Question

plz do fast .........................​

Answer 1

Answer: 5 metre and 11 metres respectively.

Step by step explanation:

Let the sides of two squares be x and y respectively.

Thus, as per the first condition,

9x = 4y + 1 __(eq.1)

As per the next condition,

6y² = 29x² + 1 __(eq.2)

x = (4y + 1)/9 ___(eq.3) [From eq.(1)]

Putting the above eq. in eq.(2)

6y² = 29 [(4y + 1)/9]² + 1

6y² = 29 [(4y + 1)² / 81] + 1

486y² = 29 (4y + 1)² / 81 + 81

486y² = 29 (16y² + 8y + 1)+ 81

486y² = 464y² + 232y + 29 + 81

486y² - 464y² = 232y + 110

22y² - 232y + 110 = 0

Dividing the eq. by 2, on both sides;

11y² - 116y - 55 = 0

Solving the above equation:

$\tt{y = \frac{ - b \pm \sqrt{b {}^{2} - 4ac }}{2a}}$

$\tt{y = \frac{ 116 \pm \sqrt{ - (116) {}^{2} - 4 \times 11 \times ( - 55) }}{22}}$

$\tt{y = \frac{ 116 \pm \sqrt{ 13456- 2420 }}{22}}$

$\tt{y = \frac{ 116 \pm \sqrt{ 15876 }}{22}}$

$\tt{y = \frac{ {116 \: \pm \: 126 }}{22}}$

$\tt{y = \frac{{ 116 + 126 }}{22}} \: \: or \: \: \tt{y = \frac{{ 116 - 126 }}{22}}$

$\tt{y = \frac{{ 242 }}{22}} or \: y = \frac{{ - 10 }}{22}{}$

$\tt{y = 11 \: \: or \: \: y = - 0.45}$

y = 11 (as length of any side can't be negative.)

Putting the value of y in eq. 1

9x = 4y + 1

9x = 44 + 1

9x = 45

x = 5

Thus,

The value of x is 5 whereas that of y is 11

Therefore, Side of first square is 5 metres, and that of second square is 11 metres.

Answer 2

Refer to the attachment.