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If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)². - Mathematics

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प्रश्न

If y (3x – y) : x (4x + y) = 5 : 12. Find (x² + y²) : (x + y)².

योग

उत्तर

If y (3x – y) : x (4x + y) = 5 : 12
Find (x2 + y2) : (x + y)2 

`(3xy - y^2)/(4x^2 + xy) = (5)/(12)`

⇒ 36xy – 12y2 = 20x2 + 5xy
⇒ 20x2 + 5xy –  36xy + 12y2 = 0
⇒ 20x2 –  31xy + 12y2 = 0

⇒ `20x^2/y^2 - 31 xy/y^2 + (12y^2)/y^2` = 0  ...(Dividing by y2)

⇒ `20(x^2/y^2) - 31((xy)/y^2) + 12` = 0

⇒ `20(x/y)^2 -15(x/y) -16(x/y) + 12` = 0

⇒ `5(x/y)[4(x/y) -3] -4[4(x/y) -3]` = 0

⇒ `[4(x/y) -3][5(x/y) -4]` = 0

Either `[4(x/y) -3]` = 0,

then `4(x/y)` = 3

⇒ `x/y = (3)/(4)`

or `[5(x/y) -4]` = 0

then `5(x/y)` = 4

⇒ `x/y = (4)/(5)`

(a) When `x/y = (3)/(4)`
then (x2 : y2) : (x+ y)2

= `(x^2 + y^2)/(x + y)^2`

= `(x^2/y^2 + y^2/y^2)/(y^2/(1/y^2(x + y)^2`  ...(Dividing by y2)

= `(x^2/y^2 + 1)/((x/y + 1)`

= `(3/4)^2/(3/4 + 1)^2`

= `((9)/(16) + 1)/(7/4)^2`

= `(25/16)/(49/16)`

= `(25)/(16) xx (16)/(49)`

= `(25)/(49)`
∵ (x2 + y2) : (x + y)2 = 25 : 49

(b) When `x/y = (4)/(5)`, then

`(x^2  y^2)/((x + y)^2`

= `(x^2/y^2 + 1)/((x/y + 1)^2`

= `((x/y)^2 + 1)/(x/y + 1)^2`

= `((4/5)^2 + 1)/(4/5 + 1)^2`

= `(16/25 + 1)/(9/5)^2`

= `(41/25)/(81/25)`

= `(41)/(25) xx (25)/(81)`

= `(41)/(81)`
∵ (x2 + y2) : (x + y)2 = 41 : 81.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 7: Ratio and Proportion - Exercise 7.1

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एमएल अग्रवाल Understanding ICSE Mathematics [English] Class 10
अध्याय 7 Ratio and Proportion
Exercise 7.1 | Q 12.2
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