Six rectangles, each with the common base width of 2, have lengths of 1, 4, 9, 16, 25 and 36. What is the sum of the areas of the six rectangles?

Hello!

The area of a rectangle with sides a and b is a*b. Here we have six rectangles, all of them have a=2 and the second sides are 1, 4, 9, 16, 25 and 36. Therefore the sum of the areas is equal to

2*1+2*4+2*9+2*16+2*25+2*36.

By the distributive law this is equal to

2*(1+4+9+16+25+36).

Note that the lengths have a form of n^2 for n from 1 to 6. There is a formula for such...

Hello!

The area of a rectangle with sides a and b is a*b. Here we have six rectangles, all of them have a=2 and the second sides are 1, 4, 9, 16, 25 and 36. Therefore the sum of the areas is equal to

2*1+2*4+2*9+2*16+2*25+2*36.

By the distributive law this is equal to

2*(1+4+9+16+25+36).

Note that the lengths have a form of n^2 for n from 1 to 6. There is a formula for such a sum, but in this case it is simpler to sum directly.

1+4+9+16+25+36=(1+9)+(4+16)+(25+36)=10+20+61=91.

And 2*91=182 (square units). This is the answer.