Question 176017
Let b=base and h=height


"The base of a triangle is 8cm greater than the height" means that {{{b=h+8}}}



{{{A=(1/2)bh}}} Start with the area of a triangle formula



{{{42=(1/2)(h+8)h}}} Plug in {{{A=42}}} (the given area) and {{{b=h+8}}}



{{{42*2=(h+8)h}}} Multiply both sides by 2.



{{{84=(h+8)h}}} Multiply



{{{84=h(h+8)}}} Rearrange the terms.



{{{84=h^2+8h}}} Distribute



{{{0=h^2+8h-84}}} Subtract 84 from both sides.



Notice we have a quadratic equation in the form of {{{ah^2+bh+c}}} where {{{a=1}}}, {{{b=8}}}, and {{{c=-84}}}



Let's use the quadratic formula to solve for h



{{{h = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{h = (-(8) +- sqrt( (8)^2-4(1)(-84) ))/(2(1))}}} Plug in  {{{a=1}}}, {{{b=8}}}, and {{{c=-84}}}



{{{h = (-8 +- sqrt( 64-4(1)(-84) ))/(2(1))}}} Square {{{8}}} to get {{{64}}}. 



{{{h = (-8 +- sqrt( 64--336 ))/(2(1))}}} Multiply {{{4(1)(-84)}}} to get {{{-336}}}



{{{h = (-8 +- sqrt( 64+336 ))/(2(1))}}} Rewrite {{{sqrt(64--336)}}} as {{{sqrt(64+336)}}}



{{{h = (-8 +- sqrt( 400 ))/(2(1))}}} Add {{{64}}} to {{{336}}} to get {{{400}}}



{{{h = (-8 +- sqrt( 400 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{h = (-8 +- 20)/(2)}}} Take the square root of {{{400}}} to get {{{20}}}. 



{{{h = (-8 + 20)/(2)}}} or {{{h = (-8 - 20)/(2)}}} Break up the expression. 



{{{h = (12)/(2)}}} or {{{h =  (-28)/(2)}}} Combine like terms. 



{{{h = 6}}} or {{{h = -14}}} Simplify. 



So the possible answers are {{{h = 6}}} or {{{h = -14}}} 



However, since a negative height is not possible, this means that {{{h = -14}}} is NOT a solution



So the only answer for the height is {{{h = 6}}} which means that the height is 6 cm.



{{{b=h+8}}} Go back to the first equation



{{{b=6+8}}} Plug in {{{h=6}}}



{{{b=14}}} Add. So the base is 14 cm.



====================================


Answer:



So the base is 14 cm and the height is 6 cm