Question

Solve the problems below. Please answer with completely simplified exact value(s) or expression(s). Given: ΔАВС, m∠ACB = 90° CD ⊥ AB , m∠ACD = 60°,BC = 6 cm Find CD, Area of ΔABC
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Answer 1

Answer:

$CD=3\sqrt{3}\ cm,\\ \\A_{ABC}=18\sqrt{3}\ cm^2$

Step-by-step explanation:

Triangle ABC is right triangle. Since  m∠ACB = 90°  and m∠ACD = 60°, you get m∠BCD = 90°-60°=30°.

Triangle BCD is rigth triangle, then the leg that is opposite to the angle of 30° is half of the hypotenuse. Thus,

$BD=\dfrac{1}{2}BC=\dfrac{1}{2}\cdot 6=6\ cm.$

By the Pythagorean theorem,

$CD^2=BC^2-BD^2=6^2-3^2=36-9=27,\\ \\CD=3\sqrt{3}\ cm.$

Then

$CD^2=AD\cdot BD,\\ \\27=3\cdot AD,\\ \\AD=9\cm.$

Hypotenuse AB of the triangle ABC is equal to

$AB=AD+BD=9+3=12\ cm.$

The area of the triangle ABC is

$A_{ABC}=\dfrac{1}{2}\cdot AB\cdot CD=\dfrac{1}{2}\cdot 12\cdot 3\sqrt{3}=18\sqrt{3}\ cm^2.$