Question

set

1. using the right triangle with altitude drawn to the hypotenuse, prove the

pythagorean theorem, $a^2 + b^2 = c^2$.

find the missing values for each right triangle including the length of the altitude.
3.
4.

Explanation:

Step1: Analyze Triangle 3

The triangle is a right triangle with legs 5 and 12, and altitude \( y \) to hypotenuse \( x \). First, find hypotenuse \( x \) using Pythagorean theorem: \( x = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \). Then, area of triangle: \( \frac{1}{2} \times 5 \times 12 = \frac{1}{2} \times x \times y \). So, \( 30 = \frac{1}{2} \times 13 \times y \), solving for \( y \): \( y = \frac{60}{13} \approx 4.62 \).

Step2: Analyze Triangle 4

The vertical side is 8, split into two parts. Let the upper segment be \( m \), lower be \( 8 - m \). For the upper triangle (leg 5, hypotenuse part \( m \)): \( 5^2 = a^2 + m^2 \). For the lower triangle (hypotenuse \( b \), leg \( a \), segment \( 8 - m \)): \( b^2 = a^2 + (8 - m)^2 \). Also, by geometric mean, \( a^2 = m(8 - m) \) and \( 5^2 = m \times 8 \)? Wait, no, correct approach: The two smaller triangles are similar to the large triangle. Let the altitude divide the vertical side into \( m \) and \( 8 - m \). For the upper triangle (right triangle with hypotenuse 5, one leg \( a \), other \( m \)): \( 5^2 = a^2 + m^2 \). For the lower triangle (hypotenuse \( b \), leg \( a \), other \( 8 - m \)): \( b^2 = a^2 + (8 - m)^2 \). Also, by geometric mean theorem (altitude to hypotenuse in right triangle: \( a^2 = m \times 8 \)? Wait, no, the vertical side is not the hypotenuse. Wait, the large triangle has vertical side 8, and the two smaller triangles are similar. Wait, maybe better: The two triangles (upper and lower) are congruent? No, the left sides are 5 and \( b \). Wait, maybe the vertical side is the hypotenuse? No, the figure shows a right triangle with vertical side 8, altitude \( a \) to it? Wait, maybe I misinterpret. Let's re-express: The triangle has vertical side length 8, split by altitude \( a \) into two segments. The upper triangle has hypotenuse 5, lower has hypotenuse \( b \). By geometric mean, for the upper triangle: \( 5^2 = a^2 + m^2 \), and for the lower: \( b^2 = a^2 + (8 - m)^2 \), and \( a^2 = m(8 - m) \). Also, from upper triangle, \( m = \frac{5^2 - a^2}{1} \), but maybe easier: The length of the altitude \( a \) can be found by geometric mean if the vertical side is the hypotenuse. Wait, no, the vertical side is 8, which is the sum of two segments. Let's assume the vertical side is the hypotenuse? No, the triangles are right-angled at \( a \). So the large triangle is right-angled, with vertical side 8, and the altitude \( a \) to the vertical side? No, the figure shows a triangle with a horizontal altitude \( a \), splitting the vertical side (length 8) into two parts. So the large triangle is right-angled, with legs: horizontal \( a \), vertical 8? No, the left sides are 5 and \( b \), so the large triangle has hypotenuse \( 5 + b \)? No, the figure is two right triangles sharing a common horizontal leg \( a \), with vertical legs summing to 8. So upper triangle: legs \( a \) and \( m \), hypotenuse 5. Lower triangle: legs \( a \) and \( 8 - m \), hypotenuse \( b \). Then, by Pythagoras: \( 5^2 = a^2 + m^2 \) and \( b^2 = a^2 + (8 - m)^2 \). Also, the two triangles are similar to each other and to the large triangle (if we consider the large triangle as combining them). So \( \frac{5}{8} = \frac{m}{5} \) (similar triangles: upper ~ large), so \( m = \frac{25}{8} = 3.125 \). Then \( a^2 = 5^2 - m^2 = 25 - \frac{625}{64} = \frac{1600 - 625}{64} = \frac{975}{64} \), \( a = \frac{\sqrt{975}}{8} = \frac{5\sqrt{39}}{8} \approx 3.90 \). Then \( 8 - m = 8 - \frac{25}{8} = \frac{39}{8} = 4.875 \), so \( b^2 = a^2 + (8 - m)^2 = \frac{975}{64} + \frac{1521}{64} = \frac{2496}{64} = 39 \), so \( b = \sqrt{39} \approx 6.24 \).

Answer:

For Problem 3: \( x = 13 \), \( y = \frac{60}{13} \approx 4.62 \)
For Problem 4: \( a = \frac{5\sqrt{39}}{8} \approx 3.90 \), \( b = \sqrt{39} \approx 6.24 \) (or exact values: \( a = \frac{5\sqrt{39}}{8} \), \( b = \sqrt{39} \), \( m = \frac{25}{8} \))