Perimeter

Question

Find the sum of the perimeters of all the rectangles with integral sides whose area is 216 sq. units.

Comments

Area=lxb=216= 8*27=cube(2)*cube(3)

but perimeter = l+b. say for example l=1 and b= 216. then l+b=217.

now put l=2 and b= square(2)(cube(3) so that lxb=area=216 holds. now l+b = 2+108=110.

for all pairs of (l,b) such that lxb=216, add l+b values which are on the right side as mentioned above.

I got the sum as 472. Dont know whether its wrong.

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power operation ^ has higher precedence over multiplication, assumed to minimize paranthesis.

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472 is wrong answer

Answer 1

$\large l*b = 216$, if both $l$ and $b$ are greater than $\sqrt{216}$ then, product will be larger than $216$. So, we need to check for first factor of $216$ till $\lceil \sqrt{216} \rceil = 15$ and if $x$ is a factor of $216$, then $\frac{216}{x}$ will be other factor.

$l*b = 216 = 2^3*3^3$ (this will help in finding factors)

Pairs of $(l,b)$ are $(1,2^3*3^3) = (1,216)$ and $(2,2^23^3) = (2,108)$. Similarly other pairs are : $(3, 72), (4,54), (6,36), (8,27), (9,24), (12, 18)$

Perimeter of a rectangle $= 2(l+b)$

So, Perimeter of all possible rectangles $\color{navy}{= \sum 2(l+b) = 2\sum l + 2 \sum b = 2( 1 + 2 + 3 + 4 + 6 + 8 + 9 + 12 + 18
+ 24 + 27 + 36 + 54 + 72 + 108 + 216) = 1200}$  

It is equivalent to finding 2 times sum of all divisors of $216$

PS:) Here $l = 12, b = 18$ and $l = 18, b = 12$ are considered as one rectangle, if considered to be different then it would be $\color{blue}{2400}$

Answer 2

Area of rectangle= l*b = 216
possible numbers to result in product 216
6,36
2,108

4,54

8,27

3,72

9,24

12,18

 

perimeter= 2(l+b)

summing up the perimeters = 2(42+110+58+35+75+33+30)=766

There can be two such rectangles for every pair , interchanging the length&breadth .. 766*2 = 1532