1 \times (s_1 + s_2) \) is wrong. Wait, the two segments of hypotenuse are \( s_1 = 9 \), \( s_2 = 36 \), so hypotenuse \( u = s_1 + s_2 = 9 + 36 = 45 \). Then, \( a^2 = s_1 \times u \)? Wait, no: \( a^2 = s_1 \times (s_1 + s_2) \)? Wait, no, the correct formula is that in right triangle with altitude \( h \), segments \( s_1, s_2 \), legs \( u_1, u_2 \), then:

\( h^2 = s_1 s_2 \)

\( u_1^2 = s_1 (s_1 + s_2) \)? No, wait, \( u_1^2 = s_1 (s_1 + s_2) \) is incorrect. Wait, actually, \( u_1^2 = s_1 (s_1 + s_2) \) is wrong. Let's rederive:

Let the right triangle be \( \triangle ABC \), right-angled at \( C \), altitude \( CD \) to hypotenuse \( AB \), with \( AD = s_1 = 9 \), \( DB = s_2 \), \( CD = h = 18 \), \( AC = u_1 \), \( BC = u_2 \), \( AB = s_1 + s_2 \).

Then, \( \triangle ACD \sim \triangle ABC \), so \( \frac{AC}{AB} = \frac{AD}{AC} \implies AC^2 = AD \times AB \)

Similarly, \( BC^2 = BD \times AB \)

And \( CD^2 = AD \times BD \) (which we used for \( s_2 \))

So, \( AB = s_1 + s_2 = 9 + 36 = 45 \)

Then, \( a^2 = s_1 \times AB = 9 \times 45 = 405 \)? Wait, no, wait, \( a \) is \( AC \), so \( AC^2 = AD \times AB = 9 \times 45 = 405 \), so \( a = \sqrt{405} = \sqrt{81 \times 5} = 9\sqrt{5} \)? Wait, no, wait, I think I messed up. Wait, no, the leg \( a \) (the one from the right angle to the vertex, the altitude's top) – wait, no, the leg \( a \) is the one that is the geometric mean of \( s_1 \) and \( s_2 \)? No, wait, \( a^2 = s_1 \times (s_1 + s_2) \) is wrong. Wait, let's use \( a^2 = s_1 \times (s_1 + s_2) \)? Wait, no, let's use the first formula: \( a^2 = s_1 \times (s_1 + s_2) \) is incorrect. Wait, let's use \( a^2 = s_1 \times (s_1 + s_2) \) – no, let's compute \( a \) as \( \sqrt{s_1^2 + h^2} \), since \( \triangle ACD \) is right-angled at \( D \), so \( AC^2 = AD^2 + CD^2 = 9^2 + 18^2 = 81 + 324 = 405 \), so \( a = \sqrt{405} = 9\sqrt{5} \)? Wait, no, 9² is 81, 18² is 324, sum is 405, square root of 405 is 9√5? Wait, 405 = 81×5, so √405 = 9√5. Wait, but let's check with the other formula: \( AC^2 = AD \times AB \), \( AB = 9 + 36 = 45 \), so \( 9×45 = 405 \), which matches. So \( a = \sqrt{405} = 9\sqrt{5} \)? Wait, no, wait, 9×45 is 405, √405 is 9√5? Wait, 9√5 is approximately 20.12, but 18² + 9² = 324 + 81 = 405, yes.

Wait, but let's find \( u_1 \) and \( u_2 \). Wait, \( u_1 \) is \( AC \), which we just found as \( 9\sqrt{5} \)? Wait, no, wait, \( u_1 \) is \( AC \), \( u_2 \) is \( BC \). Let's find \( u_2 \): \( BC^2 = BD \times AB = 36 \times 45 = 1620 \), so \( u_2 = \sqrt{1620} = \sqrt{324×5} = 18\sqrt{5} \). Wait, but let's check with \( BC^2 = CD^2 + BD^2 = 18² + 36² = 324 + 1296 = 1620 \), which matches. So \( u_2 = 18\sqrt{5} \), \( u_1 = 9\sqrt{5} \)? Wait, no, wait, \( AC \) is \( u_1 \), \( BC \) is \( u_2 \). Wait, but let's re-express:

Wait, first, we found \( s_2 = 36 \) (from \( h^2 = s_1 s_2 \implies 18² = 9 s_2 \implies s_2 = 36 \))

Then, \( a \) (the leg, which is \( CD \)? No, \( a \) is the leg, the one that's the altitude? No, \( a \) is the leg, the one from the right angle to the vertex, so \( a \) is the leg, so \( a^2 = s_1 \times (s_1 + s_2) \)? No, \( a^2 = s_1 \times (s_1 + s_2) \) is wrong. Wait, the leg \( a \) (the one adjacent to \( s_1 \)): \( a^2 = s_1 \times (s_1 + s_2) \)? No, the correct formula is \( a^2 = s_1 \times (s_1 + s_2) \) is incorrect. Wait, the two legs are \( u_1 \) and \( u_2 \), with \( u_1^2 = s_1 (s_1 + s_2) \)? No, \( u_1^2 = s_1 (s_1 + s_2) \) is wrong. Wait, let's use Pythagoras on \( \triangle ACD \): \( u_1^2 = h^2 + s_1^2 = 18² + 9² = 324 + 81 = 405 \implies u_1 = \sqrt{405} = 9\sqrt{5} \)

Similarly, \( u_2^2 = h^2 + s_2^2 = 18² + 36² = 324 + 1296 = 1620 \implies u_2 = \sqrt{1620} = 18\sqrt{5} \)

Wait, but let's check with the other formula: \( u_1^2 = s_1 \times (s_1 + s_2) \), \( s_1 + s_2 = 45 \), so \( 9×45 = 405 \), which matches. Similarly, \( u_2^2 = s_2 \times (s_1 + s_2) = 36×45 = 1620 \), which matches.

So summarizing:

\( s_2 = 36 \)

\( a = \sqrt{s_1^2 + h^2} = \sqrt{9^2 + 18^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5} \) (Wait, no, \( a \) is the leg, which is \( u_1 \)? Wait, the problem says "a" – maybe "a" is the leg, so \( a = 9\sqrt{5} \)? Wait, but let's check the problem statement: "find the other four: \( s_2, a, u_1, u_2 \)". So:

\( s_2 = 36 \)

\( a \): Let's see, in the diagram, "a" is the leg (the one perpendicular to hypotenuse, the altitude? No, the altitude is h=18. Wait, maybe "a" is the leg, so \( a^2 = s_1 \times s_2 \)? No, that's h². Wait, no, I think I made a mistake. Wait, the altitude is h=18, segments s1=9, s2=36. Then the leg "a" (the one that's the geometric mean of s1 and s2? No, h is the geometric mean of s1 and s2. The legs: each leg is the geometric mean of s1 (or s2) and the hypotenuse. Wait, hypotenuse is s1 + s2 = 45. So leg u1 (adjacent to s1) is \( u_1 = \sqrt{s_1 \times (s_1 + s_2)} = \sqrt{9 \times 45} = \sqrt{405} = 9\sqrt{5} \)

Leg u2 (adjacent to s2) is \( u_2 = \sqrt{s_2 \times (s_1 + s_2)} = \sqrt{36 \times 45} = \sqrt{1620} = 18\sqrt{5} \)

And "a" – maybe "a" is the leg, so \( a = u_1 = 9\sqrt{5} \)? Wait, but let's check with the right triangle: the leg "a" (the one from the right angle to the vertex, forming the two smaller right triangles). So:

So to recap:

1. \( h^2 = s_1 s_2 \implies 18^2 = 9 s_2 \implies s_2 = 36 \)

2. \( a^2 = s_1^2 + h^2 \) (since in the smaller right triangle with legs s1 and h, hypotenuse a) \(\implies a = \sqrt{9^2 + 18^2} = \sqrt{81 + 324} = \sqrt{405} = 9\sqrt{5}\)

3. \( u_1 = a = 9\sqrt{5} \) (since u1 is the leg adjacent to s1)

4. \( u_2 = \sqrt{s_2^2 + h^2} = \sqrt{36^2 + 18^2} = \sqrt{1296 + 324} = \sqrt{1620} = 18\sqrt{5}\)

Answer:

\( s_2 = \boldsymbol{36} \)

\( a = \boldsymbol{9\sqrt{5}} \)

\( u_1 = \boldsymbol{9\sqrt{5}} \)

\( u_2 = \boldsymbol{18\sqrt{5}} \)