Nine times the side of one square exceeds the perimeter of a second square by one meter and six times - Maths - Quadratic Equations

Question

Nine times the side of one square exceeds the perimeter of a
second square by one meter and six times the area of the second
square exceeds twenty nine times the area of the first square by
one square meter Find the side of each square

Ankush Jain · Accepted Answer

Let the side of first square be x m and that of second square be y m.

Given, nine times the side of one square exceeds the perimeter of a second square by one meter.

$\Rightarrow 9 x = 4 y + 1 \Rightarrow x = \frac{4 y + 1}{9} .... (1)$

Again, it is given that six times the area of the second square exceeds twenty nine times the area of the first square by one square meter.

$\Rightarrow 6 y^{2} = 29 x^{2} + 1 \Rightarrow 6 y^{2} = 29 (\frac{4 y + 1}{9})^{2} + 1 [u \sin g (1)] \Rightarrow 6 y^{2} = \frac{29 (16 y^{2} + 1 + 8 y) + 81}{81} \Rightarrow 486 y^{2} = 464 y^{2} + 29 + 232 y + 81 \Rightarrow 22 y^{2} - 232 y - 110 = 0 \Rightarrow 11 y^{2} - 116 y - 55 = 0 \Rightarrow 11 y^{2} - 121 y + 5 y - 55 = 0 \Rightarrow (y - 11) (11 y + 5) = 0 \Rightarrow y = 11, \frac{- 5}{11}$

Now, y ≠ -5/11, as the side of a square can't be negative.

$\Rightarrow y = 11 m$

On putting value of y = 11 in (1), we get

$x = \frac{4 \times 11 + 1}{9} = \frac{45}{9} = 5 m$

Nine times the side of one square exceeds the perimeter of a second square by one meter and six times the area of the second square exceeds twenty nine times the area of the first square by one square meter. Find the side of each square.