Math, asked by nandanyennam, 9 months ago

35.
Sum of the areas of two squares is 544 m2. If the difference of their
perimeters is 32 m, find the sides of the two squares.​

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Answers

Answered by rahul123437
6

Side of the square is 12 m and 20 m.

Given:

  • The sum of the areas of two squares is 544 sq.m.
  • Difference between their perimeters is 32 m.

To find:

The sides of two squares.

Formula used:

  • Area of square = (side)²
  • Perimeter of square = 4 × side

Explanation:

Let us consider,

Side of first square = x

Side of second square = y

From given data,      

x² + y² = 544  ..................1

4x - 4y = 32      ......................2      

4(x-y) = 32          

Dividing "4" on both the sides, we get

x  -y = 8        

x = y + 8      

From equation 1,  

(Y + 8)² + Y² = 544

Y² + 64 + 16y + Y² = 544        

2Y² + 16Y - 480 = 0        

y² + 8y - 240 = 0      

y² + 20y - 12y - 240 = 0    

y(y + 20) - 12(y + 20) = 0    

(y + 20)(y - 12) = 0        

y = 12m

From equation 2,    

x = y + 8

x = 12 + 8

x = 20m

Therefore, the side of the square is 12 m and 20 m.

To learn more....

1. 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

brainly.in/question/2163623

2. Sum of areas of two squares is 400 cm 2 .If the difference in their perimeters is 16cm. find the sides of the two squares

brainly.in/question/289826

Answered by aaqhil
0

Answer:

Let the sides of first and second square be X and Y .

Area of first square = (X)²

And,

Area of second square = (Y)²

According to question,

(X)² + (Y)² = 544 m² ------------(1).

Perimeter of first square = 4 × X

and,

Perimeter of second square = 4 × Y

According to question,

4X - 4Y = 32 -----------(2)

From equation (2) we get,

4X - 4Y = 32

4(X-Y) = 32

X - Y = 32/4

X - Y = 8

X = 8+Y ---------(3)

Putting the value of X in equation (1)

(X)² + (Y)² = 544

(8+Y)² + (Y)² = 544

(8)² + (Y)² + 2 × 8 × Y + (Y)² = 544

64 + Y² + 16Y + Y² = 544

2Y² + 16Y - 544 +64 = 0

2Y² + 16Y -480 = 0

2( Y² + 8Y - 240) = 0

Y² + 8Y - 240 = 0

Y² + 20Y - 12Y -240 = 0

Y(Y+20) - 12(Y+20) = 0

(Y+20) (Y-12) = 0

(Y+20) = 0 Or (Y-12) = 0

Y = -20 OR Y = 12

Putting Y = 12 in EQUATION (3)

X = 8+Y = 8+12 = 20

Side of first square = X = 20 m

and,

Side of second square = Y = 12 m.

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