Similarity - Maps and Models

A map of a square plot of land is drawn to a scale of $1 : 2500$. If the area of the plot on the map is $72 \; cm^2$, find

1. the actual area of the land in square kilometer;
2. the length of the diagonal in the actual plot of land in meters.

1. Scale of square plot of land $= 1 : 2500$
   
   

   On Map	Actual
   $1 \; cm$	$2500 \; cm$
   $1 \; cm^2$	$625 \times 10^4 \; cm^2$
   $72 \; cm^2$	$x \; cm^2$

   
   
   $\therefore \;$ $x = 72 \times 625 \times 10^4 = 45000 \times 10^4 = 45 \times 10^7 \; cm^2$
   
   i.e. $\;$ Actual area of the land $= 45 \times 10^7 \; cm^2$
   
   Now, $\;$ $1000 \; m \equiv 1 \; km$
   
   $\therefore \;$ $10^6 \; m^2 \equiv 1 \; km^2$
   
   $\left[\text{Note: } 1 \; m \equiv 100 \; cm \implies 1 \; m^2 = 10^4 \; cm^2 \right]$
   
   i.e. $\;$ $10^6 \times 10^4 \; cm^2 \equiv 1 \; km^2$
   
   i.e. $\;$ $10^{10} \; cm^2 \equiv 1 \; km^2$
   
   $\therefore \;$ $45 \times 10^7 \; cm^2 \equiv \dfrac{45 \times 10^7}{10^{10}} = 45 \times 10^{-3} = 0.045 \; km^2$
   
   $\therefore \;$ Actual area of the land in square kilometer $= 0.045 \; km^2$



2. Let the length of the side of the plot on the map be $= s \; cm$
   
   Then, area of the square plot on the map $= s^2 \; cm^2$
   
   Given: Area of the plot on the map $= 72 \; cm^2$
   
   i.e. $\;$ $s^2 = 72$ $\implies$ $s = \sqrt{72} \; cm$
   
   $\therefore \;$ Length of the diagonal of the square plot on the map
   
   $= \ell = \sqrt{\left(\sqrt{72}\right)^2 + \left(\sqrt{72}\right)^2} = \sqrt{72 + 72} = \sqrt{144} = 12 \; cm$
   
   

   On Map	Actual
   $1 \; cm$	$2500 \; cm$
   $12 \; cm$	$x \; cm$

   
   
   $\therefore \;$ $x = 12 \times 2500 = 30000 = 3 \times 10^4 \; cm$
   
   i.e. $\;$ length of the diagonal in the actual plot of land $= 3 \times 10^4 \; cm$
   
   Now, $\;$ $100 \; cm \equiv 1 \; m$
   
   $\therefore \;$ $3 \times 10^4 \; cm \equiv \dfrac{3 \times 10^4}{10^2} = 3 \times 10^2 = 300 \; m$
   
   $\therefore \;$ length of the diagonal (in meter) in the actual plot of land $= 300 \; m$

Trigonometry

A man in a boat rowing away from a lighthouse $150 \; m$ high takes $1$ minute to change the angle of elevation of the top of the lighthouse from $60^\circ$ to $45^\circ$. Find the speed of the boat in meter per minute. [Given: $\sqrt{3} = 1.732$]

$AB =$ Lighthouse of height $150 \; m$

$C =$ First point of observation

$D =$ Second point of observation

$CD =$ Distance traveled by the boat in $1$ minute

In $\triangle ABD$, $\;$ $\dfrac{AB}{BD} = \tan 45^\circ = 1$

$\implies$ $BD = AB = 150 \; m$

In $\triangle ABC$, $\;$ $\dfrac{AB}{BC} = \tan 60^\circ = \sqrt{3}$

$\implies$ $BC = \dfrac{AB}{\sqrt{3}} = \dfrac{150}{\sqrt{3}} = 50 \sqrt{3} = 86.6 \; m$

Now, $\;$ $CD = BD - BC = 150 - 86.6 = 63.4 \; m$

$\therefore \;$ Distance covered by the boat in $1$ minute $= 63.4 \; m$

$\therefore \;$ Speed of the boat $= 63.4 \; m/\text{min}$

Statistics

The mean of the following distribution is $24$. Find the missing frequency '$a$'.

Class Interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
Frequency	$10$	$6$	$a$	$12$	$5$

Class Interval	Mid value $x_i$	Frequency $f_i$	$f_i x_i$
$0 - 10$	$5$	$10$	$50$
$10 - 20$	$15$	$6$	$90$
$20 - 30$	$25$	$a$	$25 a$
$30 - 40$	$35$	$12$	$420$
$40 - 50$	$45$	$5$	$225$
		$\Sigma f_i = 33 + a$	$\Sigma f_i x_i =785 + 25 a$

Mean $= \dfrac{\Sigma f_i x_i}{\Sigma f_i} = \dfrac{785 + 25 a}{33 + a}$

Given: $\;$ Mean $= 24$

i.e. $\;$ $\dfrac{785 + 25 a}{33 + a} = 24$

i.e. $\;$ $785 + 25 a = 792 + 24 a$

$\implies$ $a = 7$

Matrices

Given: $\;$ $A = \begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix}$, $\;$ $B = \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix}$ $\;$ and $\;$ $C = \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$. Find matrix $X$ such that $A + 2 X = 2B + C$.

Let $X = \begin{bmatrix} p & q \\ r & s \end{bmatrix}$

Given: $\;$ $A + 2 X = 2 B + C$

i.e. $\;$ $\begin{bmatrix} 2 & -6 \\ 2 & 0 \end{bmatrix} + 2 \begin{bmatrix} p & q \\ r & s \end{bmatrix} = 2 \begin{bmatrix} -3 & 2 \\ 4 & 0 \end{bmatrix} + \begin{bmatrix} 4 & 0 \\ 0 & 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} 2 + 2p & -6 + 2q \\ 2 + 2r & 0 + 2s \end{bmatrix} = \begin{bmatrix} -6 + 4 & 4 + 0 \\ 8 + 0 & 0 + 2 \end{bmatrix}$

i.e. $\;$ $\begin{bmatrix} 2 + 2p & -6 + 2q \\ 2 + 2r & 2s \end{bmatrix} = \begin{bmatrix} -2 & 4 \\ 8 & 2 \end{bmatrix}$

Two matrices are equal when all their corresponding elements are equal.

i.e. $\;$ $2 + 2p = -2$ $\implies$ $2 p = -4$ $\implies$ $p = -2$

$-6 + 2q = 4$ $\implies$ $2 q = 10$ $\implies$ $q = 5$

$2 + 2 r = 8$ $\implies$ $2 r = 6$ $\implies$ $r = 3$

$2 s = 2$ $\implies$ $s = 1$

$\therefore \;$ $X = \begin{bmatrix} -2 & 5 \\ 3 & 1 \end{bmatrix}$

Arithmetic Progression

Find the sum of the terms of the A.P $\;$ $4, \; 9, \; 14, \; \cdots \; 89$. Also find the $4^{th}$ term from the end.

Given A.P is: $\;$ $4, \; 9, \; 14, \; \cdots \; 89$

First term $= a = 4$

Last term $= n^{th}$ term $= t_n = \ell = 89$

Common difference $= d = 5$

Let the number of terms in the given A.P be $= n$

$n^{th}$ term of A.P $= t_n = a + \left(n - 1\right)d$

i.e. $\;$ $89 = 4 + \left(n - 1\right) \times 5$

i.e. $\;$ $n - 1 = \dfrac{85}{5} = 17$ $\implies$ $n = 18$

$\therefore \;$ Number of terms in the given A.P is $= n = 18$

Sum of $n$ terms of an A.P $= S_n = \dfrac{n}{2} \left[2a + \left(n - 1\right)d\right]$

Here, $\;$ $S_{18} = \dfrac{18}{2} \left[2 \times 4 + \left(18 - 1\right) \times 5\right]$

i.e. $\;$ $S_{18} = 9 \left[8 + 85\right] = 837$

$n^{th}$ term from the end of an A.P $= \ell - \left(n - 1\right)d$

Here, $4^{th}$ term from the end of the A.P $= 89 - \left(4 - 1\right) \times 5 = 89 - 15 = 74$

Trigonometry

Prove that $\;$ $\dfrac{\left(\cot A - \text{cosec }A\right)^2 + 1}{\sec A \left(\text{cosec }A - \cot A\right)} = 2 \cot A$

$\begin{aligned} LHS & = \dfrac{\left(\cot A - \text{cosec }A\right)^2 + 1}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{\cot^2 A + \text{cosec}^2 A - 2 \cot A \text{cosec }A + 1}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & \left[\text{Note: }1 + \cot^2 \theta = \text{cosec}^2 \theta\right] \\\\ & = \dfrac{\text{cosec}^2 A + \text{cosec}^2 A - 2 \cot A \text{cosec }A}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{2 \; \text{cosec}^2 A - 2 \cot A \text{cosec }A}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{2 \; \text{cosec }A \left(\text{cosec }A - \cot A\right)}{\sec A \left(\text{cosec }A - \cot A\right)} \\\\ & = \dfrac{2 \; \text{cosec }A}{\sec A} \\\\ & = \dfrac{2 / \sin A}{1 / \cos A} \\\\ & = \dfrac{2 \; \cos A}{\sin A} \\\\ & = 2 \cot A = RHS \end{aligned}$

Hence proved.

Quadratic Equations

For what value of '$p$' does the equation $x^2 - 2px + \left(7p - 12\right) = 0$ have equal roots?

Given quadratic equation: $\;$ $x^2 - 2px + \left(7p - 12\right) = 0$

Comparing with the standard equation $\;$ $ax^2 + bx + c = 0$ $\;$ gives

$a = 1$, $\;$ $b = -2 p$, $\;$ $c = 7p - 12$

A quadratic equation has equal roots when Discriminant $= \Delta = b^2 - 4ac = 0$

Here, discriminant $\;$ $\Delta = \left(-2p\right)^2 - 4 \times 1 \times \left(7p - 12\right)$

i.e. $\;$ $\Delta = 4 \left(p^2 - 7 p + 12\right)$

$\therefore \;$ For equal roots,

$4 \left(p^2 - 7p + 12\right) = 0$

i.e. $\;$ $p^2 - 7p + 12 = 0$

i.e. $\;$ $p^2 - 3p - 4p + 12 = 0$

i.e. $\;$ $p \left(p - 3\right) - 4 \left(p - 3\right) = 0$

i.e. $\;$ $\left(p - 3\right) \left(p - 4\right) = 0$

$\implies$ $p = 3$ $\;$ or $\;$ $p = 4$

$\therefore \;$ The given quadratic equation has equal roots when $p = 3$ or $p = 4$.