Question

7 practice 7 (from unit 3, lesson 13) find the lengths of segments ad and bd. then check your answers using a different method. round to the nearest tenth if necessary. type your answers in the boxes. ad = and bd =

Explanation:

Step1: Find AB using Pythagoras

In right triangle \(ABC\), \(BC = 13\), \(AC = 12\)? Wait, no, \(DC + AD = AC\), but first find \(AB\)? Wait, no, \(BD\) is altitude to hypotenuse \(AC\)? Wait, triangle \(ABC\) is right-angled at \(B\), so \(AB^2 + BC^2 = AC^2\)? Wait, no, \(AC\) is 12? Wait, the diagram: \(D\) is on \(AC\), \(BD \perp AC\), \(BC = 13\), \(DC\)? Wait, maybe \(AC\) is not 12. Wait, no, the diagram shows \(DC = 12\)? Wait, no, the label is \(12\) next to \(DC\)? Wait, no, the segment from \(D\) to \(C\) is 12? Wait, maybe \(AC\) is \(AD + DC\), and \(BC = 13\), \(BD\) is altitude. Wait, in right triangle \(ABC\) (right-angled at \(B\)), \(BD\) is the altitude to hypotenuse \(AC\). Then we can use geometric mean theorems.

First, let's denote \(AD = x\), \(DC = 12\) (from the diagram), so \(AC = x + 12\). In right triangle \(ABC\), \(BC = 13\), and \(BD\) is altitude. By geometric mean, \(BC^2 = DC \times AC\)? Wait, no, the geometric mean theorem: in a right triangle, the square of a leg is equal to the product of the hypotenuse and the adjacent segment. So \(BC^2 = DC \times AC\)? Wait, \(BC\) is a leg, \(DC\) is the segment of hypotenuse adjacent to \(BC\). So \(BC^2 = DC \times AC\). So \(13^2 = 12 \times AC\)? Wait, that would make \(AC = 169 / 12 \approx 14.08\), but that doesn't make sense. Wait, maybe I misread the diagram. Wait, maybe \(AC = 12\)? No, \(BC = 13\), so if \(AC\) were 12, then \(AB\) would be imaginary. So maybe the diagram has \(AC = 12\) is wrong. Wait, no, the user's diagram: \(A---D---C\), with \(DC = 12\), \(BC = 13\), right angle at \(B\), \(BD \perp AC\). So triangle \(ABC\) is right-angled at \(B\), \(BD \perp AC\). Then:

By geometric mean theorem (altitude-on-hypotenuse theorem):

1. \(BD^2 = AD \times DC\)
2. \(AB^2 = AD \times AC\)
3. \(BC^2 = DC \times AC\)

So let's use \(BC^2 = DC \times AC\). \(BC = 13\), \(DC = 12\), so \(13^2 = 12 \times AC\) → \(169 = 12 \times AC\) → \(AC = 169 / 12 ≈ 14.08\). Then \(AD = AC - DC = 169/12 - 12 = 169/12 - 144/12 = 25/12 ≈ 2.1\)? Wait, no, that can't be. Wait, maybe \(AC\) is 12, and \(BC = 13\) is wrong. Wait, maybe the right angle is at \(D\), so \(BD \perp AC\), and triangle \(ABC\) is right-angled at \(B\). Wait, no, the diagram shows \(B\) with a right angle, so \(AB \perp BC\), and \(BD \perp AC\). So \(ABC\) is right-angled at \(B\), \(BD\) is altitude to \(AC\). Then \(AC\) is hypotenuse, \(AB\) and \(BC\) are legs. Let's denote \(AD = x\), \(DC = 12\), so \(AC = x + 12\). Then by Pythagoras in \(ABC\): \(AB^2 + BC^2 = AC^2\). By Pythagoras in \(ABD\): \(AB^2 = AD^2 + BD^2 = x^2 + BD^2\). By Pythagoras in \(CBD\): \(BC^2 = DC^2 + BD^2 = 12^2 + BD^2\). So substitute into \(ABC\)'s Pythagoras: \(x^2 + BD^2 + 144 + BD^2 = (x + 12)^2\). Simplify: \(x^2 + 2BD^2 + 144 = x^2 + 24x + 144\). Cancel \(x^2\) and 144: \(2BD^2 = 24x\) → \(BD^2 = 12x\). Also, by geometric mean, \(BD^2 = AD \times DC = x \times 12\), which matches. Now, also, \(BC^2 = DC \times AC\) (from geometric mean: leg squared = hypotenuse segment × hypotenuse). So \(BC^2 = 12 \times (x + 12)\). But we also have \(BC^2 = 12^2 + BD^2 = 144 + 12x\) (from triangle \(CBD\)). So \(12(x + 12) = 144 + 12x\), which is an identity. So we need another equation. Wait, maybe \(BC = 13\) is given. So \(BC^2 = 169 = 144 + 12x\) → \(12x = 169 - 144 = 25\) → \(x = 25/12 ≈ 2.1\)? Wait, no, that would be \(AD = 25/12\), but then \(BD^2 = 12x = 25\) → \(BD = 5\). Wait, that makes sense. Let's check:

If \(AD = x = 25/12 ≈ 2.1\), \(DC = 12\), \(AC = 25/12 + 12 = 25/12 + 144/12 = 169/12 ≈ 14.08\). Then \(BC^2 = 13^2 = 169\), and \(DC \times AC = 12 \times 169/12 = 169\), which matches. Then \(BD^2 = AD \times DC = (25/12) \times 12 = 25\) → \(BD = 5\). Perfect. So \(AD = 25/12 ≈ 2.1\) (wait, 25/12 is approximately 2.083..., rounded to nearest tenth is 2.1? Wait, no, 25/12 is 2.083..., so 2.1? Wait, but let's check again. Wait, maybe I mixed up the segments. Wait, maybe \(DC\) is not 12, but \(AC\) is 12? Wait, the diagram shows \(12\) next to \(DC\), so \(DC = 12\). Then \(BC = 13\), so using \(BC^2 = DC \times AC\) → \(13^2 = 12 \times AC\) → \(AC = 169/12 ≈ 14.08\), so \(AD = AC - DC = 169/12 - 144/12 = 25/12 ≈ 2.1\), and \(BD^2 = AD \times DC = (25/12) \times 12 = 25\) → \(BD = 5\). So that works.

Step2: Calculate AD and BD

From above, \(AD = \frac{25}{12} \approx 2.1\) (wait, 25/12 is 2.083..., so to nearest tenth is 2.1? Wait, no, 25 divided by 12 is 2.0833..., which is 2.1 when rounded to nearest tenth? Wait, 0.0833 is less than 0.1, so 2.1? Wait, no, 2.0833 rounded to nearest tenth is 2.1? Wait, tenth place is 0, hundredth is 8, so round up: 2.1. And \(BD = 5\) (exact).

Wait, let's verify with another method. Let's use Pythagoras in triangle \(ABD\) and \(CBD\). \(BD = 5\), \(AD = 25/12\), so \(AB^2 = AD^2 + BD^2 = (625/144) + 25 = (625 + 3600)/144 = 4225/144\) → \(AB = 65/12 ≈ 5.4\). Then in triangle \(ABC\), \(AB = 65/12\), \(BC = 13 = 156/12\), \(AC = 169/12\). Check \(AB^2 + BC^2 = (4225 + 24336)/144 = 28561/144\), and \(AC^2 = (169/12)^2 = 28561/144\). So it works.

So \(AD = \frac{25}{12} \approx 2.1\) (or exact 25/12 ≈ 2.1) and \(BD = 5\).

Answer:

\(AD = \frac{25}{12} \approx 2.1\) (or \(2.1\) when rounded to nearest tenth) and \(BD = 5\)

Wait, but 25/12 is approximately 2.083, which is 2.1 when rounded to nearest tenth. And BD is 5 exactly. So:

\(AD = \boxed{2.1}\) and \(BD = \boxed{5}\) (or \(AD = \boxed{\frac{25}{12}}\) but the problem says round to nearest tenth if necessary, so 2.1 is better).