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In parallelogram ABCD, the diagonals AC is at right angles to AB.If AB=12 & AC=13. I have to find the area of parallelogram. How can I use Pythagoras theorem here? I do not understand.

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    $\begingroup$ Are you sure you're supposed to use the Pythagorean theorem? It's not what I'd use for this problem. $\endgroup$ – Jack M Feb 18 '15 at 14:35
  • $\begingroup$ Have you drawn the parallelogram? $\endgroup$ – Gobabis Feb 18 '15 at 14:36
  • $\begingroup$ yes i have drawn but confused about it.i have to use pythagorean theorem because it is taken from pythagoras theorem exercise $\endgroup$ – tanz Feb 18 '15 at 14:41
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    $\begingroup$ I can see no reason to work out the length of BC as you have AB as the base and AC as the hieght $\endgroup$ – Gobabis Feb 18 '15 at 14:47
  • $\begingroup$ Your question doesn't make sense. Take a look at: algebra.com/algebra/homework/Pythagorean-theorem/… $\endgroup$ – user137035 Feb 18 '15 at 15:07
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The area of a parallogram is the product of base and height. In your case, the base is AB and the height is AC, so the area is just $12\cdot13=156$, so no need of Pythagoras' theorem. You can use it, however, to calculate $BC=\sqrt{AB^2+AC^2}$.

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  • $\begingroup$ thanks a lot..i will try to solve it using pythagoras theorem. $\endgroup$ – tanz Feb 18 '15 at 14:49
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As sranthrop says, there is no need to use Pythagoras thm.

Try solving this one :

In parallelogram ABCD, the diagonals AC is at right angles to DB.If AB=12 & AC=13. Find the area of parallelogram, using Pythagoras theorem. (Rhombus).

Typo in your exercise?

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