Question 1151122
The base of a triangle is {{{x}}} cm long and its altitude is quarter={{{1/4}}} the length of its base. 
then the length of its altitude is {{{a=(1/4)x}}} 

If the triangle has an area of {{{12 cm^2}}}, from an equation in {{{x}}} and solve it. 

{{{A=(1/2)base*altitude}}}

{{{12=(1/2)x*(1/4)x}}}

{{{12=(1/8)x^2}}}

{{{12*8=x^2}}}

{{{96=x^2}}}

{{{x=sqrt(96)}}}

{{{x=4sqrt(6)‭cm‬}}}-> exact solution

{{{x=9.8‭cm‬}}}-> approximate solution

 the altitude of the triangle is:

{{{a=(1/4)x}}}=>{{{a=(1/4)4sqrt(6)}}}=>{{{a=sqrt(6)}}}-> exact solution





2.

The length of a parallelogram is {{{d}}} cm and its altitude is {{{3}}} cm shorter. 

altitude is {{{d-3}}}

The side of a square is shorter by a further {{{2}}} cm=> The side of a square is {{{d-5}}}

If the sum of the areas of the parallelogram and square is {{{95}}}cm^2, prove the {{{2d^2 - 13d -70=0}}}:

{{{(d-5)^2+d(d-3)=95}}}

{{{d^2-10d+25+d^2-3d=95}}}

{{{2d^2-13d+25-95=0}}}

{{{2d^2-13d-70=0}}}...to find {{{d}}} use quadratic formula



{{{d=(-(-13)+-sqrt((-13)^2-4*2*(-70)))/(2*2)}}}

{{{d=(13+-sqrt(169+560))/4}}}


{{{d=(13+-sqrt(729))/4}}}

{{{d=(13+-27)/4}}}-> since looking for the length oh the parallelogram, disregard negative solution

{{{d=(13+27)/4}}}

{{{d=40/4}}}

{{{d=10}}} 

 find the length oh the parallelogram is {{{10}}} centimeters 

check the areas:
length of a parallelogram is {{{d=10}}} cm and its altitude is {{{d-3=10-3=7}}}cm

{{{d(d-3)=10*7=70}}}

The side of a square is {{{d-5=10-5=5}}} cm, and area is 
{{{(d-5)^2=(10-5)^2=5^2=25}}}

the sum of the areas of the parallelogram and square is {{{70+25=95}}} which confirms our solution