+2446 
This image will make it easier to name individual segments, which should make the explanation more comprehensible.
Using the Pythagorean's theorem, one can solve for BC.
| \(BC^2+CD^2=BD^2\) | This is what the Pythagorean theorem states. Substitute in the known values and solve for the unknown. |
| \(BC^2+1^2=5^2\) | Now, simplify the left and right sides of the equation. |
| \(BC^2+1=25\) | Subtract 1 from both sides of the equation. |
| \(BC^2=24\) | Take the square root of both sides to eliminate the index of 2. |
| \(BC=\sqrt{24}\) | Of course, there is no reason to consider the negative answer because distances can only be positive. There is no need to put this in simplest radical form because it is easier to work with computationally if it is in its current form. |
Knowing this information, one can then solve for the missing side, AC in the diagram.
| \(AB^2+BC^2=AC^2\) | \(\triangle ABC\) is indeed a right triangle, so we can use the special relationship of the sides indicated byt he Pythagorean Theorem again. |
| \(5^2+\left({\sqrt{24}}\right)^2=AC^2\) | Simplify the left hand side first before proceeding. |
| \(25+24=AC^2\) | |
| \(49=AC^2\) | Take the square root of both sides just like the previous problem. |
| \(AC=7\) | As aforementioned, the negative answer does not make sense in the context of geometry. |
Now that the hypotenuse of \(\triangle ABC\) has been solved for, you are done.
+2446
This image will make it easier to name individual segments, which should make the explanation more comprehensible.
Using the Pythagorean's theorem, one can solve for BC.
| \(BC^2+CD^2=BD^2\) | This is what the Pythagorean theorem states. Substitute in the known values and solve for the unknown. |
| \(BC^2+1^2=5^2\) | Now, simplify the left and right sides of the equation. |
| \(BC^2+1=25\) | Subtract 1 from both sides of the equation. |
| \(BC^2=24\) | Take the square root of both sides to eliminate the index of 2. |
| \(BC=\sqrt{24}\) | Of course, there is no reason to consider the negative answer because distances can only be positive. There is no need to put this in simplest radical form because it is easier to work with computationally if it is in its current form. |
Knowing this information, one can then solve for the missing side, AC in the diagram.
| \(AB^2+BC^2=AC^2\) | \(\triangle ABC\) is indeed a right triangle, so we can use the special relationship of the sides indicated byt he Pythagorean Theorem again. |
| \(5^2+\left({\sqrt{24}}\right)^2=AC^2\) | Simplify the left hand side first before proceeding. |
| \(25+24=AC^2\) | |
| \(49=AC^2\) | Take the square root of both sides just like the previous problem. |
| \(AC=7\) | As aforementioned, the negative answer does not make sense in the context of geometry. |
Now that the hypotenuse of \(\triangle ABC\) has been solved for, you are done.
