Question

The height of a triangle is three more than two times the base. The area of the triangle is 45 square feet. What are the dimensions of the triangle? (solved by factoring)

Answer 1

SOLUTION

Given the question in the tab, the following are the solution steps to answer the question.

STEP 1: Write the given data

$\begin{gathered} Area=45ft^2 \\ 2\text{ times the base \lparen b\rparen}\Rightarrow2b \\ three\text{ more than two times the base \lparen b\rparen}\Rightarrow2b+3 \\ height(h)=2b+3 \end{gathered}$

STEP 2: Write the formula for the are of a triangle

$Area=\frac{1}{2}\times base\times height$

STEP 3: Substitute the given values

$\begin{gathered} Area=\frac{1}{2}\times b\times h=45ft^2 \\ By\text{ substitution,} \\ \frac{1}{2}\times b\times h=\frac{bh}{2} \\ Recall\text{ from step 1 that:} \\ h=2b+3 \\ By\text{ substitution,} \\ Area=\frac{b(2b+3)}{2}=45 \end{gathered}$

STEP 4: Cross multiply

$\begin{gathered} b(2b+3)=2\times45 \\ b(2b+3)=90 \end{gathered}$

Open the bracket

$2b^2+3b=90$

Subtract 90 from both sides

$2b^2+3b-90=0$

STEP 5: Solve the derived equations using quadratic formula

$\begin{gathered} b_1,b_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ a=2,b=3,c=-90 \\ By\text{ substitution,} \\ b_{1,\:2}=\frac{-3\pm \sqrt{3^2-4\cdot \:2\left(-90\right)}}{2\cdot \:2} \\ \sqrt{3^2-4\times2(-90)}=\sqrt{9+720}=\sqrt{729}=27 \\ \\ \text{By substitution, we have:} \\ b_{1,\:2}=\frac{-3\pm \:27}{2\cdot \:2} \\ \mathrm{Separate\:the\:solutions} \\ b_1=\frac{-3+27}{2\cdot\:2}=\frac{24}{4}=6 \\ b_2=\frac{-3-27}{2\cdot\:2}=\frac{-30}{4}=-7.5 \\ \\ \therefore base=-7.5,6 \end{gathered}$

Since the value of the base of a triangle can not be negative, the base of the triangle is 6ft

STEP 6: Find the value of the height

$\begin{gathered} From\text{ step 3} \\ h=2b+3 \\ b=6ft \\ h=2(6)+3=12+3=15ft \end{gathered}$

Hence, the dimension of the triangle are:

base = 6ft

height = 15ft