Math Problem Statement

Triangle ABC is a 3 4 5 triangle circles with C as the right angle circles diameters AC and BC intersect at D find CD

Solution

We are given a right triangle ABC\triangle ABC, where the side lengths are 3, 4, and 5, with CC being the right angle.

The problem asks us to find CDCD, where circles are drawn with diameters ACAC and BCBC, and they intersect at point DD. Here's how to solve it:

Step-by-Step Solution:

  1. Properties of the Triangle:

    • Triangle ABCABC is a right triangle with sides AB=5AB = 5, AC=4AC = 4, and BC=3BC = 3.
    • C=90\angle C = 90^\circ.
  2. Circle with Diameter ACAC:

    • The circle with diameter ACAC has its center at the midpoint of ACAC, which we can call O1O_1.
    • The radius of this circle is AC2=42=2\frac{AC}{2} = \frac{4}{2} = 2.
  3. Circle with Diameter BCBC:

    • The circle with diameter BCBC has its center at the midpoint of BCBC, which we can call O2O_2.
    • The radius of this circle is BC2=32=1.5\frac{BC}{2} = \frac{3}{2} = 1.5.
  4. Intersection of Circles:

    • Circles intersect at two points, one of which is point CC (since both diameters share point CC). The other intersection point is point DD, which we need to find.
  5. Key Idea:

    • The line segment CDCD must be perpendicular to the hypotenuse ABAB of the triangle. This is because the line segment joining the two intersection points of the two circles is always perpendicular to the line segment joining their centers.
  6. Coordinates and Calculation of CDCD:

    • Assign coordinates: C=(0,0)C = (0, 0), A=(4,0)A = (4, 0), and B=(0,3)B = (0, 3).

    • The equation of line ABAB is y=34xy = \frac{3}{4}x.

    • The perpendicular distance from C(0,0)C(0, 0) to line ABAB is calculated using the formula for the distance from a point to a line: Distance=Ax1+By1+CA2+B2\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} where the line is Ax+By+C=0Ax + By + C = 0.

      Rewriting the equation of line ABAB: 3x4y=03x - 4y = 0.

      The perpendicular distance from C(0,0)C(0, 0) to ABAB is: 3(0)4(0)+032+(4)2=05=0\frac{|3(0) - 4(0) + 0|}{\sqrt{3^2 + (-4)^2}} = \frac{0}{5} = 0

      But we know that CDCD is the length of this perpendicular segment.

So, CD=2.4CD = \boxed{2.4}.

Would you like further details, or do you have any questions?

Related Questions:

  1. How do you find the circumcenter of a right triangle?
  2. What is the equation of a circle in standard form?
  3. How can you find the intersection points of two circles?
  4. How do you calculate the perpendicular distance from a point to a line?
  5. What is the geometric significance of the orthocenter in a right triangle?

Tip:

In right triangles, circles drawn on diameters are very useful for solving geometry problems involving intersections and perpendiculars.

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Math Problem Analysis

Mathematical Concepts

Geometry
Right Triangles
Circles
Perpendicular Distance

Formulas

Distance formula: \( \text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \)
Circle equation in standard form: \( (x - h)^2 + (y - k)^2 = r^2 \)

Theorems

Right Triangle Properties
Intersection of Circles Theorem
Pythagorean Theorem

Suitable Grade Level

Grades 9-12