Pythagoras' Theorem showing incorrect answer 
Given the diagram above, where I calculate y to be $11.3 = \sqrt{8.2^2 + 7.8^2}$ after square rooting the answer, to 1 d.p. Z therefore should be $10.1$, given by $11.3^2 - 5^2$, then square rooted?
This is flagging incorrect, can someone explain why?
 A: $y=\sqrt{8.2^2+7.8^2}$
$z=\sqrt{y^2-5^2}=\sqrt{8.2^2+7.8^2-5^2}=
\sqrt{67.24+60.84-25}=\sqrt{103.08}= 10.2$ (rounded to one digit).
Note that if you approximate $y$ to one digit, you get $y=11.3$; then
$$
\sqrt{11.3^2-5^2}=\sqrt{102.69}=10.1 \text{ (rounded to one digit)}
$$
It's wrong to approximate $y$ before doing the complete computation, because you have to square it back and lose accuracy: indeed, the rounding of the “true” result to one decimal digit is $10.2$ and not $10.1$.
A: By calculating $y$ first and rounding $y$, you get an error that causes the final rounding to be wrong.
If $y^2=8.2^2+7.8^2$ then $z=\sqrt{y^2-5^2}$. But you rounded $y$ so that $$\sqrt{y^2-5^2}\approx 10.13$. But using the original $y^2$ you get $z\approx 10.152$, which rounds to $10.2$.
Rounding twice in two different places is always going to introduce more error.
Note that while $y-11.3\approx 0.017$, $y^2-11.3^2=0.39$. By rounding $y$ to one decimal place, you've gotten a much bigger error in $y^2$.

Side note:
Assuming that the values $7.8, 8.2, 5.0$ were rounded to one decimal place, then the minimal value for $z$ is:
$$\sqrt{7.75^2+8.15^2-5.05^2}\approx 10.05$$
and the maximum value for $z$ is:
$$\sqrt{7.85^2+8.25^2-4.95^2}\approx 10.26$$
Of course, this does not take into account that the right angles were also just approximates :)
What this means is that, while $10.2$ is the correct rounding for this result, the error is more like $\pm 0.1$ than $\pm 0.05$. Rounding in computations is very
risky.
A: Although to the nearest tenth $y = 11.3$, and $\sqrt{(11.3)^2 - 5^2} = 10.134$ which rounds to 10.1, if you carry $y$ to more accuracy you get $y=11.317$ wgich gives $z = \sqrt{(11.317)^2 - 5^2} = 10.153$ which rounds to 10.2.
In general, unless you have a really good grasp of the sensitivity of results on the uncertainties in inputs and intermediate results, you need to carry intermediate calculations (in this case, the square root done for $y$) to more decimals than you will require in the eventual answer.
A: Note that you don't need to take a square root until the end, and you can use $(a+b)^2+(a-b)^2=2a^2+2b^2$ to take most of the pain out of the calculation $$z^2=y^2-5^2=(8.2)^2+(7.8)^2-5^2=(8+0.2)^2+(8-0.2)^2-5^2=128+0.08-25=103.08$$
Then it is easy to see that $z$ is a little more than $10.15$ because $0.15^2=\frac 9{400}(=0.0225)\lt 0.08$ so to one decimal place you get $10.2$.
Rounding half way through gives you an answer a little below $10.15$.
