Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{Is it possible to design a rectangular mango grove whose length is}$
$\displaystyle \text{twice its breadth and the area is }800\text{ m}^2\text{? If so, find its length and breadth.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the breadth of the rectangular mango grove be }x\text{ m.}$
$\displaystyle \therefore \text{Length}=2x\text{ m.}$
$\displaystyle \text{Area of rectangle}=\text{Length}\times\text{Breadth}$
$\displaystyle \therefore 2x\times x=800$
$\displaystyle \Rightarrow 2x^2=800$
$\displaystyle \Rightarrow x^2=400$
$\displaystyle \Rightarrow x=\pm20$
$\displaystyle \text{Since breadth cannot be negative, }x=-20\text{ is rejected.}$
$\displaystyle \therefore \text{Breadth}=20\text{ m.}$
$\displaystyle \therefore \text{Length}=2(20)=40\text{ m.}$
$\displaystyle \therefore \text{Yes, it is possible to design the rectangular mango grove.}$
$\displaystyle \text{Its length is }40\text{ m and breadth is }20\text{ m.}$
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$\displaystyle \textbf{Question 2: }\text{Is it possible to design a rectangular park of perimeter }$
$\displaystyle 80\text{ m and area }400\text{ m}^2\text{?} \ \text{If so, find its length and breadth.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the breadth of the rectangular park be }x\text{ m.}$
$\displaystyle \text{Since perimeter}=80\text{ m,}$
$\displaystyle 2(\text{Length}+\text{Breadth})=80$
$\displaystyle \Rightarrow \text{Length}+x=40$
$\displaystyle \Rightarrow \text{Length}=(40-x)\text{ m.}$
$\displaystyle \text{Area of rectangle}=\text{Length}\times\text{Breadth}$
$\displaystyle \therefore (40-x)x=400$
$\displaystyle \Rightarrow 40x-x^2=400$
$\displaystyle \Rightarrow x^2-40x+400=0$
$\displaystyle \Rightarrow (x-20)^2=0$
$\displaystyle \Rightarrow x=20$
$\displaystyle \therefore \text{Breadth}=20\text{ m.}$
$\displaystyle \therefore \text{Length}=40-20=20\text{ m.}$
$\displaystyle \therefore \text{Yes, it is possible to design the rectangular park.}$
$\displaystyle \text{Its length and breadth are both }20\text{ m.}$
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$\displaystyle \textbf{Question 3: }\text{Sum of the areas of two squares is }640\text{ m}^2\text{. If the difference}$
$\displaystyle \text{of their perimeters is }64\text{ m, find the sides of the two squares.}\hfill\text{[CBSE 2008, 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the side of the smaller square be }x\text{ m.}$
$\displaystyle \text{Since the difference of their perimeters is }64\text{ m,}$
$\displaystyle 4(\text{Side of larger square})-4x=64$
$\displaystyle \Rightarrow \text{Side of larger square}-x=16$
$\displaystyle \Rightarrow \text{Side of larger square}=(x+16)\text{ m.}$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2+(x+16)^2=640$
$\displaystyle \Rightarrow x^2+x^2+32x+256=640$
$\displaystyle \Rightarrow 2x^2+32x-384=0$
$\displaystyle \Rightarrow x^2+16x-192=0$
$\displaystyle \Rightarrow x^2+24x-8x-192=0$
$\displaystyle \Rightarrow x(x+24)-8(x+24)=0$
$\displaystyle \Rightarrow (x+24)(x-8)=0$
$\displaystyle \Rightarrow x=-24\text{ or }x=8$
$\displaystyle \text{Since length cannot be negative, }x=-24\text{ is rejected.}$
$\displaystyle \therefore \text{Side of the smaller square}=8\text{ m.}$
$\displaystyle \therefore \text{Side of the larger square}=8+16=24\text{ m.}$
$\displaystyle \therefore \text{The sides of the two squares are }8\text{ m and }24\text{ m.}$
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$\displaystyle \textbf{Question 4: }\text{The sum of the areas of two squares is }52\text{ cm}^2\text{ and difference of}$
$\displaystyle \text{their perimeters is }8\text{ cm. Find the lengths of the sides of the two squares.}\hfill\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the side of the smaller square be }x\text{ cm.}$
$\displaystyle \text{Since the difference of their perimeters is }8\text{ cm,}$
$\displaystyle 4(\text{Side of larger square})-4x=8$
$\displaystyle \Rightarrow \text{Side of larger square}-x=2$
$\displaystyle \Rightarrow \text{Side of larger square}=(x+2)\text{ cm.}$
$\displaystyle \text{According to the question,}$
$\displaystyle x^2+(x+2)^2=52$
$\displaystyle \Rightarrow x^2+x^2+4x+4=52$
$\displaystyle \Rightarrow 2x^2+4x-48=0$
$\displaystyle \Rightarrow x^2+2x-24=0$
$\displaystyle \Rightarrow (x+6)(x-4)=0$
$\displaystyle \Rightarrow x=-6\text{ or }x=4$
$\displaystyle \text{Since length cannot be negative, }x=-6\text{ is rejected.}$
$\displaystyle \therefore \text{Side of the smaller square}=4\text{ cm.}$
$\displaystyle \therefore \text{Side of the larger square}=4+2=6\text{ cm.}$
$\displaystyle \therefore \text{The sides of the two squares are }4\text{ cm and }6\text{ cm.}$
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$\displaystyle \textbf{Question 5: }\text{The area of a rectangular plot is }528\text{ m}^2\text{. The length of the plot (in metres) is one}$
$\displaystyle \text{metre more than twice its breadth. Find the length and the breadth of the plot.}\hfill\text{[CBSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the breadth of the rectangular plot be }x\text{ m.}$
$\displaystyle \therefore \text{Length}=(2x+1)\text{ m.}$
$\displaystyle \text{Area of rectangle}=\text{Length}\times\text{Breadth}$
$\displaystyle \therefore x(2x+1)=528$
$\displaystyle \Rightarrow 2x^2+x-528=0$
$\displaystyle \Rightarrow 2x^2+33x-32x-528=0$
$\displaystyle \Rightarrow x(2x+33)-16(2x+33)=0$
$\displaystyle \Rightarrow (2x+33)(x-16)=0$
$\displaystyle \Rightarrow x=-\frac{33}{2}\text{ or }x=16$
$\displaystyle \text{Since breadth cannot be negative, }x=-\frac{33}{2}\text{ is rejected.}$
$\displaystyle \therefore \text{Breadth}=16\text{ m.}$
$\displaystyle \therefore \text{Length}=2(16)+1=33\text{ m.}$
$\displaystyle \therefore \text{The length and breadth of the rectangular plot are }33\text{ m and }16\text{ m respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{In the centre of a rectangular lawn of dimensions }50\text{ m}\times40\text{ m, a rectangular}$
$\displaystyle \text{pond has to be constructed so that the area of the grass surrounding the pond would be}$
$\displaystyle 1184\text{ m}^2\text{. Find the length and breadth of the pond.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the uniform width of the grass surrounding the pond be }x\text{ m.}$
$\displaystyle \therefore \text{Length of the pond}=(50-2x)\text{ m.}$
$\displaystyle \text{Breadth of the pond}=(40-2x)\text{ m.}$
$\displaystyle \text{Area of the lawn}=50\times40=2000\text{ m}^2.$
$\displaystyle \therefore \text{Area of the pond}=2000-1184=816\text{ m}^2.$
$\displaystyle \text{According to the question,}$
$\displaystyle (50-2x)(40-2x)=816$
$\displaystyle \Rightarrow 2000-180x+4x^2=816$
$\displaystyle \Rightarrow 4x^2-180x+1184=0$
$\displaystyle \Rightarrow x^2-45x+296=0$
$\displaystyle \Rightarrow (x-8)(x-37)=0$
$\displaystyle \Rightarrow x=8\text{ or }x=37$
$\displaystyle \text{Since }x=37\text{ gives negative dimensions for the pond, it is rejected.}$
$\displaystyle \therefore x=8.$
$\displaystyle \therefore \text{Length of the pond}=50-2(8)=34\text{ m.}$
$\displaystyle \therefore \text{Breadth of the pond}=40-2(8)=24\text{ m.}$
$\displaystyle \therefore \text{The length and breadth of the pond are }34\text{ m and }24\text{ m respectively.}$
$\displaystyle \\$