Sum of the areas of two squares is 468 m². If the difference of their perimeters are 24 m, find the sides of the two squares.

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  Q) Sum of the areas of two squares is 468 m². If the difference of their perimeters are 24 m, find the sides of the two squares. Solution) Let the sides of the two squares be x m and y m. Thus, their perimeters will be 4x and 4y and areas will be x² and y². Hence, according to question – =4x-4y=24 ⇒x-y=6 ⇒x=y+6 And x²+y² = 468 Substituting value of x- (y+6)²+y²=468 ⇒36+y²+12y+y²=468 ⇒2y²+12y-432=0 ⇒y²+6y-216=0 ⇒y²+18y-12y-216=0 ⇒y(y+18)-12(y+18)=0 ⇒(y+18)(y-12)=0 ⇒y=-18 or 12 Since, side cannot be negative. Therefore, the sides of the square are 12 m and (12+6)m i.e 18 m.
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The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

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 Q) The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.


Solution) Let the shorter side of the rectangle be x m .

Thus , Larger side of the rectangle will be ( x + 30 ) m .

Diagonal of the rectangle be√x² + ( x + 30 ) ²

 Hence , according to question x² + ( x + 30 ) ² = x + 60

⇒ x² + ( x + 30 ) ² = ( x + 60 ) ²
= x² + x² + 900 + 60x = x² + 3600 + 120x
= x² - 60x - 2700-0
=x² - 90x + 30x - 2700 = 0
=x ( x - 90 ) + 30 ( x - 90 ) = 0
⇒ ( x - 90 ) ( x + 30 ) = 0
 ⇒x = 90 , -30

Therefore , the length of the shorter side of rectangle is 90 m .
Hence , length of the larger side of the rectangle be 120 m .
Since , side cannot be negative .





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