1
$\begingroup$

A square is divided into 7 identical rectangles, the rectangle length is 5. The perimeter of the whole shape is 34. what is the square area?

Since the shape is square and its perimeter is 34, then its length is 8.5 and then its area is 72.25 but the answer says the area is 70 what am I missing here?

$\endgroup$

closed as off-topic by José Carlos Santos, Paul Frost, Gibbs, Cesareo, Chinnapparaj R Nov 17 '18 at 3:20

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Paul Frost, Gibbs, Cesareo, Chinnapparaj R
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ The large shape can't be square and be divided into 7 identical rectangles. $\endgroup$ – Doug M Nov 17 '16 at 23:14
  • $\begingroup$ why? could you explain more? $\endgroup$ – user3188039 Nov 17 '16 at 23:15
  • 1
    $\begingroup$ Suppose it is. then if the width of the small rectangle is W and the length is L. 5W = 2L and W+L = 2L. And those two equations are incompatible. It must be a rectangle divided into 7 identical rectangles. $\endgroup$ – Doug M Nov 17 '16 at 23:17
  • $\begingroup$ why 5W = 2L ? if shape were rectangle, then the W will be 10 and hence L will be 7 (from rectangle perimeter formula), then area is 70. I think you are right, it must be a rectangle. but I would like to know what do you mean by 5W = 2L and W+L = 2L $\endgroup$ – user3188039 Nov 17 '16 at 23:31
0
$\begingroup$

Let $W$ and $L$ be the width and length of the small rectangles.

Suppose the big rectangle were actually a square.

Then the side length equals the top equals the bottom. $L+W = 2L = 5W$ and that is not possible unless $L = W = 0$

The large figure is not a square. It must be a rectangle

$5W$ = $2L$ because the width across the top of the big rectangle equals the width across the bottom. $L = \frac 52 W$

the side is $L+W = \frac 72 W$

the perimiter $2\times$top + $2\times$ side $= 17W = 34$

$W = 2, L = 5$

And the big rectagle is $10\times 7$

$\endgroup$
0
$\begingroup$

If the perimeter is 34, the side length is 8.5 and the area is (8.5)^2 = 72.25. The division into identical rectangles, while an interesting condition, is irrelevant.

(It's certainly possible to divide a square into 7 identical rectangles. If the length of one rectangle is 5, then its width is 5/7, meaning the perimeter would be 20 and the area would be 25 u^2.)

$\endgroup$
  • $\begingroup$ Welcome to Math.SE, and thanks for your first contribution! Note, however, that you have duplicated the explanation from the question itself, which isn't stating anything the questioner didn't already know. The real issue here is: Why does the official answer ($70$) differ? As DougM's answer indicates, the reason is that the word "square" is evidently a mistake. (Also, while you're correct that it's possible to divide a square into $7$ identical rectangles, I believe DougM's point is that it isn't possible to divide a square into $7$ identical rectangles using the configuration shown.) $\endgroup$ – Blue Nov 16 '18 at 17:20

Not the answer you're looking for? Browse other questions tagged or ask your own question.