Question

HELP ME!!
Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24m find the sides of the two squares .
Sol.) sum of the areas of two squares is = 468 m2
diff of their perimeter = 24m
Let the areas of first square with side xm= x2m2
(p) of 1st sqr = 4xm
the Area of 2nd sqr with side y
m =y2m2
(p) of 2nd sqr 4ym
x2+y2=468 .....eq(1)
4x-4y=24
4(x-y) =24
then x-y =24/4=6
x-y=6. ....(2)
iska baad kya kru...
HELP ME!!



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Answer 1

From eq 2 take the value of y and put it in eq 1

Answer 2

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Sum of the areas of two squares is 468 m²

∵ x² + y² = 468 . ………..(1) .[ ∵ area of square = side²] → The difference of their perimeters is 24 m.

∵ 4x – 4y = 24 [ ∵ Perimeter of square = 4 × side] ⇒ 4( x – y ) = 24

⇒ x – y = 24/4 .

⇒ x – y = 6 .

∴ y = x – 6 ……….(2)

From equation (1) and (2),

∵ x² + ( x – 6 )² = 468

⇒ x² + x² – 12x + 36 = 468

⇒ 2x² – 12x + 36 – 468 = 0

⇒ 2x² – 12x – 432 = 0

⇒ 2( x² – 6x – 216 ) = 0

⇒ x² – 6x – 216 = 0

⇒ x² – 18x + 12x – 216 = 0

⇒ x( x – 18 ) + 12( x – 18 ) = 0

⇒ ( x + 12 ) ( x – 18 ) = 0

⇒ x + 12 = 0 and x – 18 = 0

⇒ x = – 12m [ rejected ] and x = 18m

∴ x = 18 m

Put the value of ‘x’ in equation (2),

∵ y = x – 6

⇒ y = 18 – 6

∴ y = 12 m

Hence, sides of two squares are 18m and 12m respectively

Hope it's Helpful....:)