Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: } \text{The sides of a right-angled triangle containing the right}$
$\displaystyle \text{angle are }4x\text{ cm and }(2x-1)\text{ cm. If the area of the triangle}$
$\displaystyle \text{is }30\text{ cm}^2,\text{ calculate the lengths of its sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Area of a triangle}=\frac{1}{2}\times\text{base}\times\text{height}$
$\displaystyle 30=\frac{1}{2}\times4x\times(2x-1)$
$\displaystyle 60=8x^2-4x$
$\displaystyle 8x^2-4x-60=0$
$\displaystyle 2x^2-x-15=0$
$\displaystyle 2x^2+5x-6x-15=0$
$\displaystyle x(2x+5)-3(2x+5)=0$
$\displaystyle (x-3)(2x+5)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=-\frac{5}{2}$
$\displaystyle \text{Since length cannot be negative, }x\neq -\frac{5}{2}.$
$\displaystyle \therefore x=3$
$\displaystyle 4x=12\text{ cm},\qquad 2x-1=5\text{ cm}$
$\displaystyle \text{Hypotenuse}=\sqrt{12^2+5^2}=\sqrt{169}=13\text{ cm}$
$\displaystyle \therefore \text{The lengths of the sides are }5\text{ cm},12\text{ cm and }13\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 2: } \text{The hypotenuse of a right-angled triangle is }26\text{ cm}$
$\displaystyle \text{and the sum of the other two sides is }34\text{ cm. Find their lengths.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one side be }x\text{ cm.}$
$\displaystyle \therefore \text{The other side is }(34-x)\text{ cm.}$
$\displaystyle \text{Using Pythagoras theorem,}$
$\displaystyle x^2+(34-x)^2=26^2$
$\displaystyle x^2+1156-68x+x^2=676$
$\displaystyle 2x^2-68x+480=0$
$\displaystyle x^2-34x+240=0$
$\displaystyle (x-24)(x-10)=0$
$\displaystyle \Rightarrow x=24\text{ or }x=10$
$\displaystyle \therefore \text{The other side is }10\text{ cm or }24\text{ cm respectively.}$
$\displaystyle \therefore \text{The required lengths are }10\text{ cm and }24\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 3: } \text{The sides of a right-angled triangle are }(x-1)\text{ cm},$
$\displaystyle 3x\text{ cm and }(3x+1)\text{ cm. Find the value of }x,\text{ the lengths}$
$\displaystyle \text{of its sides and its area.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }3x+1\text{ is the largest side, it is the hypotenuse.}$
$\displaystyle \text{Using Pythagoras theorem,}$
$\displaystyle (x-1)^2+(3x)^2=(3x+1)^2$
$\displaystyle x^2-2x+1+9x^2=9x^2+6x+1$
$\displaystyle x^2-8x=0$
$\displaystyle x(x-8)=0$
$\displaystyle \Rightarrow x=0\text{ or }x=8$
$\displaystyle \text{Since lengths must be positive, }x\neq 0.$
$\displaystyle \therefore x=8$
$\displaystyle x-1=7\text{ cm},\quad 3x=24\text{ cm},\quad 3x+1=25\text{ cm}$
$\displaystyle \text{Area}=\frac{1}{2}\times7\times24=84\text{ cm}^2$
$\displaystyle \therefore \text{The lengths are }7\text{ cm},24\text{ cm and }25\text{ cm.}$
$\displaystyle \therefore \text{The area is }84\text{ cm}^2.$
$\\$

$\displaystyle \textbf{Question 4: } \text{The hypotenuse of a right-angled triangle exceeds one side}$
$\displaystyle \text{by }1\text{ cm and the other side by }18\text{ cm. Find the lengths of the sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the hypotenuse be }x\text{ cm.}$
$\displaystyle \therefore \text{The other two sides are }(x-1)\text{ cm and }(x-18)\text{ cm.}$
$\displaystyle \text{Using Pythagoras theorem,}$
$\displaystyle (x-1)^2+(x-18)^2=x^2$
$\displaystyle x^2-2x+1+x^2-36x+324=x^2$
$\displaystyle x^2-38x+325=0$
$\displaystyle (x-25)(x-13)=0$
$\displaystyle \Rightarrow x=25\text{ or }x=13$
$\displaystyle \text{If }x=13,\text{ then }x-18=-5,\text{ which is not possible.}$
$\displaystyle \therefore x=25$
$\displaystyle x-1=24\text{ cm},\qquad x-18=7\text{ cm}$
$\displaystyle \therefore \text{The required sides are }7\text{ cm},24\text{ cm and }25\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 5: } \text{The diagonal of a rectangle is }60\text{ m more than its shorter side}$
$\displaystyle \text{and the larger side is }30\text{ m more than the shorter side. Find the sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the shorter side be }x\text{ m.}$
$\displaystyle \therefore \text{The larger side is }(x+30)\text{ m.}$
$\displaystyle \text{The diagonal is }(x+60)\text{ m.}$
$\displaystyle \text{Using Pythagoras theorem,}$
$\displaystyle x^2+(x+30)^2=(x+60)^2$
$\displaystyle x^2+x^2+60x+900=x^2+120x+3600$
$\displaystyle x^2-60x-2700=0$
$\displaystyle (x-90)(x+30)=0$
$\displaystyle \Rightarrow x=90\text{ or }x=-30$
$\displaystyle \text{Since length cannot be negative, }x\neq -30.$
$\displaystyle \therefore x=90$
$\displaystyle \text{Shorter side}=90\text{ m}$
$\displaystyle \text{Larger side}=90+30=120\text{ m}$
$\displaystyle \therefore \text{The sides of the rectangle are }90\text{ m and }120\text{ m}.$
$\\$

$\displaystyle \textbf{Question 6: } \text{The perimeter of a rectangle is }104\text{ m and its area is }640\text{ m}^2.$
$\displaystyle \text{Find its length and breadth.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length be }x\text{ m and the breadth be }y\text{ m.}$
$\displaystyle xy=640\qquad ...\text{(i)}$
$\displaystyle 2(x+y)=104$
$\displaystyle x+y=52\qquad ...\text{(ii)}$
$\displaystyle \text{From (ii), }y=52-x$
$\displaystyle \text{Substituting in (i),}$
$\displaystyle x(52-x)=640$
$\displaystyle 52x-x^2=640$
$\displaystyle x^2-52x+640=0$
$\displaystyle (x-32)(x-20)=0$
$\displaystyle \Rightarrow x=32\text{ or }x=20$
$\displaystyle \therefore y=20\text{ or }y=32$
$\displaystyle \therefore \text{The length and breadth are }32\text{ m and }20\text{ m}.$
$\\$

$\displaystyle \textbf{Question 7: } \text{A footpath of uniform width runs around the inside of a}$
$\displaystyle \text{rectangular field }32\text{ m long and }24\text{ m wide. If the path occupies }208\text{ m}^2,$
$\displaystyle \text{find the width of the footpath.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the width of the footpath be }x\text{ m.}$
$\displaystyle \text{Area of the rectangular field}=32\times24=768\text{ m}^2$
$\displaystyle \text{Length of inner rectangle}=(32-2x)\text{ m}$
$\displaystyle \text{Breadth of inner rectangle}=(24-2x)\text{ m}$
$\displaystyle \text{Area of inner rectangle}=(32-2x)(24-2x)$
$\displaystyle \therefore 768-(32-2x)(24-2x)=208$
$\displaystyle 768-(768-112x+4x^2)=208$
$\displaystyle 112x-4x^2=208$
$\displaystyle 4x^2-112x+208=0$
$\displaystyle x^2-28x+52=0$
$\displaystyle x=\frac{28\pm\sqrt{28^2-4(1)(52)}}{2}$
$\displaystyle x=\frac{28\pm\sqrt{784-208}}{2}$
$\displaystyle x=\frac{28\pm24}{2}$
$\displaystyle x=26\text{ or }x=2$
$\displaystyle \text{Since }x=26\text{ is not possible, }x=2.$
$\displaystyle \therefore \text{The width of the footpath is }2\text{ m}.$
$\\$

$\displaystyle \textbf{Question 8: } \text{Two squares have sides }x\text{ cm and }(x+4)\text{ cm.}$
$\displaystyle \text{The sum of their areas is }656\text{ cm}^2.\text{ Express this as an algebraic}$
$\displaystyle \text{equation in }x\text{ and solve it to find the sides of the squares.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x^2+(x+4)^2=656$
$\displaystyle x^2+x^2+8x+16=656$
$\displaystyle 2x^2+8x-640=0$
$\displaystyle x^2+4x-320=0$
$\displaystyle (x+20)(x-16)=0$
$\displaystyle \Rightarrow x=-20\text{ or }x=16$
$\displaystyle \text{Since length cannot be negative, }x\neq -20.$
$\displaystyle \therefore x=16$
$\displaystyle x+4=20$
$\displaystyle \therefore \text{The sides of the squares are }16\text{ cm and }20\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 9: } \text{The dimensions of a rectangular field are }50\text{ m by }40\text{ m.}$
$\displaystyle \text{A flower bed is prepared inside this field leaving a gravel path of uniform}$
$\displaystyle \text{width all around the flower bed. The cost of preparing the flower bed}$
$\displaystyle \text{and graveling the path at Rs. }30\text{ and Rs. }20\text{ per m}^2\text{ respectively is Rs. }52000.$
$\displaystyle \text{Find the width of the gravel path.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the width of the gravel path be }x\text{ m.}$
$\displaystyle \text{Area of flower bed}=(50-2x)(40-2x)$
$\displaystyle =(2000-180x+4x^2)\text{ m}^2$
$\displaystyle \text{Area of gravel path}=50\times40-(50-2x)(40-2x)$
$\displaystyle =2000-(2000-180x+4x^2)$
$\displaystyle =(180x-4x^2)\text{ m}^2$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 30(2000-180x+4x^2)+20(180x-4x^2)=52000$
$\displaystyle 60000-5400x+120x^2+3600x-80x^2=52000$
$\displaystyle 40x^2-1800x+8000=0$
$\displaystyle x^2-45x+200=0$
$\displaystyle (x-5)(x-40)=0$
$\displaystyle \Rightarrow x=5\text{ or }x=40$
$\displaystyle \text{Since }x=40\text{ is not possible, }x=5.$
$\displaystyle \therefore \text{The width of the gravel path is }5\text{ m}.$
$\\$

$\displaystyle \textbf{Question 10: } \text{An area is paved with square tiles of a certain size}$
$\displaystyle \text{and the number required is }128.\text{ If the tiles had been }2\text{ cm smaller}$
$\displaystyle \text{each way, }200\text{ tiles would have been needed to pave the same area.}$
$\displaystyle \text{Find the size of the larger tiles.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the side of the larger square tile be }x\text{ cm.}$
$\displaystyle \therefore \text{Side of the smaller square tile}=(x-2)\text{ cm}$
$\displaystyle \text{Since the area paved is the same,}$
$\displaystyle 128x^2=200(x-2)^2$
$\displaystyle 128x^2=200(x^2-4x+4)$
$\displaystyle 128x^2=200x^2-800x+800$
$\displaystyle 72x^2-800x+800=0$
$\displaystyle 9x^2-100x+100=0$
$\displaystyle 9x^2-90x-10x+100=0$
$\displaystyle 9x(x-10)-10(x-10)=0$
$\displaystyle (x-10)(9x-10)=0$
$\displaystyle \Rightarrow x=10\text{ or }x=\frac{10}{9}$
$\displaystyle \text{Since }x=\frac{10}{9}\text{ gives }x-2<0,\text{ it is not possible.}$
$\displaystyle \therefore x=10$
$\displaystyle \therefore \text{The size of the larger tile is }10\text{ cm by }10\text{ cm}.$
$\\$

$\displaystyle \textbf{Question 11: } \text{A farmer has }70\text{ m of fencing, with which he encloses}$
$\displaystyle \text{three sides of a rectangular sheep pen; the fourth side being a wall.}$
$\displaystyle \text{If the area of the pen is }600\text{ m}^2,\text{ find the length of its shorter side.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the shorter side be }x\text{ m.}$
$\displaystyle \therefore \text{The larger side is }(70-2x)\text{ m.}$
$\displaystyle \text{Area of the pen}=600\text{ m}^2$
$\displaystyle \therefore x(70-2x)=600$
$\displaystyle 70x-2x^2=600$
$\displaystyle 2x^2-70x+600=0$
$\displaystyle x^2-35x+300=0$
$\displaystyle (x-20)(x-15)=0$
$\displaystyle \Rightarrow x=20\text{ or }x=15$
$\displaystyle \therefore \text{The shorter side can be }20\text{ m or }15\text{ m}.$
$\\$

$\displaystyle \textbf{Question 12: } \text{A square lawn is bounded on three sides by a path }4\text{ m wide.}$
$\displaystyle \text{If the area of the path is }\frac{7}{8}\text{ that of the lawn, find the dimensions of the lawn.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the side of the square lawn be }x\text{ m.}$
$\displaystyle \text{Since the path is }4\text{ m wide on three sides,}$
$\displaystyle \text{area of path}=(x+8)(x+4)-x^2$
$\displaystyle \text{Given, }(x+8)(x+4)-x^2=\frac{7}{8}x^2$
$\displaystyle x^2+12x+32-x^2=\frac{7}{8}x^2$
$\displaystyle 12x+32=\frac{7}{8}x^2$
$\displaystyle 96x+256=7x^2$
$\displaystyle 7x^2-96x-256=0$
$\displaystyle 7x^2-112x+16x-256=0$
$\displaystyle 7x(x-16)+16(x-16)=0$
$\displaystyle (x-16)(7x+16)=0$
$\displaystyle \Rightarrow x=16\text{ or }x=-\frac{16}{7}$
$\displaystyle \text{Since length cannot be negative, }x\neq -\frac{16}{7}.$
$\displaystyle \therefore x=16$
$\displaystyle \therefore \text{The dimensions of the lawn are }16\text{ m by }16\text{ m}.$
$\\$

$\displaystyle \textbf{Question 13: } \text{The area of a big rectangular room is }300\text{ m}^2.$
$\displaystyle \text{If the length were decreased by }5\text{ m and the breadth increased by }5\text{ m,}$
$\displaystyle \text{the area would be unaltered. Find the length of the room.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the room be }x\text{ m and the breadth be }y\text{ m.}$
$\displaystyle xy=300\qquad \Rightarrow \qquad y=\frac{300}{x}$
$\displaystyle \text{Also, }(x-5)(y+5)=300$
$\displaystyle \text{Substituting }y=\frac{300}{x},$
$\displaystyle \left(x-5\right)\left(\frac{300}{x}+5\right)=300$
$\displaystyle 300-\frac{1500}{x}+5x-25=300$
$\displaystyle 5x-25-\frac{1500}{x}=0$
$\displaystyle 5x^2-25x-1500=0$
$\displaystyle x^2-5x-300=0$
$\displaystyle (x-20)(x+15)=0$
$\displaystyle \Rightarrow x=20\text{ or }x=-15$
$\displaystyle \text{Since length cannot be negative, }x=20.$
$\displaystyle y=\frac{300}{20}=15\text{ m}$
$\displaystyle \therefore \text{The length of the room is }20\text{ m}.$
$\\$