Question

The area of a rectangle is 45 cm2. Two squares are constructed such that two adjacent sides of the rectangle are each also the side of one of the squares. The combined area of the two squares is 106 cm². Find the lengths of the sides of the squares.

Answer 1

Answer:

9cm and 5 cm

Step-by-step explanation:

Given that area of rectangle = lw = 45 cm square...i

where l =length and w = width

First square has side l and second side w

Area of two squares combined = sum of their areas

=$[tex](\frac{45}{w} )^{2}+w^{2} =106cm^2...ii

Eliminate l from I equation

l =[tex]\frac{45}{w}$

Substitute in II equation

\\2025+w^4-106w^2=0\\(w^2-25)(w^2-81)=0\\w=5 or 9\\l=9 or 5[/tex][/tex]

So  length of rectangle = 9 cm and width = 5 cm

Answer 2

Answer:

Thus, the sides of the squares are 5 cm and 9 cm.

Step-by-step explanation:

Let, the length and the width of the rectangle be x and y cm respectively.

It is given that, sides of the two squares are x and y cm respectively.

As, the area of the rectangle is 45 cm².

So, we have, xy = 45.

Since, the combined area of the two squares is 106 cm².

We get, $x^{2}+y^{2}=106$.

Solving the equations, we have,

$x^{2}+(\frac{45}{x})^{2}=106$

i.e. $x^{2}+(\frac{45^2}{x^2})=106$

i.e. $x^{2}+(\frac{2025}{x^2})=106$

i.e. $x^{4}+2025=106\times {x^2}$

i.e. $x^{4}-106\times {x^2}+2025=0$

i.e. $(x^2-25)(x^2-81)=0$

i.e. $(x^2-25)=0$ and $(x^2-81)=0$

i.e. $x^2=25$ and $x^2=81$

i.e. x = 5 and x = 9, because the length cannot be negative.

So, $y=\frac{45}{x}$ ⇒ $y=\frac{45}{5}$ and $y=\frac{45}{9}$ i.e. y = 9 and y = 5.

Thus, the sides of the squares are 5 cm and 9 cm.