Q. 329 - Intermediate Algebra

Question

A pennant is shaped like a right triangle, with hypotenuse $10 feet$. The length of one side of the pennant is two feet longer than the length of the other side. Find the length of the two sides of the pennant.

Step-by-Step Solution

Answer

The length of the two sides of the pennant are $6 feet ,8 feet$.

Step 1. Assumption of integers.

Assume the length of the other side is $x feet$ and length of one side of the pennant is $x+2 feet$.

We know, the Pythagoras theorem,

$a^2+b^2=c^2$

Substitute the values in the equation of Pythagoras theorem,

$$\begin{gathered} x^2+{\left(x+2\right)}^2={\left(10\right)}^2 ...... \left(i\right) \\ {x}^2+x^2+2x+4=100 \\ 2x^2+2x-96=0 \\ 2\left(x^2+x-48\right)=0 \\ {x}^2+x-48=0 \end{gathered}$$

Step 2. Factor the equation.

On factoring,

$$\begin{gathered} x^2+8x-6x-48=0 \\ x\left(x+8\right)-6\left(x+8\right)=0 \\ \left(x+8\right)\left(x-6\right)=0 \end{gathered}$$

Use the zero product property,

$$\begin{gathered} x+8=0 \\ x=-8 \end{gathered}$$

Then,

$$\begin{gathered} x-6=0 \\ x=6 \end{gathered}$$

As we know, one side of the pennant cannot be negative, so $x=6$.

Then the other side,

$$\begin{gathered} =x+2 \\ =6+2 \\ =8 \end{gathered}$$

Step 3. Check the answers.

Substitute the length of the sides in equation (i),

$$\begin{gathered} 6^2+8^2={10}^2 \\ 36+64=100 \\ 100=100 \end{gathered}$$

This is true.

Hence, the two required sides of the pennant are $6 feet, 8 feet$.