Question

the sides of the square exceeds the side of another square by 4 cm and the sum of areas of the two squares of 400 cm² . find the dimensions.​

Answer 1

Answer:

let the side of 1st square be X

so side of 2nd square = X + 4

area of square = side2

sum of area of 2 squares is 400

X² + (X+4)² = 400

x² + x² + 16 = 400

2x² + 16 = 400

2x² = 400-16

x² = 384/2

x² = 192

x = ✓192

x = 14

So first square side = 14cm

second square side = 14+4

= 18

So the side of first and second square are 14cm and 18cm respectively

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Answer 2

Let S1 and S2 be two squares. Let the side of the square S2 be x cm in length.

Then, the side of square S1 is (x+4)cm

Area of square S1=(x+4)²

and Area of square S2 =x²

It is given that,

(Area of square S1)+(Area of square S2 )=400cm ²

⇒(x+4)² +x²=400

⇒(x²+8x+16)+x =400

⇒2x²+8x−384=0

⇒x²+4x−192=0

⇒x²+16x−12x−192=0

⇒(x+16)(x−12)=0

⇒x=12 or x=−16

As the length of the side of a square cannot be negative. Therefore, x=12

Side of square S1 =x+4=12+4=16cm

Side of square S2 =12cm

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