Chapter 1, Problem 2P

To determine

To express: The length of the hypotenuse as a function of the perimeter.

Answer to Problem 2P

Solution:

The length of the hypotenuse is $h=\frac{P^2}{2P+24}$.

Explanation of Solution

Given:

The altitude perpendicular to the hypotenuse $a=12$ cm

Formula used:

$\text{Area of the triangle}$$A=\frac{1}{2}\left(x\text{y}\right)$ (1)

Pythagoras theorem $x^2+y^2=h^2$ (2)

Where, h is a hypotenuse, and x, y is the sides of triangle

Perimeter of the triangle $P=x+y+h$ (3)

${\left(x+y\right)}^2=x^2+y^2+2xy$

Calculation:

Consider the triangle as shown below in the Figure 1.

From the Figure 1, $A=\frac{1}{2}\left(ah\right)$

Substitute $a=12$ in above equation,

$\begin{array}{c}A\text{= }\frac{1}{2}\left(\text{12h}\right)\\ =6h\end{array}$

From the equation (1),

$A=\frac{1}{2}\left(xy\right)$

This implies $xy=2A$ (4)

Consider the equation ${\left(x+y\right)}^2=x^2+y^2+2xy$

Substitute the values of equation (1) and equation (2) in above equation,

${\left(x+y\right)}^2=h^2+4A$ (5)

From equation (3), $x+y=P-h$

Substitute the value of $x+y$ in equation (4),

$\begin{array}{l}{\left(P-h\right)}^2=h^2+4A\\ P^2+h^2-2Ph=h^2+4A\\ 2Ph=P^2-4A\\ h=\frac{P^2-4A}{2P}\end{array}$

Substitute $A=6h$ in above equation,

$\begin{array}{l}h=\frac{P^2-4\left(6h\right)}{2P}\\ 2Ph=P^2-24h\\ 2Ph+24h=P^2\\ h\left(2P+24\right)=P^2\end{array}$

On further simplification, the value of $h=\frac{P^2}{2P+24}$

Therefore, the length of the hypotenuse is $h=\frac{P^2}{2P+24}$.