the sum of areas of two squares is 468 m square if the difference of their perimeter is 24 find find the side of two squares
Answers
Let the side of first square be x nd second be y.
Area of first square = x^2
Area of second square = y^2
Perimeter of first square = 4*x = 4x
Perimeter of second square = 4*y = 4y.
According to question ;
=) x^2 + y^2 = 468 m^2
& 4x - 4y = 24m
=) 4(x-y) = 24m
=) x -y = 24/4 = 6m
Squaring both sides ;
=) (x-y) ^2 = 6^2
=) x^2 + y^2 - 2xy = 36
=) 468 (by 1st eq) - 2xy = 36
=) 468 - 36 = 2xy
=) 432 = 2xy
=) 432/2 = 216 = xy
Since (x+y) ^2 = x^2 +y^2 + 2xy
=) (x+y) ^2 = 468 + 2(216)
=) (x+y) ^2 = 900
=) (x+y) = 30
Final equations are;
=) x+y = 30
=) x -y = 6
Add both eq ;
=) 2x = 36m
=) x = 36/2 = 18m
Put the value of x in eq1;
=) 18 + y = 30
=) y = 30-18 = 12m.
Hence sides are 18m and 12m.
Thanks!!
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 . ....