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 Mackenzie
❥คɴᎦᴡєя 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 side of the first square be ‘a’m and that of the second be ′ A ′ m. Area of the first square =a 2 sq m. Area of the second square =A 2 sq m. Their perimeters would be 4a and 4A respectively. Given 4A−4a=24 A−a=6 –(1) A 2 +a 2 =468 –(2) From (1), A=a+6 Substituting for A in (2), we get (a+6) 2 +a 2 =468 a 2 +12a+36+a 2 =468 2a 2 +12a+36=468 a 2 +6a+18=234 a 2 +6a−216=0 a 2 +18a−12a−216=0 a(a+18)−12(a+18)=0 (a−12)(a+18)=0 a=12,−18 Reply
❥คɴᎦᴡєя
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 side of the first square be ‘a’m and that of the second be
′
A
′
m.
Area of the first square =a
2
sq m.
Area of the second square =A
2
sq m.
Their perimeters would be 4a and 4A respectively.
Given 4A−4a=24
A−a=6 –(1)
A
2
+a
2
=468 –(2)
From (1), A=a+6
Substituting for A in (2), we get
(a+6)
2
+a
2
=468
a
2
+12a+36+a
2
=468
2a
2
+12a+36=468
a
2
+6a+18=234
a
2
+6a−216=0
a
2
+18a−12a−216=0
a(a+18)−12(a+18)=0
(a−12)(a+18)=0
a=12,−18