Question

the diagonals of a rectangular field is 60metre more than the shorter side if the longer side is 30 M more than the shorter side find the sides of field​

Answer 1

Answer:

the length of the shorter side equals to 90M and that of longer side equals to 120M.

let the length of the shorter side equals to X M .

then the length of the longer side equals to (X+30)M .

Again the length of the diagonal is (X +60)M.

hence according to the law of pithagoras , we can write

(X+60)2 = (X+30)2 + X2

by solving it we can write

X= 90m

ie the length of the shorter side

length of the longer side = X+30=90+30=120m

Answer 2

Answer:

LET ABCD be rectangular field

AC is diagonal

$let \: the \: shorter \: side = ab = xm$

$let \: shorter \: side \: ab = xm$

it is given

diagonal is 60 m more than side

$ac = ab + 60 = x + 60m$

also longer side is 30m more than shorter side

$bc = ab + 30 = x + 30m$

In triangle ABC

$< b = 90$

( as all side of rectangle are 90 degree)

hence ABC is a right angle triangle

by phytagorus theorem

$hypotenuse ^{2} = base ^{2} + height ^{2}$

$ac =^{2} ab ^{2} + bc ^{2}$

$(x + 60) = {x}^{2} + (x + 30) ^{2}$

${x}^{2} + {60}^{2} + 2 \times x \times 60 = {x}^{2} + {x}^{2} + {30}^{2} + 2 + x \times 30$

${x}^{2} + 60 \times 60 + 120x = {x}^{2} + {x}^{2} + \\ 30 \times 30 + 60x$

${x}^{2} + 3600 + 120x = {x}^{2} + {x}^{2} + 900 + 60x$

${x}^{2} + 3600 + 120x - 60x + 3600 \\ - 900 = 0$

${x}^{2} - {x}^{2} - {x}^{2} + 120x - {x}^{2} - {x}^{2} - 900 \\ = 0$

${x}^{2} - {x}^{2} - {x}^{2} + 120x - 60x + 3600 - 900 = 0$

$- {x}^{2} + 60x + 2700 = 0$

$0 = {x}^{2} - 60x - 2700$

${x}^{2} - 60x - 2700 = 0$

we factories using

${x}^{2} - 90x + 30x - 2700 = 0$

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

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

$x + 30 = 0$

$x = - 30$

$x - 90 = 0$

$x = 90$

$so \: x = - 30 \: and \: x = 90 \:$

$shorter \: side \: of \: field \: = x = 90$

$lenght \: of \: field \: x + 30 = 90 + 30 = 120m$

Answer

Hope it is helpful for you