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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.:)
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:
$\displaystyle BE = f; CE = g; DE = h; AE = j.$
Now, you just use Pythagoras on each one of the four triangles, separately. This gives you:
$\displaystyle f^2 + g^2 = (5 cm)^2$
$\displaystyle g^2 + h^2 = (4 cm)^2$
$\displaystyle h^2 + j^2 = (3 cm)^2$
$\displaystyle j^2 + f^2 = x^2$
where $\displaystyle x$ 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:
$\displaystyle (f^2 + g^2) - (g^2 + h^2) + (h^2 + j^2) = f^2 + j^2 = x^2
$
and so
$\displaystyle x^2 = (5cm)^2 - (4cm)^2 + (3cm)^2 = (25 - 16 + 9)cm^2 = 18 cm^2 $
$\displaystyle x = \sqrt{18} cm \approx 4.24 cm$
Thanks for the help,
So is
$\displaystyle (f^2 + g^2) - (g^2 + h^2) + (h^2 + j^2) = f^2 + j^2 = x^2$
because that essentially is the same as
$\displaystyle f^2 + g^2 + h^2 + j^2 - g^2 - h^2 = f^2 + j^2 = x^2$
The two $\displaystyle g^2$'s and $\displaystyle h^2$'s cancel out, equalling $\displaystyle x^2$
Thanks:)
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:
$\displaystyle f^2 + g^2 = (5 cm)^2$ (1)
$\displaystyle g^2 + h^2 = (4 cm)^2$ (2)
$\displaystyle h^2 + j^2 = (3 cm)^2$ (3)
$\displaystyle f^2 + j^2 = x^2$ (4)
Start with equation 1:
$\displaystyle f^2 + g^2 = (5 cm)^2$
"Subtract equation 2". This means, subtract the LHS of equation 2 from the LHS, and subtract the RHS of equation 2 from the RHS.
$\displaystyle f^2 + g^2 - g^2 - h^2 = (5 cm)^2 - (4 cm)^2$
$\displaystyle f^2 - h^2 = (5 cm)^2 - (4 cm)^2$
(the $\displaystyle h^2$ cancels)
Now "add equation 3" to this, just like before.
$\displaystyle f^2 - h^2 + h^2 + j^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2$
$\displaystyle f^2 + j^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2$
The final step is to use equation 4: $\displaystyle x^2 = f^2 + j^2$
so $\displaystyle x^2 = (5 cm)^2 - (4 cm)^2 + (3 cm)^2$
...and there you have your answer!
Thanks loads. I think the 1st answer is easier to understand. 2 is always better than 1.:)
