Sum of the areas of two squares is 468 m square if the difference of their perimeter is 24 m find the sides of the two square . About the author Claire
✎Δnsɯer࿐ Let the sides of the first and second square be X and Y . Area of the first square = (X)² Area of the second square = (Y)² According to question, (X)² + (Y)² = 468 m² ——(1). Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y According to question, 4X – 4Y = 24 ——–(2) From equation (2) we get, 4X – 4Y = 244X–4Y=24 4(X-Y) = 244(X−Y)=24 X – Y = 24/4X–Y=24/4 X – Y = 6X–Y=6 X=6+Y———(3) Putting the value of X in equation (1) (X)² + (Y)² = 468(X)²+(Y)²=468 = > (6+Y)² + (Y)² = 468 = > (6)² + (Y)² + 2 × 6 × Y + (Y)² = 468 = > 36 + Y² + 12Y + Y² = 468 = > 2( Y² + 6Y – 216) = 0 = > Y² + 6Y – 216 = 0 = > Y² + 18Y – 12Y -216 = 0 = > Y(Y+18) – 12(Y+18) = 0 = > (Y+18) (Y-12) = 0 Putting Y = 12 in EQUATION (3) X = 6+Y = 6+12 = 18X=6+Y=6+12=18 Side of first square = X = 18 m Side of second square = Y = 12 m. Reply
Let the sides of the first and second square be X and Y . Area of the first square = (X)² Area of the second square = (Y)² According to question, (X)² + (Y)² = 468 m² ——(1). Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y According to question, 4X – 4Y = 24 ——–(2) From equation (2) we get, 4X – 4Y = 24, 4(X-Y) = 24 X – Y = 24/4 , X – Y = 6 X = 6+Y ———(3) Putting the value of X in equation (1) (X)² + (Y)² = 468, (6+Y)² + (Y)² = 468 (6)² + (Y)² + 2 × 6 × Y + (Y)² = 468 36 + Y² + 12Y + Y² = 468 2Y² + 12Y – 468 +36 = 0 2Y² + 12Y -432 = 0 2( Y² + 6Y – 216) = 0 Y² + 6Y – 216 = 0 Y² + 18Y – 12Y -216 = 0 Y(Y+18) – 12(Y+18) = 0 (Y+18) (Y-12) = 0 (Y+18) = 0 Or (Y-12) = 0 Y = -18 OR Y = 12 Putting Y = 12 in EQUATION (3) X = 6+Y = 6+12 = 18 Side of first square = X = 18 m Side of second square = Y = 12 m. [tex]\huge{\tt{\red{S}\green{H}\purple{I}\pink{N}\blue{C}\orange{H}\red{A} \green{N} \: \pink{✿༉}}}[/tex] Reply
✎Δnsɯer࿐
Let the sides of the first and second square be X and Y .
Area of the first square = (X)²
Area of the second square = (Y)²
According to question,
(X)² + (Y)² = 468 m² ——(1).
Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y
According to question,
4X – 4Y = 24 ——–(2)
From equation (2) we get,
4X – 4Y = 244X–4Y=24
4(X-Y) = 244(X−Y)=24
X – Y = 24/4X–Y=24/4
X – Y = 6X–Y=6
X=6+Y———(3)
Putting the value of X in equation (1)
(X)² + (Y)² = 468(X)²+(Y)²=468
= > (6+Y)² + (Y)² = 468
= > (6)² + (Y)² + 2 × 6 × Y + (Y)² = 468
= > 36 + Y² + 12Y + Y² = 468
= > 2( Y² + 6Y – 216) = 0
= > Y² + 6Y – 216 = 0
= > Y² + 18Y – 12Y -216 = 0
= > Y(Y+18) – 12(Y+18) = 0
= > (Y+18) (Y-12) = 0
Putting Y = 12 in EQUATION (3)
X = 6+Y = 6+12 = 18X=6+Y=6+12=18
Side of first square = X = 18 m
Side of second square = Y = 12 m.
Let the sides of the first and second square be X and Y . Area of the first square = (X)²
Area of the second square = (Y)²
According to question, (X)² + (Y)² = 468 m² ——(1).
Perimeter of first square = 4 × X and Perimeter of second square = 4 × Y
According to question,
4X – 4Y = 24 ——–(2)
From equation (2) we get,
4X – 4Y = 24, 4(X-Y) = 24
X – Y = 24/4 , X – Y = 6
X = 6+Y ———(3)
Putting the value of X in equation (1)
(X)² + (Y)² = 468, (6+Y)² + (Y)² = 468
(6)² + (Y)² + 2 × 6 × Y + (Y)² = 468
36 + Y² + 12Y + Y² = 468
2Y² + 12Y – 468 +36 = 0
2Y² + 12Y -432 = 0
2( Y² + 6Y – 216) = 0
Y² + 6Y – 216 = 0
Y² + 18Y – 12Y -216 = 0
Y(Y+18) – 12(Y+18) = 0 (Y+18) (Y-12) = 0
(Y+18) = 0 Or (Y-12) = 0 Y = -18 OR Y = 12
Putting Y = 12 in EQUATION (3)
X = 6+Y = 6+12 = 18
Side of first square = X = 18 m
Side of second square = Y = 12 m.
[tex]\huge{\tt{\red{S}\green{H}\purple{I}\pink{N}\blue{C}\orange{H}\red{A} \green{N} \: \pink{✿༉}}}[/tex]