Math, asked by sonali88, 1 year ago

Sum of the areas of two square is468 meter square .If the difference of their perimeter is 24 m. Find the sodes of the two squares

Answers

Answered by Panzer786
1
Hii friend,

Let , the sides of first and second square be X and Y.

THEREFORE,

Area of first square = (side)² = (X)² = X²

Area of second square = (side)² = (Y)² = Y².

A/Q,

X² + Y² = 468.....(1)

And,

4X² - 4Y² = 24 .....(2)

From equation (1) we get,

X² = 468-Y² .......(3)

PUTTING THE VALUE OF X² IN EQUATION (2)

4X² - 4Y² = 24

4×(468-Y²) - 4Y² = 24

1872 - 4Y² - 4Y² = 24

-8Y² = 24-1872

-8Y² = -1848

Y² = 1848/8

Y² = 231

Y = ✓231 = 15.18 CM.

Putting the value of Y in equation (3)

X² = 468-Y²

X² = 468 - (15.18)²

X² = 468 - 230.5

X² = 237.5

X= ✓237.5 = 15.4 CM.

HOPE IT WILL HELP YOU... :-)

sonali88: thanks
Answered by Anonymous
0

Step-by-step explanation:

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 .

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