Question

the side of a square exceeds the side of anither square by 4 cm and the sum of the areas of the two square is 400 sq. cm . find the dimensions of the squares one more question can we solve this using linear equations ???

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Let the side of the smaller square = x cm

Then the side of larger square = (x+4)cm

Sum of the areas of the two squares = $\sf{{400cm}^{2}}$

$\longrightarrow$$\sf\red{ {x}^{2} + {(x + 4)}^{2} = 400}$

$\longrightarrow$$\sf\red{ {x}^{2} + {(x}^{2} + 8x + 16 )= 400}$

$\longrightarrow$$\sf\red{{2x}^{2} + 8x - 384 = 0}$

$\longrightarrow$$\sf\red{{x}^{2} + 4x - 192 = 0}$

$\longrightarrow$$\sf\red{(x + 16)(x - 12) = 0}$

$\longrightarrow$$\sf\red{x = - 16 \: or \: x = 12}$

As side of square cannot be negative,

x ≠ -16 , so n = 12

Hence, side of smaller square = 12cm

and side of larger square = 12 + 4 = 16cm