Math, asked by comando123, 1 year ago

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Plz answer my question ,

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

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Answers

Answered by siddhartharao77
57

Answer:

18m, 12m

Step-by-step explanation:

Let the sides of the squares be x and y. {x > y}

(i) Area of the larger square = x².

Then,perimeter of larger square = 4x.

(ii) Area of the smaller square = y²

Then,perimeter of smaller square = 4y.


Now,

(1) Sum of areas:

Sum of areas of two squares is 468.

x² + y² = 468


(ii) Difference of their perimeters is 24.

4x - 4y = 24

x - y = 6

x = 6 + y


Substitute x = 6 + y in (i), we get

⇒ (6 + y)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 - 12)(y + 18) = 0

⇒ y = 12,-18{Length of the side of the square cannot be negative}.

⇒ y = 12.


Then:

x = 6 + y

x = 18


Therefore:

Length of the sides of the two squares is 18m, 12m.


Hope it helps!


comando123: Nice answer
siddhartharao77: Thank you!
siddhartharao77: Thank you sis!
SmallMiniDoraemon: Wonderful explanation Bro :-)
trisha10433: :) nyc bro
siddhartharao77: Great thanks Dora, Trisha sis!
Answered by Anonymous
73

Question:

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

Answer:

18m and 12 m .

Step-by-step explanation:


Let the sides of two squares be x m and y m respectively .


Case 1 .

→ Sum of the areas of two squares is 468 m² .

A/Q,

∵ x² + y² = 468 . ...........(1) .

[ ∵ area of square = side² . ]

Case 2 .

→ The difference of their perimeters is 24 m .

A/Q,

∵ 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) , we get

∵ 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), we get

∵ y = x - 6 .

⇒ y = 18 - 6 .

∴ y = 12 m .

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

THANKS .


comando123: Great answer .
trisha10433: :)
SmallMiniDoraemon: Wonderful answer :-)
Anonymous: thanks 2 all of you
Anonymous: thanks
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