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Sum of the areas of two …

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Sum of the areas of two squares is 400cm². If the difference of their perimeters is 16cm, find the sides of the two squares?

Posted by Charu Arora 5 years, 1 month ago

CBSE > Class 10 > Mathematics

2 answers

Manthan Singhal 5 years, 1 month ago

Square 1 side x cm Square 2 side y cm ATQ 4x - 4y = 16 4( x - y ) = 16 x- y = 4 x= y+ 4 x² + y² = 400 ( y+4)² + y² = 400 y² + 16 + 8y + y² = 400 2y² + 8y - 384 = 0 y² + 4y - 192 = 0 y² + 16y - 12y - 192 = 0 y( y + 16 ) - 12( y + 16 )= 0 (y - 12 )( y + 16 ) = 0 y-12 = 0 y+16=0 y= 12 cm y= -16 cm Rejected bcz side Cannot be negative x = y + 4 x = 12 + 4 = 16 cm Thus sides are 12 cm and 16 cm

Gaurav Seth 5 years, 1 month ago

Let the sides of the two squares be x cm and y cm where x > y.

Then, their areas are x² and y² and their perimeters are 4x and 4y.

By the given condition:

x² + y² = 400...(1)

and 4x - 4y = 16

4(x - y) = 16 x - y = 4

x = y + 4... (2)

Substituting the value of x from (2) in (1), we get:

(y + 4)² + y² = 400

y² + 16 + 8y + y² = 400

2y² + 16 + 8y = 400

y² + 4y - 192 = 0

y² + 16y - 12y - 192 = 0

y(y + 16) - 12 (y + 16) = 0

(y + 16) (y - 12) = 0

y = -16 or y = 12

Since, y cannot be negative, y = 12.

So, x = y + 4 = 12 + 4 = 16

Thus, the sides of the two squares are 16 cm and 12 cm.

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