Let $\triangle ABC$ be a right-angled triangle. The length of its sides are $a$, $b$, and $c$, such that $ a + b + c =22 $ and $ a^{2} + b^{2} + c^{2} = 200 $.
$1.$ Find out the length of $a$, $b$, and $c$.
$2.$ Without using question $1$, find out the area of $\triangle ABC$.
On
Let's say $c$ is the hypotenuse (so that $C$ is the right angle). By the Pythagorean Theorem, $a^2+b^2=c^2$, which together with the second given condition (when substituted into it) gives $2c^2=200$, so you now know your $c$, which is $c=10$. Substituting this value of $c$ into both given conditions, we now have $$a+b=12 \quad \text{and} \quad a^2+b^2=100.$$ Solve the first equation for $b$ in terms of $a$ (or the other way around), substitute into the second equation, and solve the resulting quadratic equation for $a$. That pretty much completes question 1.
I'm not sure what exactly is expected for question 2. I suspect that we still need to find and eliminate $c$. But once we have those two equations $a+b=12$ and $a^2+b^2=100$, we can find the area, which is $\frac{1}{2}ab$, without finding explicitly both $a$ and $b$. Hint: square the first equation and subtract the second equation from that.
ok assuming $c$ is the hypotenuse, then we have $$a^2+b^2=c^2$$ thus we get $$2c^2=200$$ and $$c=10$$. Now we have the system $$a+b=12$$ $$a^2+b^2=100$$ plugging $$b=12-a$$ in the second equation then we have to solve $$a^2-12a+22=0$$ can you proceed?