Question

The side of a square exceeds the side of the another square by 3 cm and the sum of the areas of the two

squares is 225 sq. cm. Find the dimensions of the square. ​

Answer 1

SOLUTION :-

Let,

Side of the first square = x

Side of the second square = x + 3

Area of the first square = x²

Area of the second square = ( x + 3 )²

According to the question,

$\longrightarrow\sf x^2+(x+3)^2= 225 \\\\\longrightarrow\sf x^2+x^2+3^2+6x=225 \\\\\longrightarrow\sf2x^2+6x+9 = 225 \\\\\longrightarrow 2x^2+6x- 216=0 \\\\\longrightarrow x^2+3x-108 =0 \\\\\longrightarrow x^2-9x+12x-108 = 0 \\\\\longrightarrow x(x-9)+12(x-9)=0 \\\\\longrightarrow (x+12)(x-9)= 0 \\\\\longrightarrow \bf x = -12 \ , \ x = 9$

Side of the square cannot be negative.

Side of the first square = 9 cm

Side of the second square = 12 cm

Answer 2

Given ,

• The side of a square exceeds the side of the another square by 3 cm

• The sum of the areas of the two squares is 225 cm²

Let

• The side of one square = " a "
• The side of other square = " a' "

According to the question ,

➡a - a' = 3

➡a' = (a - 3) --- (i)

And

(a)² + (a')² = 225

(a)² + (a - 3)² = 225

(a)² + (a)² + (3)² - 2(a)(3) = 225

2(a)² + 9 - 6a = 225

2(a)² - 6a - 216 = 0

(a)² - 3a - 108 = 0

(a)² - 12a + 9a - 108 = 0

a(a - 12) + 9(a - 12) = 0

(a + 9)(a - 12) = 0

a = -9 cm or a = 12 cm

Put a = 12 cm in eq (i) , we get

a' = 12 - 3 = 9

$\therefore$ The dimensions of squares are

• 12 cm and 9 cm