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kakasveta [241]
3 years ago
14

A pennant is shaped like a right triangle, with hypotenuse 10 feet. the length of one side of the pennant is two feet longer tha

n the length of the other side. find the length of the two sides of the pennant.
Mathematics
2 answers:
svetoff [14.1K]3 years ago
6 0
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

         
Zarrin [17]3 years ago
5 0
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
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