✏️TRIANGLE
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Problem: The length of the base of a triangle is three times its height. The area of a triangle is 24cm². Find the base and the height of the triangle.
Solution: Represent b and h as the base and the height of a triangle respectively. Make two equations of the given statement.
- [tex] \begin{cases}b = 3h& \green{(eq. \: 1)} \\ \begin{gathered}24 = \frac{1}{2}bh \end{gathered} & \green{(eq. \: 2)}\end{cases}[/tex]
- Substitute b from the first equation to the second equation in terms of h.
- [tex] \begin{cases}b = 3h\\ \begin{gathered}24 = \frac{1}{2}(3h)h \end{gathered}\end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ \begin{gathered}24 = \frac{1}{2}( {3h}^{2}) \end{gathered}\end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ \begin{gathered}24 = \frac{ {3h}^{2} }{2} \end{gathered}\end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ \begin{gathered}24 \cdot2 = \frac{ {3h}^{2} }{ \cancel2} \cdot \cancel2 \end{gathered}\end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ 24 \cdot2 = {3h}^{2} \end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ 48 = {3h}^{2} \end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ \begin{gathered} \frac{48}{3} = \frac{ { \cancel3h}^{2} }{ \cancel3} \end{gathered}\end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ {h}^{2} = 16 \end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ \sqrt{ {h}^{2} } = \sqrt{16} \end{cases}[/tex]
- [tex] \begin{cases}b = 3h\\ h = 4 \end{cases}[/tex]
- The height of a triangle is 4 cm, Substitute it to the first equation to find the base of a triangle.
- [tex] \begin{cases}b = 3(4)\\ h = 4 \end{cases}[/tex]
- [tex] \begin{cases}b = 12\\ h = 4 \end{cases}[/tex]
- Therefore, the base and the height of a triangle are:
- [tex] \large \rm Base = \boxed{ \rm \green{ \: 12 \: cm \: }}[/tex]
- [tex] \large \rm Height = \boxed{ \rm \green{ \: \: 4 \: cm \: \: }}[/tex]
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