Question 1044245

The base of a triangle is 4 dm longer than the altitude, and its area is 16 dm^2. Find the lengths of the base and the altitude.
<pre>Let altitude be A
Then base = A + 4
We then get: {{{(1/2) * (A + 4)A = 16}}}
{{{(1/2) * (A^2 + 4A) = 16}}}
{{{(A^2 + 4A)/2 = 16}}}
{{{A^2 + 4A = 32}}} ------ Cross-multiplying 
{{{A^2 + 4A - 32 = 0}}}
(A - 4)(A + 8) = 0
{{{highlight_green(matrix(1,9, A, or, altitude = 4, dm, and, Base = 4 + 4, or, 8, dm))}}}