Chapter 11.2, Problem 31WE

(a)

To determine

To Find: The area of triangle in terms of $aandb$ .

Answer to Problem 31WE

  $A=\frac{ab}{2}$

Explanation of Solution

Given:

  

  

Area of triangle $ABC$

  $\begin{array}{l}b=a\\ h=b\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(a)(b)\\ A=\frac{ab}{2}\end{array}$

Hence, $A=\frac{ab}{2}$

(b)

To determine

To Find: The area of triangle in terms of $candh$ .

Answer to Problem 31WE

  $A=\frac{ch}{2}$

Explanation of Solution

Given:

  

  

Area of triangle $ABC$

  $\begin{array}{l}b=c\\ h=h\\ A=\frac{1}{2}bh\\ A=\frac{1}{2}(c)(h)\\ A=\frac{ch}{2}\end{array}$

Hence, $A=\frac{ch}{2}$

(c)

To determine

To Solve: $h$ in terms of other variables.

Answer to Problem 31WE

  $h=\frac{ab}{c}$

Explanation of Solution

Given:

  

  

Equating the area of triangle

  $\begin{array}{l}A=\frac{ab}{2}\\ A=\frac{ch}{2}\\ ab=ch\\ h=\frac{ab}{c}\end{array}$

Hence, $h=\frac{ab}{c}$

(d)

To determine

To Find: The length of altitude and median of the hypotenuse.

Answer to Problem 31WE

  $4.8$

Explanation of Solution

Given: A right triangle has leg $6and8$

  

  

Using Pythagoras theorem

  $\begin{array}{l}{a}^2+b^2=c^2\\ 6^2+8^2=c^2\\ c^2=100\\ c=10\end{array}$

Substitute $a=6,b=8,c=10$ in $h=\frac{ab}{c}$

We get

  $\begin{array}{l}h=\frac{6(8)}{10}\\ h=4.8\end{array}$

Hence, the length of the hypotenuse is $4.8$