Question

# a) The radius of a roller, $1.2\;{\rm{m}}$ long is $0.42{\rm{ }}m$. If it makes $200$ complete revolutions, find the area of the playground.b) Using the quadratic formula, solve: $\dfrac{{x - 1}}{{x - 2}} \div \dfrac{{x - 2}}{{x - 3}} = 4$c) If $A = \left[ {\begin{array}{*{20}{c}}4&{ - 5}\\3&2\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&4\end{array}} \right]$, find $6A - 3B.$

Hint:Let us consider, the radius and the length of the roller is $r\,\&\, h$ respectively. So, the lateral surface of the roller is $2\pi rh$. Using the formula we will find the area covered by roller in one revolution, and with the found area we multiply the total number of revolutions and get the area of the playground.

$\text{a})$
It is given that; the radius of roller $0.42{\rm{ }}m$ and the length of the roller is $1.2{\rm{ }}m$.
The roller makes $200$ complete revolutions, to cover the playground.
The roller covers the area of its lateral surface area in one revolution.
Let us consider, the radius and the length of the roller is $r\,\&\, h$ respectively. So, the lateral surface of the roller is $2\pi rh$.
Substitute, the values of the radius and the length of the roller in the above formula we get,
The lateral surface area is $2 \times \dfrac{{22}}{7} \times 0.42 \times 1.2 = 0.144{\rm{ }}sq.{\rm{ }}m$
So, in one revolution it covers $0.144{\rm{ }}sq.{\rm{ }}m$ of the area of the playground.
Therefore to find the area of ground, we should multiply the area of the roller by the number of revolutions to cover the ground.
So, the area of the playground is $0.144 \times 200\,sq.{\rm{ }}m$.
On simplifying the above multiplication we get,
The area of the playground is $28.8{\rm{ }}sq.{\rm{ }}m$
Hence, the area of the playground is $28.8{\rm{ }}sq.{\rm{ }}m$.

Note:The roller covers the area of its lateral surface area in one revolution.Students should remember the formula of lateral surface of cylindrical objects to solve the types of questions.

$\text{b})$
Hint:The solution of any quadratic equation $a{x^2} + bx + c = 0$ is $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. This is known as Sreedharacharya’s formula. We first solve the left hand side of the given equation and form a quadratic equation and solve it using Sreedharacharya’s formula we get the value of x.

To find the solution of $\dfrac{{x - 1}}{{x - 2}} \div \dfrac{{x - 2}}{{x - 3}} = 4$
When we divide two fractions we can write them as a multiplication of numerator and the inverse of denominator.
That is $\dfrac{{\dfrac{a}{b}}}{{\dfrac{c}{d}}} = \dfrac{a}{b} \times \dfrac{d}{c}$
Hence the left hand side of the given equation is,
$\dfrac{{\dfrac{{x - 1}}{{x - 2}}}}{{\dfrac{{x - 2}}{{x - 3}}}} = \dfrac{{x - 1}}{{x - 2}} \times \dfrac{{x - 3}}{{x - 2}}$
On substituting it in the given equation we get,
$\dfrac{{x - 1}}{{x - 2}} \times \dfrac{{x - 3}}{{x - 2}} = 4$
Let us simplify the numerator and denominator we get,
$\dfrac{{{x^2} - 4x + 3}}{{{x^2} - 4x + 4}} = 4$
By cross-multiplication we get,
${x^2} - 4x + 3 = 4{x^2} - 16x + 16$
Let us simplify the above equation we get,
$3{x^2} - 12x + 13 = 0$
By applying Sreedharacharya’s formula we get,
$x = \dfrac{{-(-12) \pm \sqrt {{{12}^2} - 4 \times 12 \times 13} }}{{2.3}}$
Let us simplify the terms in the square root we get,
$x = \dfrac{{12 \pm \sqrt {144 - 624} }}{{6}}$
On simplification we get the following roots,
$x = \dfrac{{12 + \sqrt { - 480} }}{{6}}\,\&\,x = \dfrac{{12 - \sqrt { - 480} }}{{6}}$

Note:There will be two solutions for a quadratic equation.Students should remember the formula of Sridharacharya i.e $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ in order to find solution of quadratic equation.In above question we can further simplify the numerator part where we get complex values.

$\text{c})$
Hint:We know that $A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right],kA = \left[ {\begin{array}{*{20}{c}}{ka}&{kb}\\{kc}&{kd}\end{array}} \right]$using this we will find the values of 6A and 3B initially and subtract them to get the required answer.

It is given that; $A = \left[ {\begin{array}{*{20}{c}}4&{ - 5}\\3&2\end{array}} \right]$ and $B = \left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&4\end{array}} \right]$
We have to find the value of $6A - 3B.$
The value of $6A = 6\left[ {\begin{array}{*{20}{c}}4&{ - 5}\\3&2\end{array}} \right]$
We know that $A = \left[ {\begin{array}{*{20}{c}}a&b\\c&d\end{array}} \right],kA = \left[ {\begin{array}{*{20}{c}}{ka}&{kb}\\{kc}&{kd}\end{array}} \right]$using this in the above term we get,
$6\left[ {\begin{array}{*{20}{c}}4&{ - 5}\\3&2\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{6(4)}&{ - 5(6)}\\{3(6)}&{2(6)}\end{array}} \right]$
$\left[ {\begin{array}{*{20}{c}}{6(4)}&{ - 5(6)}\\{3(6)}&{2(6)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{24}&{ - 30}\\{18}&{12}\end{array}} \right]$
Now let us find the value of 3B,
The value of $3B = 3\left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&4\end{array}} \right]$
$3\left[ {\begin{array}{*{20}{c}}2&{ - 3}\\{ - 1}&4\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{2(3)}&{ - 3(3)}\\{ - 1(3)}&{4(3)}\end{array}} \right]$
On simplifying we get,
$\left[ {\begin{array}{*{20}{c}}{2(3)}&{ - 3(3)}\\{ - 1(3)}&{4(3)}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}6&{ - 9}\\{ - 3}&{12}\end{array}} \right]$
Let us find$6A - 3B$, we get,
$6A - 3B = \left[ {\begin{array}{*{20}{c}}{24}&{ - 30}\\{18}&{12}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}6&{ - 9}\\{ - 3}&{12}\end{array}} \right]$
On further solving we get,
$\left[ {\begin{array}{*{20}{c}}{24}&{ - 30}\\{18}&{12}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}6&{ - 9}\\{ - 3}&{12}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{18}&{ - 21}\\{21}&0\end{array}} \right]$
Hence,
$6A - 3B = \left[ {\begin{array}{*{20}{c}}{18}&{ - 21}\\{21}&0\end{array}} \right]$

Note:Multiplying any constant value with any matrix means multiplying each element of the matrix with that constant.