Quadrilaterals and Pythagoras Theorem

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June 4th 2007, 12:17 AM
eDLam

Quadrilaterals and Pythagoras Theorem

Hi, new to this,

The diagonals AC and BD, of a quadrilateral ABCD, are perpendicular. Find the length of AB, if BC=5cm, DC=4cm and AD=3cm. Let the intersection of the diagonals be E.

Now I'm thinking the answer will be in surd form. I've been thinking about the question for a while but i don't really know how to go about answering it.

Thanks in advance.:)
June 4th 2007, 05:01 AM
Pterid

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OK... here is a solution, using only Pythagoras' Theorem. The attached diagram shows your quadrilateral ABCD, with the diagonals intersecting at point E (all points labelled in blue capitals).

Now, if the diagonals intersect at right angles, then all four of the angles labelled in red are right angles. So, the quadrilateral becomes a system of four right-angled triangles stuck together!

Notice that I have labelled the lengths of four lines:

$BE = f; CE = g; DE = h; AE = j.$

Now, you just use Pythagoras on each one of the four triangles, separately. This gives you:

$f^2 + g^2 = (5 cm)^2$
$g^2 + h^2 = (4 cm)^2$
$h^2 + j^2 = (3 cm)^2$
$j^2 + f^2 = x^2$

where $x$ is the distance you want to measure. Now, do a little bit of algebra. Start with the first equation, subtract the second one, and then add the third one. You get the following:

$(f^2 + g^2) - (g^2 + h^2) + (h^2 + j^2) = f^2 + j^2 = x^2<br />$

and so

$x^2 = (5cm)^2 - (4cm)^2 + (3cm)^2 = (25 - 16 + 9)cm^2 = 18 cm^2$
$x = \sqrt{18} cm \approx 4.24 cm$
June 5th 2007, 02:54 AM
eDLam

Thanks, want to clear something up.

Thanks for the help,

So is
$(f^2 + g^2) - (g^2 + h^2) + (h^2 + j^2) = f^2 + j^2 = x^2$

because that essentially is the same as
$f^2 + g^2 + h^2 + j^2 - g^2 - h^2 = f^2 + j^2 = x^2$

The two $g^2$ 's and $h^2$ 's cancel out, equalling $x^2$

Thanks:)
June 5th 2007, 05:18 AM
Pterid

Yes, that's right. :)

There is a better way to explain this, which another forum member suggested to me:

Label the equations, 1 to 4:

$f^2 + g^2 = (5 cm)^2$ (1)
$g^2 + h^2 = (4 cm)^2$ (2)
$h^2 + j^2 = (3 cm)^2$ (3)
$f^2 + j^2 = x^2$ (4)

Start with equation 1:

$f^2 + g^2 = (5 cm)^2$

"Subtract equation 2". This means, subtract the LHS of equation 2 from the LHS, and subtract the RHS of equation 2 from the RHS.

$f^2 + g^2 - g^2 - h^2 = (5 cm)^2 - (4 cm)^2$
$f^2 - h^2 = (5 cm)^2 - (4 cm)^2$

(the $h^2$ cancels)

Now "add equation 3" to this, just like before.

$f^2 - h^2 + h^2 + j^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2$
$f^2 + j^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2$

The final step is to use equation 4: $x^2 = f^2 + j^2$

so $x^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2$

...and there you have your answer!
June 6th 2007, 01:58 AM
eDLam

Thanks

Thanks loads. I think the 1st answer is easier to understand. 2 is always better than 1.:)