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# weird geometry

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4 Jul 20, 2020

#1
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Using Pythagorean theorem find BD and then AD Jul 20, 2020
#2
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Look carefully, if that side were 5 then the side AB should be 4, not $$\sqrt{5}$$. How you should solve this:

Divide ABCD in half diagonally. We have two triangles now. We apply the pythagorean theorem to triangle ABC. Then AC is $$\sqrt{30}$$. Now we know that for triangle ACD, that AC is $$\sqrt{30}$$ and DC is 4. Apply the pythagoren theorem again to find that AD is $$\sqrt{14}$$.

Jul 20, 2020
#3
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You've got it all wrong! Two right triangles are  ABD  and  BCD

First, find a diagonal BD             BD = sqrt( 42 + 52 )

Then, using BD and AB find side AD       AD = sqrt( BD2 - AB2 ) Guest Jul 20, 2020
#4
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We know angles A and C are right angles=90 degrees. However, we don't know that angles B and D are 90 degrees. They could very well be 89 and 91 degrees!
Therefore, it is prudent to extend the diagonal from B to D. Now, we can use Pythagoras's Theorem on triangle BCD. BD^2 =5^2 + 4^2 =41 and BD =Sqrt(41)
Again, we can use Pythagoras's Theorem on triangle BAD. Sqrt(41)^2 - sqrt(5)^2 =AD^2. AD^2 =41 - 5 =36.