Question

the length of a rectangle exceeds the width by 7cm. if the area is 60sq.cm, find the length of the rectangle

solve this sum on quadratic equation step by step please ​

Answer 1

Answer:

⋆ DIAGRAM :

$\setlength{\unitlength}{1.5cm}\begin{picture}(8,2)\linethickness{0.4mm}\put(7.7,3){\large\sf{A}}\put(7.6,2){\sf{\large{x}}}\put(7.7,1){\large\sf{B}}\put(9.2,0.7){\sf{\large{(x + 7)}}}\put(11.1,1){\large\sf{C}}\put(11.1,3){\large\sf{D}}\put(8,1){\line(1,0){3}}\put(8,1){\line(0,2){2}}\put(11,1){\line(0,3){2}}\put(8,3){\line(3,0){3}}\end{picture}$

$\rule{100}{0.8}$

Let the width of rectangle be x and length of rectangle be (x + 7) respectively.

☯ $\underline{\boldsymbol{According\: to \:the\: Question\:now :}}$

$:\implies\sf Area_{(Rectangle)}=Length \times Width\\\\\\:\implies\sf 60 = (x + 7) \times x\\\\\\:\implies\sf 60 = x^2 + 7x\\\\\\:\implies\sf x^2 + 7x - 60 = 0\\\\\\:\implies\sf x^2 + (12-5)x - 60 = 0\\\\\\:\implies\sf x^2 + 12x - 5x - 60 = 0\\\\\\:\implies\sf x(x + 12) - 5(x + 12) = 0\\\\\\:\implies\sf (x - 5)(x + 12) = 0\\\\\\:\implies\green{\sf x = 5} \quad \sf or \quad \red{x = - \:12}$

⠀

✩ Ignoring Negative , width be x = 5

$\rule{150}{1.5}$

☯ $\underline{\sf{Dimensions\: of \:the\: Rectangle :}}$

$\bullet\:\:\textsf{Length = (x + 7) = \textbf{12 cm}}\\\bullet\:\:\textsf{Width = x = \textbf{5 cm}}$

Answer 2

S O L U T I O N :

Assume the width of the rectangle be r cm

Assume the length of the rectangle be (r+7) cm

We know that formula of the area of rectangle :

$\boxed{\bf{Area\:of\:rectangle=Length\times breadth\:\:\:\:(sq.unit)}}}}$

A/q

$\longrightarrow\rm{(r+7)\times r=60}\\\\\longrightarrow\rm{r^{2} +7r=60}\\\\\longrightarrow\rm{r^{2} +7r-60=0}$

$\boxed{\bf{By \:using\:quadratic\:formula\::}}}}$

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

• a = 1
• b = 7
• c = -60

Now;

$\longrightarrow\rm{x=\dfrac{-b\pm\sqrt{b^{2} -4ac} }{2a} }\\\\\\\longrightarrow\rm{x=\dfrac{-7\pm\sqrt{(7)^{2} -4\times 1\times (-60)} }{2\times 1} }\\\\\\\longrightarrow\rm{x=\dfrac{-7\pm\sqrt{49+240} }{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-7\pm\sqrt{289} }{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-7\pm17}{2} }\\\\\\\longrightarrow\rm{x=\dfrac{-7+17}{2} \:\:\:Or\:\:\:x=\dfrac{-7-17}{2} }\\\\\\\longrightarrow\rm{x=\cancel{\dfrac{10}{2}} \:\:\;Or\:\:\:x=\cancel{\dfrac{-24}{2} }}$

$\longrightarrow\bf{x=5\:\:\:Or\:\:\:x\neq -12}$

We know that negative value isn't acceptable.

Thus;

The width of the rectangle = r = 5 cm

The length of the rectangle = (r+7) cm = (5+7) cm = 12 cm