# Estimating square numbers

A large dice has a side length of 9.2 cm. Estimate the surface area of the cube.

What I did:

$$9^2$$ = 6 × 81 = 6 × 80 = 480

But the answer says that $$9.2^2$$ is 85 as an estimate.

How do I get that?

Thank You

• It probably depends on how you're being expected to estimate these sorts of things, sadly. So it'll be difficult to get an answer that's assuredly compatible with you and also in the spirit of the exercise. For example, the exercise could've simply meant "compute $9.2^2$ (which is $84.64$) and round up to $85$". Or there might be some more formal, general method underlying it. – Eevee Trainer May 28 '19 at 6:22
• Weird. It's easy to calculate the exact surface area (unless maybe if you have to do it without a calculator), which is $6\times 9.2^2 = 507.84$. So anyway, the answer should be in around $500$. But if the "correct" answer is $85$, maybe it's only looking at only one side of the cube? Weird ... – Matti P. May 28 '19 at 6:24
• I'll just try and estimate using what $9^2$ and $10^2$ is and roughly work it out as I can't use a calculator. – user12345 May 28 '19 at 6:25
• It might help if you can say where you saw this problem. Is it as part of a course? If so, what course, and what particular topic are you studying? If it's general estimation, you might interpolate between $9^2$ and $10^2$; if this is a first-year calculus course, you might be expected to use the derivative of $x^2$ at $x = 9$ to attempt a linear approximation. Or, as Eevee Trainer indicated, you might just be expected to round the actual value of $9.2^2$. – Brian Tung May 28 '19 at 6:30
• It's from a GCSE Book so like sophomore – user12345 May 28 '19 at 6:31

Surface area of dice $$=6a^2=6\cdot (9.2)^2=6(\frac{92}{10})^2=6\cdot \frac{92\cdot92}{100}=\frac{50784}{100}=507.84\approx508$$ sq. units

However, if you divide this by 6 you will get roughly $$85$$ sq. units but that is the surface area of one side of dice.

The estimation for one square face could be like done this: $$9.2^2 = (9+0.2)^2 = 9^2 + 2\cdot 9\cdot 0.2 + 0.2^2 \approx 81 + 10\cdot 0.4 + 0 = 81 + 4 = 85$$

You can use the binomial expansion : $$(a+b)^{2} = a^{2} + 2ab + b^{2}$$ or maybe $$(a-b)^{2} = a^{2} + 2a(-b) + (-b)^{2} = a^{2}-2ab+b^{2}$$ , so $$9.2^{2}$$ can be written as $$(9+0.2)^{2}$$ or $$(10-0.8)^{2}$$.

But wait ! what is the value of $$0.2^{2}$$ or $$0.8^{2}$$ ? that's easy, for most decimal between 0 and 1 (in example $$0.5$$ or $$0.1$$, but in case of maybe difficult repeating decimal like $$0.11111111$$ and so on so on this wont work) you can write them as a number divided by $$10$$

so then : $$0.2^{2}$$ = $$(\frac{2}{10})^{2} = \frac{2^{2}}{10^{2}} = \frac{4}{100} = 0.4$$. You can also do the same for $$0.8$$, but i recommend having decimal as small as possible so the calculation would be easy

• You definitely mean $0.2^2 = 0.04$ right? In addition, I don't understand your second paragraph at all. – Toby Mak Oct 12 '19 at 7:09