the sum of area two square is 468.if the different of perimeter is 24.find the side of two square
Answers
AND SIDES OF SECOND SQUARE=b
i.e. AREA OF FIRST SQUARE=a^2
AND AREA OF SECOND SQUARE=b^2
i.e. a^2+b^2=468...................1
AND 4a-4b=24
i.e. a-b=6.................................2 AND b=a-6
i.e. (a-b)^2=6^2
i.e. a^2+b^2-2ab=36
i.e. 2ab=432
i.e. ab=216
i.e. a(a-6)=216
i.e. a^2-6a-216=0
i.e. EITHER a=18 or a=-12
i.e. a=18 BECAUSE a CAN'T BE NEGATIVE
BY PUTTING a=18 IN EQUATION NO. 2,WE GET
18-b=6
i.e. b=12
i.e. SIDES OF FIRST SQUARE=18
AND SIDES OF SECOND SQUARE=12
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 .