Chapter 4.3, Problem 44PPS

To determine

To calculate: The dimensions of the triangle.

Answer to Problem 44PPS

The dimensions of the triangle are $7,24,25$ .

Explanation of Solution

Given information:

The hypotenuse of a right triangle is 1 centimeter longer than one side and 4 centimeters longer than three times the other side.

Formula used:

In a right triangle, square of hypotenuse is equal to the sum of the squares of other two sides. This is called as Pythagoras theorem. If h is the hypotenuse, b is the base and p is the perpendicular of the right triangle then,

  $h^2=b^2+p^2$

Calculation:

Consider the given information the hypotenuse of a right triangle is 1 centimeter longer than one side and 4 centimeters longer than three times the other side.

Let the hypotenuse of right triangle be h and other two sides be b and p .

According to the question, the mathematical equations so formed are,

  $\begin{array}{c}h=b+1\\ h=3p+4\end{array}$

On equating both the equations,

  $\begin{array}{c}b+1=3p+4\\ b=3p+4-1\\ b=3p+3\end{array}$

Recall that in a right triangle, square of hypotenuse is equal to the sum of the squares of other two sides. This is called as Pythagoras theorem. If h is the hypotenuse, b is the base and p is the perpendicular of the right triangle then,

  $h^2=b^2+p^2$

Here, hypotenuse h is $h=3p+4$ ,value of b is $b=3p+3$ and p is p .

So, Pythagoras theorem will be applied as,

  $\begin{array}{c}{\left(3p+4\right)}^2={\left(3p+3\right)}^2+p^2\\ 9p^2+24p+16=9p^2+18p+9+p^2\\ p^2-6p-7=0\\ \left(p-7\right)\left(p+1\right)=0\end{array}$ .

Therefore, either $\begin{array}{c}p-7=0\text{ or }p+1=0\\ p=7\text{ or }p=-1\end{array}$ .

Since, the side of the triangle cannot be negative; the value of p is 7 units.

Now, the value of hypotenuse and other side is calculated as,

  $\begin{array}{c}h=3p+4\\ =3\left(7\right)+4\\ =21+4\\ =25\end{array}$

  $\begin{array}{c}b=3p+3\\ =3\left(7\right)+3\\ =21+3\\ =24\end{array}$

Thus, the dimensions of the triangle are $7,24,25$ .