Some wire is used to make 3 rectangles: a, b, and c. rectangle b’s dimensions are 3/5 cm larger than rectangle a’s dimensions, and rectangle c’s dimensions are 3/5 cm larger than rectangle b’s dimensions. rectangle a is 2 cm by 3 1/5 cm.
a. what is the total area of all three rectangles?
b. if a 40-cm coil of wire was used to form the rectangles, how much wire is left?

a) The total area of the three rectagles is 10,08 cm²

b) The wire left is 28,6 cm.

Step-by-step explanation:

  According to the information of the problem rectangle B is 3/5 cm > than rectangle A. And rectangle C is 3/5 cm > than rectagle B.

Deducing the area of rectangle A we can solve the others two areas of rectangle B and C.

  Using the area formula of a rectangle: Area= b*a ( base. height), we calculate the area of rectangule A ⇒ Area A= 2cm *3/5 cm= 1,2cm².

  Now rectangle B is 3/5 (0,6) cm larger than rectangle A dimentions. So If we add the 3/5 cm to the dimensions of rectangle A

 ⇒ 2cm+0,6cm= 2,6 cm and 3/5 cm+ 0,6cm= 6/5 (1,2) cm.

   So now we can calculate the area of B

             ⇒ Area B= 2,6cm* 1,2 cm= 3,12cm²

   Doing the same proceeding, rectangle C is 3/5 bigger than rectangle b dimensions, so the dimensionts of rectangle C = 3,2 cm* 9/5 cm

     And the area of rectangle C is ⇒ Area C = 5,76 cm².

 Now we have the three diferents areas so if we do the summation of the three areas

      ⇒ Area A+ Area B +Area C= 1,2cm² + 3,12 cm²+ 5,76cm² = 10,08 cm² and this is the total area of the three rectangles.

 The total summatory al the dimensions of the three rectangles is equal to 11, 4 cm.

 In conclusion, if they used to form the rectangles a 40 cm coil the wire left is going to be equal to 40 cm - 11,4 cm = 28.6 cm

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