0
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $BC=\sqrt{AB^2+AC^2}$.

$\endgroup$
  • $\begingroup$ thanks a lot..i will try to solve it using pythagoras theorem. $\endgroup$ – tanz Feb 18 '15 at 14:49
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.