Question

explaining methods
what are two different ways you could find the value of
a? explain these methods.
image of a triangle with segments 15, 9, 16, and side a, and a right angle segment

Explanation:

Step1: Method 1 - Geometric Mean (Altitude-on-Hypotenuse)

In a right triangle, when an altitude is drawn to the hypotenuse, the length of a leg is the geometric mean of the hypotenuse segment adjacent to it and the hypotenuse. First, find the length of the altitude (let's call it \( h \)) using the left leg: \( h = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12 \). Then, for the right leg \( a \), use the geometric mean with the adjacent hypotenuse segment (16) and the altitude? Wait, no, correct geometric mean: leg \( a = \sqrt{16 \times (9 + 16)} \)? Wait, no, the two segments of the hypotenuse are 9 and 16, so hypotenuse \( c = 9 + 16 = 25 \). Then, by geometric mean, \( a = \sqrt{16 \times 25} \)? Wait, no, the formula is leg \( = \sqrt{\text{segment adjacent} \times \text{hypotenuse}} \). So adjacent segment to \( a \) is 16, hypotenuse is 25, so \( a = \sqrt{16 \times 25} = \sqrt{400} = 20 \). Wait, but also, using Pythagoras on the right triangle: \( a^2 + h^2 = c^2 \), but \( h = 12 \), so \( a^2 + 12^2 = 25^2 \)? No, wait, the big triangle is right-angled? Wait, the diagram shows a triangle with an altitude, so it's a right triangle divided into two smaller right triangles. So the original triangle is right-angled, with legs 15 and \( a \), hypotenuse 25 (9 + 16). Wait, maybe I messed up. Let's re-express:

Wait, the left small triangle: legs 9 and \( h \), hypotenuse 15. So \( 9^2 + h^2 = 15^2 \implies h = 12 \). Then the right small triangle: legs 16 and \( h = 12 \), hypotenuse \( a \). So \( a = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \). That's Method 2.

Step2: Method 2 - Pythagorean Theorem on the Big Triangle

First, find the hypotenuse of the big triangle: \( c = 9 + 16 = 25 \). The left small triangle: legs 9 and \( h \), hypotenuse 15. So \( h = \sqrt{15^2 - 9^2} = 12 \) (as above). Then, the right small triangle has legs 16 and \( h = 12 \), so by Pythagoras: \( a = \sqrt{16^2 + 12^2} = \sqrt{256 + 144} = \sqrt{400} = 20 \). Alternatively, use Pythagoras on the big triangle: if the big triangle is right-angled (since altitude is drawn, so original triangle is right), then legs are 15 and \( a \), hypotenuse 25. So \( 15^2 + a^2 = 25^2 \implies 225 + a^2 = 625 \implies a^2 = 400 \implies a = 20 \). Wait, that's a better Method 1: Pythagorean Theorem on the big right triangle.

So Method 1: Pythagorean Theorem on the large right triangle.

• Step 1: Find hypotenuse of large triangle: \( 9 + 16 = 25 \).
• Step 2: Apply Pythagoras: \( 15^2 + a^2 = 25^2 \).
• Step 3: Solve for \( a \): \( a^2 = 25^2 - 15^2 = (25 - 15)(25 + 15) = 10 \times 40 = 400 \), so \( a = \sqrt{400} = 20 \).

Method 2: Use the two smaller right triangles.

• Step 1: Find the length of the altitude \( h \) in the left small triangle: \( h^2 + 9^2 = 15^2 \implies h^2 = 225 - 81 = 144 \implies h = 12 \).
• Step 2: Now, in the right small triangle, legs are \( h = 12 \) and 16, hypotenuse is \( a \). Apply Pythagoras: \( a^2 = 12^2 + 16^2 = 144 + 256 = 400 \implies a = 20 \).

Step2: Method 2 Explanation (Alternative: Geometric Mean)

In a right triangle, when an altitude is drawn to the hypotenuse, each leg is the geometric mean of the hypotenuse segment adjacent to it and the hypotenuse. The hypotenuse segments are 9 and 16, so hypotenuse \( c = 9 + 16 = 25 \). For leg \( a \), the adjacent segment is 16, so \( a = \sqrt{16 \times 25} = \sqrt{400} = 20 \). For the other leg (15), adjacent segment is 9, so \( 15 = \sqrt{9 \times 25} = \sqrt{225} = 15 \), which checks out.

Answer:

Two methods:

1. Pythagorean Theorem on the Large Triangle: The original triangle is right - angled with hypotenuse \( 9 + 16 = 25 \) and one leg \( 15 \). Using \( a^{2}+15^{2}=25^{2} \), we solve \( a^{2}=25^{2}-15^{2}=400 \), so \( a = 20 \).
2. Using Smaller Right Triangles (Altitude): First, find the altitude \( h \) in the left small triangle: \( h=\sqrt{15^{2}-9^{2}} = 12 \). Then, in the right small triangle with legs \( h = 12 \) and \( 16 \), use Pythagoras: \( a=\sqrt{12^{2}+16^{2}}=\sqrt{400}=20 \). (Or use geometric mean: \( a=\sqrt{16\times25}=20 \))