Question 165697
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I'll leave the sketch with you okay, & hopefully we can follow together:
We find first the {{{Area=(1/2)bh}}}
where, {{{BC=base=9cm}}}, & {{{AD=height=6cm}}}
So, {{{A=(1/2)(9)(6)}}}
{{{A=27cm^2}}}
Now , next is interesting in finding CE?
We first get {{{angle(B)}}} by trigo function of right triangle. How?
Since AD split the base(BC) in to half, {{{side(BD)=9/2=4.5cm}}} right? Same as {{{side(DC)}}}.
That forms a right triangle (ADB). 
{{{sin(beta)=opp/hyp=AD/AB=6/7.5=0.80}}}
{{{(beta)=sin^-1(0.80)}}}
{{{highlight((beta)=53.13^o)}}}
.
Now we know line CE(?) cuts line AB into half:{{{7.5/2=3.75cm=BE}}}, right?
That forms a triangle {{{BCE}}} and we'll use Cosine Law in getting line CE.
{{{c^2=a^2+b^2-2abcos(beta)}}}
where ---{{{system(a=BC=9cm,b=BE=3.75cm,c=CE)}}}
Continuing,
{{{c^2=9^2+3.75^2-2(9)(3.75)cos(53.13)}}}
{{{c^2=81+14.0625-40.50=95.0625-40.50}}}
{{{c=sqrt(54.5625)}}}
{{{highlight(c=7.38cm=CE)}}}
Thank you,
Jojo</pre>