Question

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

Answer 1

Answer:

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

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

Skills required: HS Geo, Special Triangles

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