Question

2 - 53. the area of the triangle at right is 25 square feet. determine the value of h. then calculate the perimeter of the entire triangle. show all work.

Explanation:

Step1: Use area formula to find h

The area formula for a triangle is $A=\frac{1}{2}bh$. Here, $b = 6$ feet and $A = 25$ square - feet. Substitute into the formula: $25=\frac{1}{2}\times6\times h$.
Solve for $h$:
\[

$$\begin{align*} 25&=\frac{1}{2}\times6\times h\\ 25& = 3h\\ h&=\frac{25}{3}\text{ feet} \end{align*}$$

\]

Step2: Use the Pythagorean theorem to find the two non - given sides

Let the base of the right - triangle formed by the height be $x$ and $y$, where $x + y=6$.
Using the Pythagorean theorem, for the two right - triangles formed by the height:
Let the two non - given sides of the large triangle be $a$ and $b$.
\[a=\sqrt{h^{2}+x^{2}}\] and \[b=\sqrt{h^{2}+y^{2}}\]
First, consider the right - triangle with base $x$ and height $h=\frac{25}{3}$ and the other with base $y$ and height $h=\frac{25}{3}$.
Let's assume one of the sub - right - triangles has base $x$ and the other has base $y$ such that $x + y = 6$.
For the right - triangle with base $x$ and height $h=\frac{25}{3}$, if we assume $x = 4$ (from the figure's hint of a split base), then the length of one non - given side $s_1$:
\[s_1=\sqrt{(\frac{25}{3})^{2}+4^{2}}=\sqrt{\frac{625}{9}+16}=\sqrt{\frac{625 + 144}{9}}=\sqrt{\frac{769}{9}}=\frac{\sqrt{769}}{3}\]
For the other sub - right - triangle with base $y=6 - 4 = 2$ and height $h=\frac{25}{3}$, the length of the other non - given side $s_2$:
\[s_2=\sqrt{(\frac{25}{3})^{2}+2^{2}}=\sqrt{\frac{625}{9}+4}=\sqrt{\frac{625+36}{9}}=\sqrt{\frac{661}{9}}=\frac{\sqrt{661}}{3}\]

Step3: Calculate the perimeter

The perimeter $P$ of the triangle is $P=s_1 + s_2+6$
\[P=\frac{\sqrt{769}}{3}+\frac{\sqrt{661}}{3}+6=\frac{\sqrt{769}+\sqrt{661}+18}{3}\text{ feet}\]

Answer:

$h=\frac{25}{3}$ feet, Perimeter $=\frac{\sqrt{769}+\sqrt{661}+18}{3}$ feet