Question

The diagonal of a rectangular field is 16 metres more than the shorter side. If the longer side is 14 metres more than the shorter side, then find the lengths of the sides of the field.

Answer 1

Answer:

Let the shorter side of the field be x meters.

Then, longer side = (x+14) m.

And, diagonal = (x+16) m.

$\sf \therefore {(x + 16)}^{2} - {(x + 14)}^{2} = {x}^{2}$

$\sf \dashrightarrow (2x + 30) \times 2 = {x}^{2}$

$\sf \dashrightarrow {x}^{2} - 4x - 60 = 0$

$\sf \dashrightarrow {x}^{2} - 10x + 6x - 60 = 0$

$\sf \dashrightarrow x(x - 10) + 6(x - 10) = 0$

$\sf \dashrightarrow (x - 10)(x + 6) = 0$

$\sf \dashrightarrow x - 10 = 0 \: \: or \: \: x + 6 = 0$

$\sf \dashrightarrow x = 10 \: \: or \: \: x = - 6$

$\sf \dashrightarrow x = 10$

ㅤㅤㅤㅤㅤㅤ[ Breadth cannot be negative.]

$\sf \therefore breadth = 10 \: m$

$\sf length = (10 + 14)m$

$\sf = 24 \: m$

Answer 2

: SOLUTION

$\looparrowright$ let the length of the shorter side be x m.

$\looparrowright$ length of diagonal = (x+16) m

$\looparrowright$ And the length of longer side =(x+14) m

A/q

(Using pythyorous therom)

$\implies$ $x {}^{2} + (x + 14) {}^{2} = (x + 16) {}^{2}$

$\implies$ $x {}^{2} - 4x - 60 = 0$

$\implies$ $x {}^{2} (x + 6x) - 10(x + 16) {}^{2}$

$\implies$ $(x + 6)(x - 10) = 0$

$\implies$ $x = - 6 \: ,\: \: 10 \: (as \: x \: can't \: be \: negative)$

$\implies$ $x = 10 \: m$

$\sf \boxed{hence, \: the \: length \: of \: the \: side \: is \: 10 \: m \: and \: 24 \: m \: }$