Solving for a+b+c in $\triangle ABC$

[Albert1]

$\triangle ABC$ (with side length $a,b,c$)

given :

$(1)\angle A=60^o$

$(2)$ the area of $\triangle ABC=10\sqrt 3$

$(3) a^2+b^2+c^2=138$

please find :$a+b+c=?$

 

[I like Serena]

Re: find a+b+c

$\triangle ABC$ (with side length $a,b,c$)

given :

$(1)\angle A=60^o$

$(2)$ the area of $\triangle ABC=10\sqrt 3$

$(3) a^2+b^2+c^2=138$

please find :$a+b+c=?$

Spoiler

Area of $\triangle ABC = \frac 1 2 b c \sin 60^\circ = 10\sqrt 3$.
So:
$$bc = 40 \qquad \qquad \qquad \qquad [1]$$

Cosine rule, using [1]:
$$a^2 = b^2+c^2 - 2bc \cos 60^\circ$$
$$a^2 = b^2+c^2 - 2\cdot 40 \cdot \frac 1 2$$
$$b^2+c^2 = a^2 + 40 \qquad \qquad [2]$$

From the given statement with [2]:
$$a^2+b^2+c^2=138$$
$$a^2+(a^2+40)=138$$
$$a=7$$

Back substituting in [2]:
$$b^2+c^2 = 7^2 + 40 = 89 \qquad [3]$$

Note that with [1] and [3]:
$$(b+c)^2 = b^2 + 2bc + c^2 = 89 + 2 \cdot 40 = 169$$

It follows that:
$$b+c = 13$$
And therefore:
$$a+b+c = 7 + 13 = 20$$