Question

given that n = 48 and a = 60, what is the value of b? (1 point) \\(\bigcirc\\) b = 75 \\(\bigcirc\\) b = 36 \\(\bigcirc\\) b = 27 \\(\bigcirc\\) b = 45

Explanation:

Step1: Identify the geometric theorem

This is a right triangle with an altitude to the hypotenuse, so we use the geometric mean theorem (or altitude-on-hypotenuse theorem), which states that 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 hypotenuse. Wait, actually, the correct formula from the geometric mean theorem for leg \( a \), leg \( b \), segment \( n \) (adjacent to \( a \)), and segment \( m \) (adjacent to \( b \)) is \( a^2 = n \times c \) (but actually, more accurately, if \( CD \) is the altitude to hypotenuse \( AB \), then \( a^2 = n \times AB \) and \( b^2 = m \times AB \), but also, we know that \( a^2 + b^2 = (m + n)^2 \), and also \( a^2 = n \times (m + n) \)? Wait, no, let's recall: In a right triangle, when you draw an altitude from the right angle to the hypotenuse, then:

1. \( a^2 = n \times c \) (where \( c = m + n \), the hypotenuse)
2. \( b^2 = m \times c \)
3. \( CD^2 = m \times n \)

But in this problem, we are given \( n = 48 \) and \( a = 60 \). Let's denote the hypotenuse as \( c = m + n \). From the first formula, \( a^2 = n \times c \), so we can solve for \( c \):

\( c = \frac{a^2}{n} \)

Step2: Calculate the hypotenuse \( c \)

Substitute \( a = 60 \) and \( n = 48 \):

\( c = \frac{60^2}{48} = \frac{3600}{48} = 75 \)

Wait, but that's the hypotenuse? Wait, no, maybe I mixed up. Wait, actually, the leg \( a \) is opposite to segment \( n \)? Wait, no, let's look at the diagram. Triangle \( ABC \) is right-angled at \( C \), and \( CD \) is perpendicular to \( AB \), so \( AD = m \), \( DB = n = 48 \), \( AC = b \), \( BC = a = 60 \), \( AB = c = m + n \).

Then by the geometric mean theorem, \( BC^2 = DB \times AB \), so \( a^2 = n \times c \). So \( 60^2 = 48 \times c \), so \( c = \frac{3600}{48} = 75 \). Then, since \( AB = c = 75 \), and \( DB = n = 48 \), then \( AD = m = c - n = 75 - 48 = 27 \). Then, for leg \( b = AC \), by the geometric mean theorem, \( b^2 = m \times c \), so \( b^2 = 27 \times 75 \)? Wait, no, that can't be. Wait, no, the correct formula is \( AC^2 = AD \times AB \), so \( b^2 = m \times c \). But we can also use Pythagoras: \( a^2 + b^2 = c^2 \). We know \( a = 60 \), \( c = 75 \), so:

\( b^2 = c^2 - a^2 = 75^2 - 60^2 \)

Step3: Calculate \( b \) using Pythagoras

Calculate \( 75^2 - 60^2 \):

\( 75^2 = 5625 \), \( 60^2 = 3600 \)

\( b^2 = 5625 - 3600 = 2025 \)

Then \( b = \sqrt{2025} = 45 \)? Wait, no, that's not matching. Wait, maybe I mixed up \( a \) and \( b \). Wait, maybe \( a \) is \( AC \) and \( b \) is \( BC \)? No, the diagram shows \( BC = a \), \( AC = b \). Wait, let's re-express:

Wait, if \( BC = a = 60 \), \( DB = n = 48 \), and by geometric mean, \( BC^2 = DB \times AB \), so \( AB = \frac{BC^2}{DB} = \frac{60^2}{48} = 75 \), so \( AB = 75 \). Then \( AD = AB - DB = 75 - 48 = 27 \). Then, by geometric mean, \( AC^2 = AD \times AB \), so \( b^2 = 27 \times 75 = 2025 \), so \( b = \sqrt{2025} = 45 \)? Wait, but the options include \( b = 45 \), \( b = 36 \), etc. Wait, but let's check with Pythagoras: \( AC^2 + BC^2 = AB^2 \). If \( AC = b \), \( BC = 60 \), \( AB = 75 \), then \( b^2 + 60^2 = 75^2 \), so \( b^2 = 75^2 - 60^2 = 5625 - 3600 = 2025 \), so \( b = 45 \). Wait, but earlier when I thought \( c = 75 \), that's the hypotenuse \( AB \). So then \( b = 45 \)? But let's check the options: the options are \( b = 75 \), \( 36 \), \( 27 \), \( 45 \). So \( b = 45 \) is an option. Wait, but let's verify again.

Wait, maybe I made a mistake in the theorem. Let's recall: In right triangle \( ABC \), right-angled at \( C \), altitude \( CD \) to hypotenuse \( AB \), then:

- \( AC^2 = AD \times AB \)
- \( BC^2 = BD \times AB \)
- \( CD^2 = AD \times BD \)

So here, \( BC = a = 60 \), \( BD = n = 48 \), so \( BC^2 = BD \times AB \implies 60^2 = 48 \times AB \implies AB = \frac{3600}{48} = 75 \). Then \( AB = 75 \), so \( AD = AB - BD = 75 - 48 = 27 \). Then \( AC^2 = AD \times AB \implies b^2 = 27 \times 75 = 2025 \implies b = \sqrt{2025} = 45 \). Yes, that's correct. So \( b = 45 \).

Wait, but let's check with Pythagoras: \( AC^2 + BC^2 = AB^2 \implies b^2 + 60^2 = 75^2 \implies b^2 = 5625 - 3600 = 2025 \implies b = 45 \). Correct.

Answer:

\( b = 45 \) (so the option is \( b = 45 \))