Given △ABC, AB=15, BD=9, AD ⊥ BC , m∠C=30° Find: The perimeter of △ABC.

Mathematics

1 answer:

Ann [662]3 years ago

7 0

Answer:

48 + 12√3 units

Step-by-step explanation:

ALL LENGTHS MUST BE POSITIVE (I did this to avoid writing ± and showing that only + works)

1. Find AD (you'll find why later): 15² = 9² + AD² --> AD² = 144 --> AD = 12

2. 30-60-90: 12-DC-AC --> AC = 24, DC = 12√3 (side opposite of 90 is double side opposite of 30, side opposite of 60 is √3 times the opposite of 30)

P = AB + BD + DC + AC = 15 + 9 + 12√3 + 24 = 48 + 12√3 units

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Ok, so dy/dx=0 at the point (0,3) that is where x=0 and y=3.

$\int { 6x+6dx } \\ \\ =\frac { 6{ x }^{ 2 } }{ 2 } +6x+C\\ \\ =3{ x }^{ 2 }+6x+C$

$\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x+C$

Now, f'(x)=0 when x=0.

Therefore:

$0=C\\ \\ \therefore \quad { f }^{ ' }\left( x \right) =3{ x }^{ 2 }+6x$

Now:

$\int { 3{ x }^{ 2 } } +6xdx\\ \\ =\frac { 3{ x }^{ 3 } }{ 3 } +\frac { 6x^{ 2 } }{ 2 } +C$

$={ x }^{ 3 }+3{ x }^{ 2 }+C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+C$

But when x=0, y=3, therefore:

$3=C\\ \\ \therefore \quad f\left( x \right) ={ x }^{ 3 }+3{ x }^{ 2 }+3$

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