Question 1129377
 the triangle {{{ABC}}} has sides:
 
{{{AB=22cm}}}
{{{AC=8cm}}}
and {{{A=34.98cm^2}}}


 If the area of the triangle {{{A=34.98cm^2}}} and {{{AB=22cm}}}, we have


{{{34.98cm^cross(2)=(22cross(cm)*h)/2}}}

{{{34.98cm=11*h}}}

{{{34.98cm/11=h}}}

{{{h=3.18cm}}}


find {{{BC}}}-> note that  {{{h}}} divides triangle in two right triangles and side {{{AB=22cm}}} in two parts {{{x}}} and {{{y}}}


since we know {{{AC=8}}}, {{{h=3.18cm}}}, 
use Pythagorean theorem to find {{{x}}}
 {{{x=sqrt(8^2-3.18^2)}}}
 {{{x=7.34}}}

then,  {{{y=22-7.34}}}->{{{y=14.66}}}

now use Pythagorean theorem  to find {{{BC}}}

{{{y=14.66}}}, and {{{h=3.18}}}

 {{{BC=sqrt(14.66^2+3.18^2)}}}

{{{BC=15.0009)}}}

{{{BC=15cm}}}



what is it's perimeter? 


{{{P=AB+AC+BC}}}

{{{P=22cm+8cm+15cm}}}

{{{P=45cm}}}


or, this way:


Use the formula 


    {{{Area =(1/2)*a*b*sin(alpha)}}} for the area of a triangle, 


where "a" and "b" are two sides lengths and  {{{alpha}}}  is the angle between them.


Using given data, find {{{sin(alpha)}}}.

 {{{34.98=(1/2)*22*8*sin(alpha)}}}

{{{34.98=88*sin(alpha)}}}

{{{sin(alpha)=34.98/88}}}
{{{sin(alpha)=0.3975}}}
{{{alpha=sin^-1(0.3975)}}}
{{{alpha=sin^-1(0.3975)}}}
{{{alpha=23.42}}}°

Having  {{{(alpha)}}},  find {{{cos(alpha)}}}.
{{{cos(23.42)=0.917616}}}

Then find the length of the third side using  the cosine law.
 {{{c^2 = a^2 + b^2 − 2ab cos(alpha)}}}
 {{{c^2 = 22^2 + 8^2 − 2*22*8*0.917616}}}
 {{{c^2 = 224.999168}}}
 {{{c = sqrt(225)}}}
 {{{c = 15}}}

Then find the perimeter,

{{{P=a+b+c}}}

{{{P=22cm+8cm+15cm}}}

{{{P=45cm}}}