Class 10: Quadratic Equations

$\displaystyle \textbf{Question 1: }\text{The hypotenuse of a right triangle is }25\text{ cm. The difference between the}$
$\displaystyle \text{lengths of the other two sides of the triangle is }5\text{ cm. Find the lengths of these sides.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the length of the shorter side be }x\text{ cm.}$
$\displaystyle \therefore \text{Length of the other side}=(x+5)\text{ cm.}$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle x^2+(x+5)^2=25^2$
$\displaystyle \Rightarrow x^2+x^2+10x+25=625$
$\displaystyle \Rightarrow 2x^2+10x-600=0$
$\displaystyle \Rightarrow x^2+5x-300=0$
$\displaystyle \Rightarrow x^2+20x-15x-300=0$
$\displaystyle \Rightarrow x(x+20)-15(x+20)=0$
$\displaystyle \Rightarrow (x+20)(x-15)=0$
$\displaystyle \Rightarrow x=-20\text{ or }x=15$
$\displaystyle \text{Since length cannot be negative, }x=-20\text{ is rejected.}$
$\displaystyle \therefore \text{The other two sides are }15\text{ cm and }20\text{ cm.}$
$\displaystyle \therefore \text{The sides of the triangle are }15\text{ cm, }20\text{ cm and }25\text{ cm.}$
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$\displaystyle \textbf{Question 2: }\text{The diagonal of a rectangular field is }60\text{ metres more than the shorter side.}$
$\displaystyle \text{If the longer side is }30\text{ metres more than the shorter side, find the sides of the field.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the shorter side of the rectangular field be }x\text{ m.}$
$\displaystyle \therefore \text{Longer side}=(x+30)\text{ m.}$
$\displaystyle \text{Diagonal}=(x+60)\text{ m.}$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle x^2+(x+30)^2=(x+60)^2$
$\displaystyle \Rightarrow x^2+x^2+60x+900=x^2+120x+3600$
$\displaystyle \Rightarrow x^2-60x-2700=0$
$\displaystyle \Rightarrow x^2-90x+30x-2700=0$
$\displaystyle \Rightarrow x(x-90)+30(x-90)=0$
$\displaystyle \Rightarrow (x-90)(x+30)=0$
$\displaystyle \Rightarrow x=90\text{ or }x=-30$
$\displaystyle \text{Since length cannot be negative, }x=-30\text{ is rejected.}$
$\displaystyle \therefore \text{Shorter side}=90\text{ m.}$
$\displaystyle \therefore \text{Longer side}=90+30=120\text{ m.}$
$\displaystyle \therefore \text{The sides of the rectangular field are }90\text{ m and }120\text{ m.}$
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$\displaystyle \textbf{Question 3: }\text{The hypotenuse of a right triangle is }3\sqrt{10}\text{ cm. If the smaller leg is tripled and}$
$\displaystyle \text{the longer leg doubled, new hypotenuse will be }9\sqrt{5}\text{ cm. How long are the legs of the triangle?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the smaller leg of the right triangle be }x\text{ cm and the longer leg be }y\text{ cm.}$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle x^2+y^2=(3\sqrt{10})^2$
$\displaystyle \Rightarrow x^2+y^2=90 \qquad ...(i)$
$\displaystyle \text{When the smaller leg is tripled and the longer leg is doubled,}$
$\displaystyle \text{the new legs are }3x\text{ cm and }2y\text{ cm.}$
$\displaystyle \text{Again, by Pythagoras theorem,}$
$\displaystyle (3x)^2+(2y)^2=(9\sqrt{5})^2$
$\displaystyle \Rightarrow 9x^2+4y^2=405 \qquad ...(ii)$
$\displaystyle \text{Multiplying equation }(i)\text{ by }4,\text{ we get}$
$\displaystyle 4x^2+4y^2=360 \qquad ...(iii)$
$\displaystyle \text{Subtracting }(iii)\text{ from }(ii),$
$\displaystyle 5x^2=45$
$\displaystyle \Rightarrow x^2=9$
$\displaystyle \Rightarrow x=3 \qquad [\because x>0]$
$\displaystyle \text{Putting }x^2=9\text{ in equation }(i),$
$\displaystyle 9+y^2=90$
$\displaystyle \Rightarrow y^2=81$
$\displaystyle \Rightarrow y=9 \qquad [\because y>0]$
$\displaystyle \therefore \text{The legs of the right triangle are }3\text{ cm and }9\text{ cm.}$
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$\displaystyle \textbf{Question 4: }\text{A pole has to be erected at a point on the boundary of a circular park }$
$\displaystyle \text{of diameter 13 metres in such a way that the difference of its distances from two diametrically }$
$\displaystyle \text{opposite fixed gates }A\text{ and }B\text{ on the boundary is }7\text{ metres. Is it possible}$
$\displaystyle \text{to do so? If yes, at what distances from the two gates should the pole be erected?}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the pole be erected at point }P\text{ on the boundary of the circular park.}$
$\displaystyle \text{Since }AB\text{ is a diameter, }\angle APB=90^\circ.$
$\displaystyle \text{Let the shorter distance from the pole to a gate be }x\text{ m.}$
$\displaystyle \therefore \text{The longer distance}=(x+7)\text{ m.}$
$\displaystyle \text{Also, }AB=13\text{ m.}$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle x^2+(x+7)^2=13^2$
$\displaystyle \Rightarrow x^2+x^2+14x+49=169$
$\displaystyle \Rightarrow 2x^2+14x-120=0$
$\displaystyle \Rightarrow x^2+7x-60=0$
$\displaystyle \Rightarrow (x+12)(x-5)=0$
$\displaystyle \Rightarrow x=-12\text{ or }x=5$
$\displaystyle \text{Since distance cannot be negative, }x=-12\text{ is rejected.}$
$\displaystyle \therefore x=5,\quad x+7=12.$
$\displaystyle \therefore \text{Yes, it is possible to erect the pole.}$
$\displaystyle \therefore \text{It should be }5\text{ m from one gate and }12\text{ m from the other gate.}$
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