Question

The diagonals of a rectangular field is 60 m more than the shorter side if the largest side is 30 m more than the shorter side find the sides of the field

Answer 1

are at a square = s2

Answer 2

Let ABCD be the rectangular field.

Let

• $\textbf{BC = X \:metres }$
• $\textbf{AB= (x+30)\: m }$
• $\textbf{diagonal \:AC = (x+60)\:m. }$

$\:\:$

${\bold { \underline{\small{ From \:right \:∆ABC, \:we\: have:}}}}$

$\:\:$

$\sf\large \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:AC² = AB² + BC²$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf (x+60)²=(x+30)²+x²$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf x²+120x+3600$

$\:\:$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:2x²+60x+900$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf x²-60x -2700=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf x²-90x+30x-2700=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf x(x-90)+30(x-90)=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf (x-90)(x+30)=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\hookrightarrow$$\sf x-90=0\:or\:x+30=0$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\underline{\boxed{\sf{x=\frak{\orange{90}}}}}$

$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:[\because \:length\: of\: any \:side\: of\: a\: rectangle$$\sf \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:cannot\: be\: negative]$

$\:\:$

$\sf \therefore\: BC = 90 m\:and \: AB=(90+30)\: m= 120m$

$\:\:$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\huge\pink{ \huge\star}$$\sf \:\: length=120\:m$

$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:$$\huge\purple{ \huge\star}$ $\sf \:\:breadth\:=90m.$