**Here’s the ANSWER to to last week’s math challenge**

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Last week, we touted Chicago’s Noble Network of Charter Schools … specifically, its intensive math curriculum

And, we presented a challenge question (taken from the 10th grade curriculum) …

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**Noble Charter HS – Math Challenge Question**

The rectangle shown below is divided into four green squares, seven gold squares, four orange squares, and one blue rectangle.

If the perimeter of the blue rectangle is 20 cm, what is the perimeter of the larger rectangle?

Explain your reasoning.

*Recommended: **click to download **and print PDF*

**Here’s the answer …. and a method for get it.**

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__One way to work through the problem …__

1) Assume that each side of a gold square is 10 cm long (an arbitrary round number that is easy to scale)

2) Note that the length of the long side of the blue rectangle is the same as the length of a gold square’s side. Since we’re assuming that a gold square’s side is 10 cm, then the length of **the long side of the blue rectangle is 10**,** too.**

3) Note that 4 orange squares fit horizontally against 5 gold squares. Said differently, the combined length of 4 orange square sides is equal to the combined length of 5 gold square sides. **If the length of a gold square side = 10 cm (as we’re assuming) then each side of an orange square = 12.5 cm** (5 x 10 = 50, 50 / 4 = 12.5)

3) Note that the length of the short side of the blue rectangle is an orange side (12.5 cm) minus a gold side (10 cm). ** So, the length of the short side of the blue rectangle is 2.5 cm.**

4) So, __assuming that a gold square’s side is 10 cm__, ** the perimeter of the blue rectangle is 25 cm** (2 x 10 plus 2 x 2.5). Don’t worry that the problem says that the perimeter of the blue rectangle is 20 cm – we’ll fix that in a moment.

5) Similar to step 2, note that 4 green squares fit horizontally against 6 gold squares. Said differently, the combined length of 6 gold square sides is equal to the combined length of 4 green square sides. So, if the length of a gold square side = 10 cm then **each side of an green square = 15 cm **(6 x 10 = 60; 60 / 4 = 15)

6) Note that the long side of the enclosing rectangle is equal to 4 green square sides. So, **the long side of the enclosing rectangle is 60 cm** (4 x 15).

7) Note that the short side of the enclosing rectangle is equal to a green (15) plus a gold (10) plus an orange (12.5). So, t**he short side of the enclosing rectangle is 37.5 cm** (15 + 10 + 12.5 = 37.5).

8) So, **the perimeter of the enclosing rectangle** **is 195** – the sum of 2 long sides (2 x 60 = 120) plus the sum of 2 short sides (2 x 37.5 = 75).

9) **But, we’re told that the perimeter of the blue rectangle is 20 cm**. No problem, we just need to rescale everything by the ratio of 20 cm (what the problem specifies) to 25 cm (based on the working assumption that a gold = 10 cm). That is, **we need multiply all of our answers by 80%** ( 20 / 23 = 80%).

**10) **Rescaling everything to conform by 80%: **a gold square side is really 8 cm**, not 10 … **an orange square side is really 10 cm**, not 12.5 cm … a** green square side is really 12**. not 15.

11) Most important, to answer the question: t**he perimeter of the big rectangle is 156 **(195 cm times 80% = 156 cm).

Done !

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**Well, are you at least as smart as a Noble 10th grader?**

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P.S. If you have a simpler, faster method, please email it to me at **

**HomaK@msb.edu**

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**Thanks to SGC for feeding the lead and the problem**

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