Question

Given: AABC, m_ACB=90°, CD 1 AB, m ZACD=60º, BC = 6 cm Find: CD, Area of AABC

Answer 1

Answer:

Look below

Step-by-step explanation:

Given that CDB is 90 degrees, ACB is 90 degrees, and ACD is 60 degrees, we can determine that DCB = 90-60 = 30 degrees.

This means triangle BCD is a 30-60-90 (angle measures) right triangle

The proportions of the sides (from smallest to largest) is

x:x√3:2x

We are given that BC = 6 cm. This means...

2x=6

x=3

This means DB is 3 cm and CD is 3√3 cm

Using the linear pair theorem, we can find that Angle CDA is 90 degrees. This means ACD is also a 30-60-90 triangle.

x=3√3

x√3=9

2x=6√3

Now we need to find AB

AB = AD + DB

AB = 9 + 3

AB = 12 cm

$Area = \frac{1}{2} bh = \frac{1}{2}(12)(3\sqrt3)=6(3\sqrt3)=18\sqrt3$