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.

Answer 1

Hey there :)

Right triangle pennant with 
Hypotenuse = 10 ft
One side is 2 ft longer than other = 2 + p
Other side = p 

Since it is a right triangle, we can solve by Pythagoras Theorem:
( Side 1 )² + ( Side 2 )² = ( Hypotenuse )²
( p )² + ( 2 + p )² = ( 10 )²
                 ↓ Solve by FOIL ( First, Outer, Inner, Last )
     2 ( 2 ) + 2 ( p ) + 2 ( p ) + p ( p ) = 4 + 4p + p²

p² + 4 + 4p + p² = 100

Take 100 to left hand side and equate to 0 and add like-terms together
2p² + 4p + 4 - 100 = 0
2p² + 4p - 96
Divide the whole equation by 2 for simplified form
$\frac{2p^2+4p-96}{2} = 0$
p² + 2p - 48

Factorise { Sum = 2 , Product = - 48 } Therefore, possible factors 8 and - 6

( p - 6 )( p + 8 ) = 0
p = 6 , p = - 8 
    ↑           ↑ 
    ↑   A length can never be negative
One length = 6 ft
Other length = 2 + 6 = 8 ft

         

Answer 2

Let the shorter side be x
The other side is x + 2

a² + b² = c²
x² + (x + 2)² = 10²
x² + x² + 4x + 4 = 100
2x² + 4x - 96 = 0
x² + 2x - 48 = 0
(x+8)(x-6) = 0  
 x = - 8 (rejected, length cannot be negative), x = 6

x = 6
x + 2 = 6 + 2 = 8

Answer: The length are 6 ft and 8 ft