Question

The diagnol of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side find the sides of the field? From ch quadratic equation

Answer 1

$<i>$Let the shorter side = x

•°•

Longer side = ( x + 30 ) m

Diagonal = ( x + 60 ) m

In ∆ ADC , /_ADC = 90°

{Applying Pythagoras Theorem}

( AC )² = ( AD )² + ( DC )²

( x + 60 )² = ( x )² + ( x + 30 )²

{ Using identity -

( a + b )² = a² + b² + 2ab }

( x )² + ( 60 )² + ( 2 ) ( x ) ( 60 ) = ( x )² + ( x )² + ( 30 )² + ( 2 ) ( x ) ( 30 )

x² + 3600 + 120x = x² + x² + 900 + 60x

On solving further, we get

x² - 60x - 2700 = 0

By Middle Term Factorisation

x² - 90x + 30x - 2700 = 0

x ( x - 90 ) + 30 ( x - 90 ) = 0

( x + 30 ) = 0 or ( x - 90 ) = 0

x = - 30 or x = 90

x = 90 ( As x ≠ - 90 , neglecting - ve value )

Now,

Longer Side = x + 30 = 90 + 30 = 120 m

Shorter Side = x = 90 m