3 Marks Question

Question 13 Marks

Is it possible to design a rectangular park of perimeter $80 \ m$ and area $400 \ m^2$? If so, find its length and breadth.

Answer

Let the length and breadth of the park be l and b.
Perimeter = $2(l+ b) = 80$
$l + b = 40 ~or,~ b = 40 - l$
Area = $ l \times b = l(40 - l)$
$= 40l - l^{2} = 400$ Given
$l^{2}-40l+400=0$
Comparing this equation with $al^2 + bl + c = 0$, we obtain
$a = 1, b = -40, c = 400$
Discriminant $D = b^2 - 4ac = (-40)^2 - 4(1) (400) = 1600 - 1600 = 0$
$As b^2 - 4ac = 0$
Therefore, this equation has equal real roots and hence, this situation is possible.
Root of this equation,
$l=-\frac{b}{2 a}$
$l=-\frac{(-40)}{2(1)}=\frac{40}{2}=20$
Therefore, length of park, $l$ = 20m
And breadth of park, b = 40 - $l$ = 40 - 20 = 20m

View full question & answer→

Question 23 Marks

A pole has to be erected at a point on the boundary of a circular park of diameter $13$ metres in such a way that the difference of its distances from two diametrically opposite fixed gates $A$ and $B$ on the boundary is $7$ metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?

Answer

Let P be the position of the pole and A & B be the opposite fixed gates. Let, BP = x metres.
$\therefore$ AP = x + 7
In right triangle APB,

$AP^2 + BP^2 = AB^2$^
$\Rightarrow (x + 7)^2 + x^2 = 13^2$
$\Rightarrow x^2 + 49 + 14x + x^2 = 169$
$\Rightarrow 2x^2 + 14x + 49 - 169 = 0$
$\Rightarrow 2x^2 + 14x - 120 = 0$
$\Rightarrow2 (x^2 + 7x - 60) = 0$
$\Rightarrow x^2 + 7x - 60 = 0$
$\Rightarrow x^2 + 12x - 5x - 60 = 0$
$\Rightarrow x(x + 12) - 5(x + 12) = 0$
$\Rightarrow (x + 12)(x - 5) = 0$
Either x+12 = 0, then x = -12 which is not possible being negative or x - 5 = 0, then x = 5.
Thus P is at a distance of 5m from B and 5+7 = 12m from A.

View full question & answer→

Question 33 Marks

Find the roots of equation:$\frac { 1 } { x } - \frac { 1 } { ( x - 2 ) } = 3 , x \neq 0,2$

Answer

The given equation may be written as
$\frac { ( x - 2 ) - x } { x ( x - 2 ) } = 3$

$ \Rightarrow$$3x(x - 2) = -2$
$\Rightarrow$ $3x^2 - 6x + 2 = 0$ ..(i)
This equation is of the form $ax^2 + bx + c = 0$, where $a = 3,\ b = -6\ and\ c = 2$.
$\therefore$ $D = (b^2 - 4ac) = (-6)^2 - 4$ $\times$ 3 $\times$ 2 = 36 - 24 = 12 > 0.
So, the given equation has real roots.
Now, $\sqrt { D } = \sqrt { 12 } = 2 \sqrt { 3 }$
$\therefore \quad \alpha = \frac { - b + \sqrt { D } } { 2 a } = \frac { 6 + 2 \sqrt { 3 } } { 2 \times 3 } = \frac { 6 + 2 \sqrt { 3 } } { 6 } = \frac { 3 + \sqrt { 3 } } { 3 }$
$\beta = \frac { - b - \sqrt { D } } { 2 a } = \frac { 6 - 2 \sqrt { 3 } } { 2 \times 3 } = \frac { 6 - 2 \sqrt { 3 } } { 6 } = \frac { 3 - \sqrt { 3 } } { 3 }$
Hence, the required values of x are $\frac { ( 3 + \sqrt { 3 } ) } { 3 }$ and $\frac { ( 3 - \sqrt { 3 } ) } { 3 }$

View full question & answer→

MATHS 3 Marks Question

Test yourself on this topic