Question 471943
The height of a triangle {{{h}}} is {{{5}}} more than half the base of the triangle {{{b}}}. 

{{{h=b/2+5}}}


The area {{{A}}} of the triangle is {{{14}}} squared centimeters:


{{{A=14cm^2}}}


also


{{{A=hb/2}}}


so {{{hb/2=14cm^2}}}....replace {{{h}}} with {{{b/2+5}}}


{{{(b/2+5)b/2=14cm^2}}}


{{{((b+10)/2)b/2=14cm^2}}}


{{{(b^2+10b)/4=14cm^2}}}


{{{b^2+10b=14cm^2*4}}}


{{{b^2+10b=56cm^2}}}


{{{b^2+10b-56cm^2=0}}}........use quadratic formula to find {{{b}}}


{{{b = (-10 +- sqrt( 10^2-4*1*(-56) ))/(2*1) }}} 


{{{b = (-10 +- sqrt( 100+224 ))/2 }}}


{{{b = (-10 +- 18)/2 }}} ......use only positive solution because the base cannot be negative


{{{b = (-10 + 18)/2 }}}


{{{b = 8/2 }}}


{{{b = 4cm}}}


now find {{{h}}}


{{{h=b/2+5}}}


{{{h=4/2+5}}}


{{{h=2+5}}}


{{{h=7cm}}}



 
check:

{{{A=hb/2}}}


{{{A=7cm*4cm/2}}}


{{{A=7cm*2cm}}}


{{{A=14cm^2}}}