Right Triangle Segment

$\angle ABC=90^{\circ}$ , hence, $AC$ is the hypotenuse. Since, $BD=12$ and $DC=18$ , we have...

$BC=BD+DC=12+18=30$

Since $AC=34$ , from the Pythagorean Theorem, we have...

$AC^2=AB^2+BC^2\Longrightarrow AB=\sqrt{AC^2-BC^2}=\sqrt{34^2-30^2}=\sqrt{256}=16$

$\Delta ABD$ is also a right triangle since $\angle ABC=90^{\circ}$ . Here, the hypotenuse is $AD$ . Hence, from the Pythagorean Theorem, we again have...

$AD^2=AB^2+BD^2=16^2+12^2=400 \Longrightarrow AD=\sqrt{400}=\fbox{20}$

here BC=30 then using Pythagorean theorem we get AB=16, now ABD is another right angle triangle , we can found AD by AD^2= AB^2+BD^2.= 400, AD= 20.

This sum entirely relies of Pythagoras Theorem

To find AD, we first need to find AB To find AB^2, we need to subtract BC^2 from AC^2, ie, 34^2-30^2. We know that BC = 30 because BC(12) +DC(18) = 30

Using Pythagoras, we find that AB = 16cm [AC^2(1156) - BC^2(900) = 256, sqrt.256 = 16)

Now to find AD, we again use Pythagoras AD is a diagonal, hence it is the hypotenuse of triangle ABD To find AD, we simply add AB^2(16^2) and BD^2(12^2) ----------> 256+144------>400 and sqrt.400 = 20 = AD = answer

we need AB to find AD
AB=BC(30) square - AC(34) square =square root 256 =16 AD=(AB)16 square + (BD)12 square =square root 400 =20

Use Pythagorean Theorem