Determine the value of 'a' which satisfies the equation 9a + 40a = 41a

l₁ = 20 cm, A₁ = 200 sq cm
? b₁ = 200/20 = 10 cm
Now, A₂ = 200 x 6/5 = 240 sq cm
b₂ = 10 cm
? l₂ = 240/10 = 24 cm

? Perimeter of new rectangle = 2(l₂ + b₂)
= 2(24 + 10) = 2 x 34 = 68 cm

Let the third side be x.
According to the question,
x + 20 = 26
? x = 26 - 20 = 6 cm
? Each equal side = 20/2 = 10 cm

Perimeter of rhombus
= 2 ? d₁ ² + d₂ ²
= 2 ?(4.8)² + (1.4)²
= 2 ?23.04 + 1.96
= 2 ?25
= 2 x 5 = 10 cm

Average = ( 3 + 6 + 9 + 12 + 15 + 18 + 21 + 24 + 27 ) / 9
= 3(1+2+3+4+5+6+7+8+9)/9
= ( 45 / 3 )
= 15

Let us assume the second number is a and the difference between consecutive numbers is d.
According to Arithmetic progression,
First number = a - d
Second number = a
Third number = a + d
According to question,
Sum of the all three numbers = 15
a - d + a + a + d = 15
3a = 15
a = 5
Again according to given question,
sum of square of the 3 numbers = 83
(a - d) ² + a ² + (a + d) ² = 83
apply the algebra formula
a ² + d ² - 2ad + a ² + a ² + d ² + 2ad = 83
3a ² + 2d ² = 83
Put the value of a in above equation.
3 x 5 ² + 2d² = 83
3 x 25 + 2d² = 83
75 + 2d² = 83
2d ² = 83 - 75
2d ² = 8
d ² = 8/2
d ² = 4
d = 2
Put the value of a and d in below equation.
First number = a - d = 5 - 2 = 3
Second number = a = 5
Third number = a + d = 5 + 2 = 7
The smallest number is 3.

The required number of combinations, when one fruit of each kind is taken
= ⁵C₁ x ⁴C₁ x ³C₁ = 5 x 4 x 3 = 60

Let the diagonals of the squares be 3x and 2x.
? Ratio of their areas = [(1/2) (3x²)] / [(1/2) (2x²)] = 9/4

Diagonal of square = ?2a [a = side]
4?2 = ?2 a
a = 4 cm
Now, area of square = a² = (4²) = 16
Side of a square whose area is 2 x 16.
a₁² = 32
? a₁ = ?32 ?a₁4?2

Now, diagonal of new square = ?2a
= ?2x 4 ?2
= 8 cm

We know that in any triangle
"the sum of two sides is always greater than its third side" and
"the difference of two sides is always less than its third side".
(i) 2 + 3 is not greater than 5
(ii) |5 - 2| not less than 3