What is the length of AD?
$$A.\ \ 2\sqrt{10}$$
$$B.\ \ 4\sqrt{10}$$
$$C.\ \ 2\sqrt{5}$$
$$D.\ \ 4\sqrt{5}$$
$$E.\ \ \sqrt{10}$$
The OA is A.
Please, can any expert explain this PS question for me? I tried to solve it but I'm not sure and I can't get the correct answer. I need your help. Thanks.
Hi swerve,
Let's take a look at your question.
Draw a horizontal line parallel to CB, from point D that intersects the line Ab at point E, such that AED is a right triangle.
Since,
$$DE=CB$$
Therefore,
$$DE=6$$
Now, let's find AE.
$$AE=AB-EB$$
$$AE=AB-DC$$
$$AE=12-10$$
$$AE=2$$
Now, we can find AD using Pythagoras Theorem,
$$AD=\sqrt{\left(AE\right)^2+\left(DE\right)^2}$$
$$AD=\sqrt{\left(2\right)^2+\left(6\right)^2}$$
$$AD=\sqrt{4+36}$$
$$AD=\sqrt{40}$$
$$AD=2\sqrt{10}$$
Therefore, option
A is correct.
Hope it helps.
I am available if you'd like any follow up.
