1) Since the height of this triangle is "less than three times the measure of its base" We can call it simply by 3b-6 in which b is the measure of the base.
2) The area is 36 in² so let's plug both pieces of information into the area of a triangle formula and solve it:
[tex]\begin{gathered} A_{\Delta}=\frac{b\cdot h}{2} \\ 36=\frac{b(3b-6)}{2} \\ 2\times36=\frac{b(3b-6)}{2}\times2 \\ 72=3b^2-6b \\ 0=3b^2-6b-72 \\ b_=\frac{-\left(-6\right)\pm\sqrt{\left(-6\right)^2-4\cdot\:3\left(-72\right)}}{2\cdot\:3} \\ b_1=\frac{-\left(-6\right)+30}{2\cdot\:3}=\frac{30+6}{6}=6 \\ b_2=\frac{-\left(-6\right)-30}{2\cdot\:3}=\frac{-30+6}{6}=-4 \end{gathered}[/tex]We can discard negative 4 as a measurement for there are no negative measurements. So, we can tell the length of the base is 6 inches
3) Now, let's plug it back into the area formula so that we can get to know the measurement of the height:
[tex]\begin{gathered} A_{\Delta}=\frac{bh}{2} \\ 36=\frac{6h}{2} \\ 72=6h \\ 6h=72 \\ 6h=72 \\ \frac{6h}{6}=\frac{72}{6} \\ h=12 \end{gathered}[/tex]Thus, the dimensions of that triangle are:
[tex]b=6",h=12"[/tex]
