Question

The diagonal of a rectangular field is 60m more than the shorter side. If the longer is 30m more than the shorter side , Find the sides of the field


Please answer this with full explanation (◍•ᴗ•◍)❤✨✨​

Answer 1

Refer to the attached image ☺️

Answer 2

S O L U T I O N :

$\bf{\large{\underline{\bf{Given\::}}}}}$

Let the shorter side of rectangular field = r m.

The longer side of rectangular field = (30+r) m.

The diagonal of rectangular field = (60+r) m.

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\thicklines\put(7.7,3){\large{A}}\put(9.6,1.7){\sf{\large{(60+r)m}}}\put(7.7,1){\large{B}}\put(9.1,0.7){\sf{\large{(30+r)m}}}\put(11.1,1){\large{C}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\put(11.1,2){\sf{\large{r}}}\put(8,1){\line(3,2){3}}\put(11.1,3){\large{D}}\put(10.8,1){\line(0,2){0.2}}\put(10.8,1.2){\line(2,0){0.2}}\end{picture}$

$\bf{\large{\underline{\bf{To\:find\::}}}}}$

The sides of the field.

$\bf{\large{\underline{\bf{Explanation\::}}}}}$

By Using Pythagoras theorem :

ΔDCB is a right angled triangle.

So;

$\longrightarrow\sf(Hypotenuse)^{2} =(perpendicular)^{2} +(Base)^{2} }\\\\\longrightarrow\sf{(BD)^{2} =(DC)^{2} +(BC)^{2} }\\\\\longrightarrow\sf{(60+r)^{2} =(r)^{2} +(30+r)^{2} }\\\\\longrightarrow\sf{(60)^{2} +(r)^{2} +2\times 60r=r^{2} +(30)^{2} +(r)^{2} +2\times 30r}\\\\\longrightarrow\sf{3600+r^{2} +120r=r^{2} +900+r^{2} +60r}\\\\\longrightarrow\sf{3600+r^{2} +120r=2r^{2} +900+60r}\\\\\longrightarrow\sf{2r^{2} -r^{2} +60r-120r+900-3600=0}\\\\\longrightarrow\sf{r^{2} -60r-2700=0}$

$\boxed{\bf{Using\:quadratic\:formula\::}}}}}$

As the get polynomial compared with ax² + bx + c;

• a = 1
• b = -60
• c = -2700

So;

$\mapsto\sf{r=\dfrac{-b\pm\sqrt{b^{2} -4ac} }{2a} }\\\\\\\mapsto\sf{r=\dfrac{-(-60)\pm\sqrt{(-60)^{2} -4\times 1\times (-2700)} }{2\times 1} }\\\\\\\mapsto\sf{r=\dfrac{60\pm\sqrt{3600+10800} }{2} }\\\\\\\mapsto\sf{r=\dfrac{60\pm\sqrt{14400} }{2} }\\\\\\\mapsto\sf{r=\dfrac{60\pm120}{2} }\\\\\\\mapsto\sf{r=\dfrac{60+120}{2}\:\:Or\:\:r=\dfrac{60-120}{2}} \\\\\\\mapsto\sf{r=\cancel{\dfrac{180}{2}} \:\:\:Or\:\:\:r=\cancel{\dfrac{-60}{2}}} \\\\\\\mapsto\bf{r=90\:\:Or\:\:r\neq -30}$

We know that negative value isn't acceptable.

Thus;

$\bullet\:\sf{The\:width\:of\:the\:rectangular\:field\:will\:be\:r=\boxed{\bf{90\:m}}}}\\\bullet\sf{The\:length\:of\:the\:rectangular\:field\:will\:be\:(30+r)=(30+90) = \boxed{\bf{120\:m}}}}$