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9966 [12]
3 years ago
12

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

Mathematics
1 answer:
just olya [345]3 years ago
8 0

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

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Yes. No graph paper.

Step-by-step explanation:

Let's say the 3 points are A, B, C.

If A, B and C lie on one line then m_{AB} = m_{BC} = m_{AC}

m_{AB} = \frac{y_{B}-y_{A} }{x_{B}-x_{A}} = \frac{2-6}{3-5} = \frac{-4}{-2} = 2\\m_{BC} = \frac{y_{C}-y_{B} }{x_{C}-x_{B}} = \frac{8-2}{6-3} = \frac{6}{3} = 2\\\\m_{AC} = \frac{y_{C}-y_{A} }{x_{C}-x_{A}} = \frac{8-6}{6-5} = \frac{2}{1} = 2\\\\

Hence, they lie on one line.

You don't need to use a graph paper to prove it.

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3 years ago
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