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erik [133]
3 years ago
15

Given: △ABC, m∠A=60°, m∠C=45°, AB=9 Find: Perimeter of △ABC, Area of △ABC

Mathematics
1 answer:
stellarik [79]3 years ago
6 0

Answer:

Perimeter of ΔABC: \frac{27}{2} + \frac{9}{2} * \sqrt6 units

Area of ΔABC: \frac{81}{8}*\sqrt3 + \frac{243}{8} units

<u>Skills required: HS Geo, Special Triangles</u>

Step-by-step explanation:

1) The best option is to break down this triangle. Let's draw an altitude from Point B down to Segment AC. The point from the altitude that intersects AC is Point D. BD is the height of our triangle, AC is the base.

2) Angle A is 60 degrees, and since Angle BDA is 90 degrees, Angle ABD is 30 degrees. We can use the 30-60-90 degree right triangle property for the triangle BDA.

  • This states that if the side opposite the 30 degree angle is x, the side opposite the 60 degree angle is x*\sqrt3, and the side opposite the 90 degree angle is 2x.

AB is 9 units, and it is opposite the 90 degree angle. This means that 2x=9, x = \frac{9}{2} ==> This then means that AD, the segment opposite the 30 degree angle in this triangle is \frac{9}{2} units. Segment BD (the height) is \frac{9}{2} * \sqrt3.

3) Angle C is 45 degrees, and Angle BDC is 90 degrees, which means that Angle CBD is 45 degrees. We can use the 45-45-90 degree right triangle property for the triangle BCD.

  • This states that if the side opposite the 45 degree angle is x, the other side opposite a 45 degree angle is also x, but the hypotenuse (side opposite the right (90 degree) angle) is \sqrt{2}*x.

BD is \frac{9}{2} * \sqrt3, which means DC is the same. BC, which is the hypotenuse is BD multiplied by square-root-2, which is \frac{9}{2} * \sqrt6.

4) Area is \frac{1}{2}*b*h, the base (b) is AC (which is \frac{9}{2}+\frac{9}{2}*\sqrt3), the height is BD (\frac{9}{2}*\sqrt3). When multiple you will get \frac{81}{4}*\sqrt3 + \frac{243}{4}, then this multiplied by 1/2 is

\frac{81}{8}*\sqrt3 + \frac{243}{8} <--> this is the area!

5) Perimeter is just the sum of all side: 9 + \frac{9}{2} + \frac{9}{2} * \sqrt6 = \frac{27}{2} + \frac{9}{2} * \sqrt6 unit

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