Answer:
The length of sides of one square is 5cm and length of sides of another square is 9cm
Step-by-step explanation:
Let the length of rectangle be x and width of rectangle be y.
We have given,
Area of rectangle = 45 cm²
i.e. Area of rectangle = x·y = 45 or xy = 45 ---------(1)
Next, two squares are constructed from two adjacent sides of rectangle.
i.e Side length of one square will be x and side length of another square will be y.
Area of one square = x²
And area of another square = y²
According to problem,
Sum of area of two squares is 106 cm²
∴ x²+y² = 106 ---------------(2)
From equation (1) and (2) , we can find x and y.
xy = 45 or x = [tex]\frac{45}{y}[/tex] , Plug this in equation (2).
We get,
x² + y² = 106
or [tex](\frac{45}{y} )^{2} + y^{2} =106[/tex]
or [tex]\frac{45^{2} +y^{4} }{y^{2} } =106[/tex]
or [tex]45^{2} + y^{4} = 106y^{2}[/tex]
On solving this equation we get ,
y²=81 or 25
or y = 9 or 5
for y =9 , x= [tex]\frac{45}{9} =5[/tex]
or for y = 5 , x = [tex]\frac{45}{5} =9[/tex]
Hence the length of sides of one square is 5cm and length of sides of another square is 9cm
Answer:
AB = 9 cm and BC = 5 cm.
Step-by-step explanation:
Given the area of a rectangle ABCD is 45 cm². Let's assume its lengths are AB = x and BC = y such that xy = 45.
Two squares are constructed such that two adjacent sides of the rectangle are each also the side of one of the squares. It means there are two squares ABPQ and BCST where AB = x and BC = y.
The combined area of the two squares is 106 cm². It means AB² + BC² = 106 i.e. x² + y² = 106.
So we have the following equations to solve:-
x² + y² = 106 and xy = 45.
x+y = √(x²+y² + 2xy) = √(106 + 2*45) = √(106 + 90) = √(196) = 14.
x-y = √(x²+y² - 2xy) = √(106 - 2*45) = √(106 - 90) = √(16) = 4.
So, x+y=14 and x-y=4.
then x = (x+y)+(x-y) / 2 = (14+4)/2 = 18/2 = 9.
and y = (x+y)-(x-y) / 2 = (14-4)/2 = 10/2 = 5.
Hence, AB = 9 cm and BC = 5 cm.
