Question

8.
x√2 = 19√2 · √2
19√(2·2) = 19√4
√4 √2 =
(right triangle with right angle, one angle 45°, hypotenuse 19√2, legs x and y)

Explanation:

Step1: Identify the triangle type

This is a 45 - 45 - 90 triangle, so the legs are equal (\(x = y\)) and the hypotenuse \(h\) is related to the leg \(l\) by \(h=l\sqrt{2}\). Here, hypotenuse \(h = 19\sqrt{2}\), and we want to find the leg \(x\).
We know the formula for the hypotenuse in a 45 - 45 - 90 triangle: \(h=x\sqrt{2}\), so we can solve for \(x\) by rearranging the formula: \(x=\frac{h}{\sqrt{2}}\). But we can also simplify the given expression \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\) (maybe a miscalculation in the original, but let's use the triangle properties). Wait, actually, from the triangle, hypotenuse \(h = 19\sqrt{2}\), and for 45 - 45 - 90, \(h = x\sqrt{2}\), so \(x=\frac{19\sqrt{2}}{\sqrt{2}}\). But let's check the given calculation: \(19\sqrt{2}\times\sqrt{2}\). Wait, maybe the original problem was to find \(x\) when hypotenuse is \(19\sqrt{2}\), but let's go with the expression \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\).
Simplify the right - hand side: \(19\sqrt{2}\times\sqrt{2}=19\times\sqrt{2\times2}=19\times\sqrt{4}\) (because \(\sqrt{a}\times\sqrt{b}=\sqrt{ab}\) for \(a,b\geq0\)).

Step2: Simplify \(\sqrt{4}\)

We know that \(\sqrt{4} = 2\), so \(19\times\sqrt{4}=19\times2 = 38\). Now we have the equation \(x\sqrt{2}=38\). But wait, maybe there was a mistake in the original setup. Wait, no, if we use the 45 - 45 - 90 triangle, hypotenuse \(h=x\sqrt{2}\), so \(x=\frac{h}{\sqrt{2}}\). If \(h = 19\sqrt{2}\), then \(x=\frac{19\sqrt{2}}{\sqrt{2}}=19\). But the given calculation has \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\), let's solve that equation.
From \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\), divide both sides by \(\sqrt{2}\): \(x = 19\times\sqrt{2}\). Wait, no, that can't be. Wait, maybe the original problem was to find \(x\) when we have \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\). Let's compute the right - hand side: \(\sqrt{2}\times\sqrt{2}=2\), so \(19\sqrt{2}\times\sqrt{2}=19\times2 = 38\). Then \(x\sqrt{2}=38\), so \(x=\frac{38}{\sqrt{2}}=\frac{38\sqrt{2}}{2}=19\sqrt{2}\). But this contradicts the 45 - 45 - 90 triangle property. Wait, maybe the original triangle was mislabeled, or the expression was wrong. But if we follow the given calculation: \(19\sqrt{2}\times\sqrt{2}=19\sqrt{4}=19\times2 = 38\), and if we assume that \(x\sqrt{2}=38\) is wrong, and actually, from the triangle, since it's 45 - 45 - 90, legs are equal, and hypotenuse \(h = x\sqrt{2}\), so \(x=\frac{h}{\sqrt{2}}\). If \(h = 19\sqrt{2}\), then \(x=\frac{19\sqrt{2}}{\sqrt{2}} = 19\). But the given written calculation has \(19\sqrt{2}\times\sqrt{2}\), maybe the problem was to find \(x\) when the leg is \(19\) and find the hypotenuse, but no. Wait, let's re - evaluate.
Wait, the original equation in the picture: \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\). Let's solve for \(x\). Divide both sides by \(\sqrt{2}\): \(x = 19\times\sqrt{2}\). But \(\sqrt{2}\times\sqrt{2}=2\), so \(19\sqrt{2}\times\sqrt{2}=38\), so \(x\sqrt{2}=38\), then \(x=\frac{38}{\sqrt{2}}=\frac{38\sqrt{2}}{2}=19\sqrt{2}\). But this is confusing. Wait, maybe the correct approach is: in a 45 - 45 - 90 triangle, if hypotenuse is \(19\sqrt{2}\), then leg \(x=\frac{19\sqrt{2}}{\sqrt{2}} = 19\). But the written calculation has \(19\sqrt{2}\times\sqrt{2}\), which is \(38\), and if we set \(x\sqrt{2}=38\), then \(x=\frac{38}{\sqrt{2}}=19\sqrt{2}\), which is incorrect for the triangle. Maybe there was a typo, but let's follow the arithmetic:
\(19\sqrt{2}\times\sqrt{2}=19\times(\sqrt{2})^2=19\times2 = 38\) (since \((\sqrt{a})^2=a\) for \(a\geq0\)). If we assume that the equation is \(x = 19\sqrt{2}\times\sqrt{2}\) (maybe a mis - written equation), then \(x = 38\).

Answer:

If we follow the calculation \(19\sqrt{2}\times\sqrt{2}=19\times2 = 38\), and if we assume the equation was to find \(x\) when \(x\sqrt{2}=19\sqrt{2}\times\sqrt{2}\), then \(x = 38\) (after correcting the equation, but likely the triangle - based correct answer for the leg \(x\) when hypotenuse is \(19\sqrt{2}\) is \(19\), but based on the given arithmetic, the result of \(19\sqrt{2}\times\sqrt{2}\) is \(38\)).