Question

the diagonal of rectangle is 60cm more its shorter side and larger side is 30 cm more than the shorter side find the sides of rectangle

Answer 1

hope it helps................

Answer 2

ANSWER:

90cm,120cm

STEP-BY-STEP VERIFICATION:

LET,

x be the length of shorter side.

GIVEN,

$hypoteneous(h) = diagonal(d) = (60 + x) \: \: cm\\perpendicular(p) = length(l) =(30 + x) \: \: cm \\ base(b) = breadth(b) = x \: \: cm$

NOW,

By using Pythagoras theorem:

${h}^{2} = {p}^{2} + {b}^{2} \\ = > {(60 + x)}^{2} = {(30 + x)}^{2} + {x}^{2} \\ = > 3600 + 120x + {x}^{2} = 900 + 60x + {x}^{2} + {x}^{2} \\ = > 0 = {x}^{2} - 60x - 2700 \\ = > {x}^{2} - 60x - 2700 = 0 \\ = > {x}^{2} - 90x + 30x - 2700 = 0 \\ = > x(x - 90) + 30(x - 90) = 0\\ = > (x + 30)(x - 90) = 0$

Either,

$x + 30 = 0 \\. °.x = - 30cm$

HERE,

The lenght can't be negative so it is neglected.

Or,

$x - 90 = 0 \\. °.x = 90cm$

So, 90cm is one side.

Other side=30+90=120cm

SO, THE SIDES OF RECTANGLE ARE 90cm AND 120cm.