Question

Rectangles are drawn on line segment of fixed lengths. When the breadths are 6 m and 5 m respectively the sum of the ares of the rectangles is 83m2. But if the breadths are 5m and 4m respectively the sum of the areas is 68m2. Find the sum of the areas of the square drawn on the line segments.​

Answer 1

The sum of the areas of the square drawn on the line segments is 113 $m^2$.

Step-by-step explanation:

Given:

Let length of first rectangle  =  x m  

Length of Second rectangle  =  y m  

We know Area of rectangle  = $Length \times Breadth$

So,

Area of first rectangle = 6x

Area of second rectangle = 5y  

According to first condition, we get  

6x  + 5y  = 83                                                [ 1 ]

Now breadth of first rectangle  =  5 m and breadth of second rectangle = 4 m  

So, according to second condition we get  

5x + 4y = 68                                                [ 2 ]

Now we multiply equation (1) by 4 , we get  

$4 (6x + 5y) = 83\times 4$

24x + 20y = 332                                         [3]

And Mutiply equation (2) by 5, we get

$5 (5x+4y)=68\times 5$

25x + 20y = 340                                           [4]

Now we subtract equation(3) from equation (4), we get  

25x + 20y - (24x + 20y) = 340 - 332

25x + 20y - 24x - 20y = 8

x = 8    

Substitute this value in equation (2), we get  

5 ( 8 ) + 4y = 68  

40 + 4y = 68

4y = 68 - 40

4y = 28  

$y=\frac{28}{4}$

y = 7  

So,  length of first rectangle = 8m  and we draw a square on this base so, Area of this square = $(side)^2 = 8^2 = 64 m^2$  

And  length of Second rectangle = 7m and we draw a square on this base so,

Area of this square = $(side)^2=7^2= 49 m^2$  

Then sum of area of these two square = 64 + 49 = 113 $m^2$

Hence the sum of the areas of the square drawn on the line segments is 113 $m^2$.