117 = 13 {tex}\times{/tex} 3 {tex}\times{/tex} 3
65 = 13 {tex}\times{/tex} 5
HCF (117, 65) = 13
LCM(117,65) = 13 {tex}\times{/tex} 5 {tex}\times{/tex} 3 {tex}\times{/tex} 3 = 585

Here is given that:
{tex}HCF =65m-117 {/tex}

{tex}13=65m-117{/tex}

{tex}65m=130{/tex}
m = {tex}\frac { 130 } { 6 5 } ={/tex}2

Here, {tex}\frac{AB}{DF}=\frac{4}{6}=\frac{2}{3}{/tex} , {tex}\frac{BC}{EF}=\frac{5}{7.5}=\frac{2}{3}{/tex} , {tex}\frac{AC}{DE}=\frac{3}{4.5}=\frac{2}{3}{/tex}

As, {tex}\frac{AB}{DF}=\frac{BC}{EF}=\frac{AC}{DE}{/tex}

So, {tex}\Delta ABC \sim \Delta DFE{/tex} [by SSS similarity criterion]

Hence ABC and DFE are similar triangles, but no other pairs of triangles in the given figure are similar.

D = b² - 4ac
{tex}= ( - 9 ( a + b ) ) ^ { 2 } - 4 \times 9 \times \left( 2 a ^ { 2 } + 5 a b + 2 a b ^ { 2 } \right){/tex}
= 81(a + b)² - 36(2a² + 5ab + 2b²)
= 9[9(a² + b² + 2ab - 8a² - 20ab - 8b²)]
= 9[a² + b² - 2ab]
= 9(a - b)²
{tex}x = \frac { - b \pm \sqrt { D } } { 2 a } = \frac { 9 ( a + b ) \pm \sqrt { 9 ( a - b ) ^ { 2 } } } { 2 \times 9 }{/tex}
{tex}{ \Rightarrow x = 3 \frac { [ 3 ( a + b ) \pm ( a - b ) ] } { 2 \times 9 } }{/tex}
{tex}{ \Rightarrow x = \frac { ( 3 a + 3 b ) \pm ( a - b ) } { 6 } }{/tex}
{tex}\Rightarrow x = \frac { 3 a + 3 b + a - b } { 6 } \text { or } x = \frac { 3 a + 3 b - a + b } { 6 }{/tex}
{tex}\Rightarrow x = \frac { 4 a + 2 b } { 6 } \text { or } x = \frac { 2 a + 4 b } { 6 }{/tex}
{tex}\Rightarrow x = \frac { 2 a + b } { 3 } \text { or } x = \frac { a + 2 b } { 3 }{/tex}

LHS = ( 3 - √5 )/(3 + 2√5 )

= [(3-√5)(2√5-3)]/[(2√5+3)(2√5-3)]

= [6√5 - 9 -10 + 3√5 ]/[ ( 2√5 )² - 3² ]

= ( 9√5 - 19 )/( 20 - 9 )

= ( 9√5 - 19 )/11

= ( 9/11 )√5 - 19/11 ---( 1 )

= a√5 - b ----------( 2 ) [ RHS ]

from ( 1 ) and ( 2 ) ,

a = 9/11 ,

b = 19/11

Let {tex} \alpha{/tex} and {tex} \frac { 1 } { \alpha }{/tex} be the zeros of (a² + 9)x² + 13x + 6a.
Then, we have
{tex} \alpha \times \frac { 1 } { \alpha } = \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ 1 = {tex} \frac { 6 a } { a ^ { 2 } + 9 }{/tex}
⇒ a² + 9 = 6a
⇒ a² - 6a + 9 = 0
⇒ a² - 3a - 3a + 9 = 0
⇒ a(a - 3)  - 3(a - 3) = 0
⇒ (a - 3) (a - 3) = 0
⇒ (a - 3)² = 0
⇒ a - 3 = 0
⇒ a = 3
So, the value of a in given polynomial is 3.

Let the speeds of the cars from A and B be x km/hr and y km/hr respectively.
Case I When the two cars move in the same direction:
In this case, let them meet at M after 5 hours

Then, AM = 5x km.
And, BM = 5y km.
{tex}\therefore{/tex} AM -BM = AB
{tex}\Rightarrow{/tex} 5x - 5y = 100
{tex}\Rightarrow{/tex} 5(x - y) = 100
{tex}\Rightarrow{/tex} x - y = 20 ...............(i)
Case II When the two cars move in the opposite directions:

Let one car move from A to B  and let the other move from B to A.
Let them meet at N after 1 hour.
Then, AN = x km and BN = y km.
{tex}\therefore{/tex}  AN + BN = AB
{tex}\Rightarrow{/tex} x + y = 100. ............. (ii)
Adding (i) and (ii), we get
(x - y) + ( x + y) = 100 + 20
{tex}\Rightarrow{/tex} x - y + x + y = 120
{tex}\Rightarrow{/tex} 2x = 120
{tex}\Rightarrow{/tex} x = 60.
Putting x = 60 in (ii), we get
x+ y = 100
{tex}\Rightarrow{/tex} 60 + y = 100
{tex}\Rightarrow{/tex} y= 100 - 60
{tex}\Rightarrow{/tex} y = 40.
So, x = 60 and y = 40
Hence , speed of the car from A = 60 km/hr,
and speed of the car from B = 40 km/hr

We have,
{tex}\sin \theta = \frac { \text { Perpendicular } } { \text { Hypotenuse } } = \frac { a ^ { 2 } - b ^ { 2 } } { a ^ { 2 } + b ^ { 2 } }{/tex}
So, Let us draw a right triangle ABC in which {tex}\angle B{/tex} is right angle, we have
Perpendicular = BC = a² - b²  Hypotenuse = AC = a² + b²  and, {tex}\angle B A C = \theta{/tex}
By Pythagoras theorem, we have
AC² = AB² + BC²
{tex}\Rightarrow \quad \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } = A B ^ { 2 } + \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B ^ { 2 } = \left( a ^ { 2 } + b ^ { 2 } \right) ^ { 2 } - \left( a ^ { 2 } - b ^ { 2 } \right) ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B ^ { 2 } = \left( a ^ { 4 } + b ^ { 4 } + 2 a ^ { 2 } b ^ { 2 } \right) - \left( a ^ { 4 } + b ^ { 4 } - 2 a ^ { 2 } b ^ { 2 } \right){/tex}
{tex}\Rightarrow \quad A B ^ { 2 } = 4 a ^ { 2 } b ^ { 2 } = ( 2 a b ) ^ { 2 }{/tex}
{tex}\Rightarrow \quad A B = 2 a b{/tex} 
Now, Let{tex}\angle B A C = \theta{/tex} 
We have
Base = AB = 2ab, Perpendicular {tex}= \mathrm { BC } = a ^ { 2 } - b ^ { 2 }{/tex}, Hypotenuse = AC = a² + b²
Therefore, {tex}\quad \cos \theta = \frac { \text { Base } } { \text { Hypotenuse } } = \frac { 2 a b } { a ^ { 2 } + b ^ { 2 } }{/tex}, {tex}\tan \theta = \frac { \text { Perpendicular } } { \text { Base } } = \frac { a ^ { 2 } - b ^ { 2 } } { 2 a b }{/tex}
{tex}\quad cosec \;\theta = \frac { \text { Hypotenuse } } { \text { Perpendicular } } = \frac { a ^ { 2 } + b ^ { 2 } } { a ^ { 2 } - b ^ { 2 } } , \quad \sec \theta = \frac { \text { Hypotenuse } } { \text { Base } } = \frac { a ^ { 2 } + b ^ { 2 } } { 2 a b }{/tex}
and, {tex}\cot \theta = \frac { \text { Base } } { \text { Perpendicular } } = \frac { 2 a b } { a ^ { 2 } - b ^ { 2 } }{/tex}

Composite Numbers -

A composite number is a positive integer that has a factor other than 1 and itself. For instance, 12 is a composite number because it is a positive integer and it has factors other than 1 and itself. 2, 3, 4, 6 are its factors other than 1 and itself. All even numbers greater than 2 are composite numbers. 2 is a prime number. The smallest composite number is 4.

Prime numbers:

Numbers which have only two factors 1 and the number itself are called prime numbers.

E.g 2 ,3 ,5, 7 ,11 ,13, 17 and so on are all prime numbers.

Of the first 100 numbers 25 are prime numbers.

Only 2 is an even prime number all prime numbers are odd.

Co prime numbers:

Numbers which do not have any common factor between them other than one are called coprime numbers.

Two natural numbers( not necessarily Prime) are coprime if their HCF is 1.
For e.g 
(1,2) , (1,3), (3,4), (3,10), (3,8), (5,6), (17,23)

even two composite numbers can also be Prime. for e.g
(16,25), (84,65)

Let side of first square be x and the side of another square be y.
difference in perimeter of two squares:-
4x-4y=16
4(x-y)=16
x-y=16/4
therefore, x-y=4
and x=4+y---take this(1)
 
sum of areas of two squares:-
 x²+y²=400
Put (1) here
(4+y)²+y²=400
(4)²+(y)²+2×4×y+y²=400
16+y²+8y+y²=400
2y²+8y+16-400=0
2y²+8y-384=0
2(y²+4y-192)=0 (here we have taken 2 common and take to another side. then it becomes:-)
y²+4y-192=0
we have done with factorisation method:-
y²-12y+16y-192=0
y(y-12)+16(y-12)=0
(y+16) (y-12)=0
either:-                             |             or:-
y+16=0                            |                   y-12=0
y= -16                              |                   y=12

we will take y=12 because y being side cannot be negative.
So, y=12
put y=12 in (1)
x=4+y
x=4+12
therefore, x=16
and hence, Side of first square= x = 16
and Side of another square= y = 12.