Question

Q..sum of the areas of two squares is 850sq.m. if the difference of their perimeter is 40m, find the sides of the two squares.

bye, :(​

Answer 1

Answer:

Let the side of first square = x metre

side of second square = y meter

According to question

$x^2 +y^2$ = 850 Equation(1)

4x - 4y =40

x - y =10 Equation (2)

x = 10 +y

$( 10 +y)^2 + y^2$ = 850

$100 + 20y +y^2 +y^2$=850

$2y^2 + 20y -750$= 0

$y^2 +10y -375$=0

$y^2 + 25y -15y -375$ =0

y(y -25) -15(y -25) =0

(y-15)(y -25)=0

y = 15, 25

One square with side 15 m and other with side 25 m..

Answer 2

Answer:

Answer:The sides of two squares are 25 m & 15 m.

Step-by-step explanation:

Given:

Sum of the areas of two squares = 850 m²

Difference of their perimeters = 40 m

Solution:

Let the side of one square be x and other be y.

Since, area of square = (side)²

Sum of the areas = x² + y² = 850 m². …(i)

also,

Difference of their perimeters = 4x - 4y = 40 m

→ 4( x - y) = 40

→ x - y = 40/4 = 10

→ y = x - 10 …(ii)

Substituting the value of y in (i)

we get,

x² + (x-10)² = 850

x² + x² - 2(10)(x) + 10² = 850

2(x²) - 20x + 100 = 850

2x² - 20x = 850 - 100

2x² - 20x - 750 = 0

dividing each term by 2

x² - 10x - 375 = 0

x² + 15x - 25x - 375 = 0

x(x+15) - 25(x+15) = 0

therefore,

x - 25 = 0 or x + 15 = 0

x = 25 or x = -15

but, since side can't be negative

→ x = 25 m

Substituting the value of x in (ii)

y = (25) - 10

y = 15 m

Thus, the sides of two squares are 25 m & 15 m.