have rt. triangle, ABD.

Question

                                             Need x which is a segment of line BCD.         Rt.   Triangle   ABD,  Altitude=21, Base=28.  Hypoteenuse is 35 with a section of it (BC=x).                                     

                                              

 

Answer 1

Do you have an image of the problem? Otherwise, here is my best interpretation:

We have a right triangle ABD with the right angle at A. The side lengths are given as: base AB = 28 units, vertical leg AD = 21 units, and hypotenuse BD = 35 units. (Notice these satisfy 212+282=352 or 21^2 + 28^2 = 35^2212+282=352, confirming a right triangle by the Pythagorean theorem.) Point C is defined on the hypotenuse BD such that C is the foot of the perpendicular from A onto BD. In other words, AC is an altitude dropped from the right angle to the hypotenuse, meeting BD at point C. This splits the hypotenuse into two segments: BC (from B to C) and CD (from C to D). Our goal is to determine the length of BC.

Using Similar Triangles (Geometric Projection Method)

When we draw the altitude AC to the hypotenuse, it creates two smaller right triangles ABC and ADC inside the original triangle. Importantly, △ABC and △ADC are each similar to the original △ABD (by AA similarity, since they all share acute angles). This means the ratios of corresponding sides are equal. In particular, focusing on triangle ABC (which is similar to ABD):

1. Angle B is common to both △ABD and △ABC.
2. ∠A in △ABD is 90°, and ∠C in △ABC is also 90° (since AC ⟂ BD by construction).
3. Therefore, ∠D in △ABD corresponds to ∠A in △ABC.

From this similarity, we can set up a proportion involving side lengths of ABD and ABC. Let BCBCBC be the projection of leg AB onto the hypotenuse (segment BC lies adjacent to AB on BD). The similarity gives:

ABBD=BCAB.\frac{AB}{BD} = \frac{BC}{AB}.BDAB​=ABBC​.

This proportion states that the ratio of AB to the full hypotenuse BD equals the ratio of BC (the segment of the hypotenuse adjacent to AB) to AB itself. Cross-multiplying, we get:

AB2=(BD)⋅(BC).AB^2 = (BD) \cdot (BC).AB2=(BD)⋅(BC).

In words, “the leg of a right triangle is the mean proportional between the hypotenuse and the projection of that leg on the hypotenuse." -Leg Geometric Theorem

Using the given values for the triangle:

1. AB=28AB = 28AB=28 and BD=35BD = 35BD=35.

Plug these into the relation AB2=BD⋅BCAB^2 = BD \cdot BCAB2=BD⋅BC:

282=35⋅BC.28^2 = 35 \cdot BC.282=35⋅BC.

This gives:

784=35⋅BC,784 = 35 \cdot BC,784=35⋅BC,

so solving for BCBCBC:

BC=78435=784÷735÷7=1125=22.4 units.BC = \frac{784}{35} = \frac{784 \div 7}{35 \div 7} = \frac{112}{5} = 22.4~\text{units}.BC=35784​=35÷7784÷7​=5112​=22.4 units.

Thus, the length of segment BC is 22.4 units.