Math Problem Statement

DO NOT USE TRIG

Solve the problems below. Please answer with completely simplified exact value(s) or expression(s): Given ΔABC, ∠ACB = 90°, CD ⊥ AB, ∠ACD = 60°, BC = 6 cm. Find: CD, Area of ΔABC. DO NOT USE TRIG.

Solution

Solution:

We are given the following:

  1. ABC\triangle ABC is a right triangle with ACB=90\angle ACB = 90^\circ,
  2. CDABCD \perp AB,
  3. ACD=60\angle ACD = 60^\circ,
  4. BC=6cmBC = 6 \, \text{cm}.

We need to find:

  1. The length of CDCD,
  2. The area of ABC\triangle ABC.

Step 1: Relationship Between Angles and Sides

Observations:

  1. Since ACB\triangle ACB is a right triangle and ACB=90\angle ACB = 90^\circ, ABAB is the hypotenuse.
  2. In ACD\triangle ACD, ACD=60\angle ACD = 60^\circ, and CAD=30\angle CAD = 30^\circ because the angles of a triangle sum to 180180^\circ.

From the geometry of 3030^\circ-6060^\circ-9090^\circ triangles:

  • The side opposite 3030^\circ (in this case CDCD) is half the hypotenuse of ACD\triangle ACD (which is ACAC).
  • The side opposite 6060^\circ (in this case ADAD) is 3\sqrt{3} times the side opposite 3030^\circ.

Applying the 3030^\circ-6060^\circ-9090^\circ Triangle Rule:

Let AC=xAC = x. Then:

  • CD=x2CD = \frac{x}{2},
  • AD=x32AD = \frac{x\sqrt{3}}{2}.

Step 2: Relating BCBC and ABAB Using Pythagoras in ABC\triangle ABC

From ABC\triangle ABC: AB2=AC2+BC2.AB^2 = AC^2 + BC^2. Substituting BC=6BC = 6: AB2=x2+62    AB2=x2+36.AB^2 = x^2 + 6^2 \implies AB^2 = x^2 + 36.

Step 3: Solve for xx Using AB=AD+DBAB = AD + DB

In ACB\triangle ACB, the full hypotenuse ABAB can be expressed as: AB=AD+DB.AB = AD + DB. From Step 1:

  • AD=x32AD = \frac{x\sqrt{3}}{2},
  • DB=6DB = 6 (since DB=BCDB = BC).

Thus: AB=x32+6.AB = \frac{x\sqrt{3}}{2} + 6.

Substitute ABAB into the Pythagorean equation: (x32+6)2=x2+36.\left( \frac{x\sqrt{3}}{2} + 6 \right)^2 = x^2 + 36.

Step 4: Expand and Simplify

Expand (x32+6)2\left( \frac{x\sqrt{3}}{2} + 6 \right)^2: (x32+6)2=(x32)2+2x326+62,\left( \frac{x\sqrt{3}}{2} + 6 \right)^2 = \left( \frac{x\sqrt{3}}{2} \right)^2 + 2 \cdot \frac{x\sqrt{3}}{2} \cdot 6 + 6^2, =3x24+6x3+36.= \frac{3x^2}{4} + 6x\sqrt{3} + 36.

Equating this to x2+36x^2 + 36: 3x24+6x3+36=x2+36.\frac{3x^2}{4} + 6x\sqrt{3} + 36 = x^2 + 36.

Cancel 3636 from both sides: 3x24+6x3=x2.\frac{3x^2}{4} + 6x\sqrt{3} = x^2.

Multiply through by 4 to eliminate the fraction: 3x2+24x3=4x2.3x^2 + 24x\sqrt{3} = 4x^2.

Simplify: x2=24x3.x^2 = 24x\sqrt{3}.

Divide through by xx (since x0x \neq 0): x=243.x = 24\sqrt{3}.

Step 5: Find CDCD and the Area

  1. CD=x2=2432=123cm.CD = \frac{x}{2} = \frac{24\sqrt{3}}{2} = 12\sqrt{3} \, \text{cm}.

  2. Area of ABC\triangle ABC: Area=12ACBC=122436=723cm2.\text{Area} = \frac{1}{2} \cdot AC \cdot BC = \frac{1}{2} \cdot 24\sqrt{3} \cdot 6 = 72\sqrt{3} \, \text{cm}^2.


Final Answers:

  1. CD=123cmCD = 12\sqrt{3} \, \text{cm},
  2. Area of ABC=723cm2.\triangle ABC = 72\sqrt{3} \, \text{cm}^2.

Do you have any questions or want further clarification?
Here are related questions to expand your understanding:

  1. How do the properties of 3030^\circ-6060^\circ-9090^\circ triangles simplify solving geometric problems?
  2. What are the conditions for a perpendicular dropped from a vertex in a triangle?
  3. How does the Pythagorean Theorem apply in solving problems involving multiple triangles?
  4. What other methods could be used to find the area of a triangle besides base-height relationships?
  5. Can this solution be adapted for a triangle with different angles (not 3030^\circ-6060^\circ-9090^\circ)?

Tip: Always simplify triangle problems by analyzing smaller triangles within the given shape!

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Math Problem Analysis

Mathematical Concepts

Geometry
Right Triangles
Special Triangles
30-60-90 Triangle Properties

Formulas

Area of a triangle = (1/2) × base × height
Properties of 30-60-90 triangles

Theorems

Pythagorean Theorem
Special Right Triangle Theorem (30°-60°-90°)

Suitable Grade Level

Grade 9-11