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November 8th, 2012, 05:03 PM   #1
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Calculation of sides and area of right-angled trapezoid

I have to calculate sides and area of right-angled trapezoid if I know only diagonals.
Quote:
 You have a right-angled trapezoid ABCD with right angle at the point B. You know that diagonal |AC| = 12cm and diagonal |BD| = 9cm. Diagonals are perpendicular each other. Calculate the sides and area of trapezoid.
I know that area is (|AC|*|BD|)/2. But what about the sides?
(Sorry for my English, I am not from English-speaking country.)

 November 9th, 2012, 01:23 AM #2 Math Team   Joined: Apr 2010 Posts: 2,780 Thanks: 361 Re: Calculation of sides and area of right-angled trapezoid Here is a sketch Triangles ABS, BCS and CDS are similar since angle BAS = angle CBS = angle DCS (see colored dots) and all these triangles are right-angled. So we have |DS|/|CS| = |CS|/|BS| = |BS|/|AS| (1), |BS| + |DS| = |BD| = 9 and |AS| + |CS| = |AC| = 12 Shifting |AC| a length of CD along CD gives us (|CD| + |AB|)^2 = 9^2 + 12^2 = 225. By Pythagoras: |CS|^2 + |DS|^2 = |CD|^2 |CS|^2 + |BS|^2 = |BC|^2 |BS|^2 + |AS|^2 = |AB|^2 |AS|^2 + |DS|^2 = |AD|^2 Now, let |DS| = a and |CS|/|DS| = b. Then |CS| = ab, |BS| = ab^2 and |AS| = ab^3. Gives us |BS| + |DS| = ab^2 + a = 9 = a(b^2 + 1) and |AS| + |CS| = ab^3 + ab = b(ab^2 + a) = 9b = 12 i.e. b = 4/3, so a = 9/(b^2 + 1) = 9/((4/3)^2 + 1) = 9/(5/3)^2 = 81/25. Can you use this to find the sides?
 November 9th, 2012, 06:56 PM #3 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 I've corrected some slips in the above post.
November 10th, 2012, 04:26 AM   #4
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Re: Calculation of sides and area of right-angled trapezoid

Hello, thanks very much for your response. But I don't understand this:
Quote:
 Originally Posted by Hoempa ... Then |CS| = ab, |BS| = ab^2 and |AS| = ab^3. ...
What's the rule for this?
---
I was thinking about my solution, here it is:
Let AB = a, BC = b, CD = c and DA = d.
Then
Code:
d = ?((a-c)² - b²)
Due to Pythagoras:
Code:
a² + b² = 12² = 144
b² + c² = 9² = 81
Then
Code:
b² = 144 - a² = 81 - c²
b² = a² - c² = 63
So
Code:
a² + b² = 144
a² + 63 = 144
a² = 81
a = 9
And
Code:
b² + c² = 81
63 + c² = 81
c² = 18
... etc.
Next thing, if AS = x, CS = 12-x, DS = y and BS = 9-y
then due to Pythagoras
Code:
a² = 12x
b² = 9*(9-y) = 12*(12-x)
c² = 9y
and
Code:
12-x = y(9-y)
9-y = x(12-x)
12x - 9y = x² - y²
so
Code:
c² - y² = a² - x²
x² - y² = a² -c²
... b² = a² -c²
But it doesn't fit, where am I doing a mistake?
Thanks very much.

 November 10th, 2012, 12:42 PM #5 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 There's more than one error. If you try to justify everything you have done, you should be able to spot the errors.
November 10th, 2012, 12:50 PM   #6
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Re: Calculation of sides and area of right-angled trapezoid

Quote:
Originally Posted by perwin
I have to calculate sides and area of right-angled trapezoid if I know only diagonals.
Quote:
 You have a right-angled trapezoid ABCD with right angle at the point B. You know that diagonal |AC| = 12cm and diagonal |BD| = 9cm. Diagonals are perpendicular each other. Calculate the sides and area of trapezoid.
I know that area is (|AC|*|BD|)/2. But what about the sides?
?? Isn't the area BC * (AB + CD) / 2 ?

 November 10th, 2012, 01:46 PM #7 Global Moderator   Joined: Dec 2006 Posts: 21,028 Thanks: 2259 Both expressions are valid.
November 10th, 2012, 05:46 PM   #8
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Re:

Quote:
 Originally Posted by skipjack Both expressions are valid.
My bad...because diagonals cross at right angles, right?

November 11th, 2012, 03:00 AM   #9
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Re:

Quote:
 Originally Posted by [b [color=#00AA00]skipjack[/color][/b]]I've corrected some slips in the above post.
Thank you. I hope, there were no major slips.

Here is what I used:
Earlier, we found:

|DS|/|CS| = |CS|/|BS| = |BS|/|AS| which gives |CS|/|DS| = |BS|/|CS| = |AS|/|BS|
I set |DS| = a and |CS|/|DS| = b, yields,
|CS| = |CS| * 1 = |CS| * (|DS|/|DS|) = |DS| * (|CS|/|DS|) = a * b = ab. (you don't need the bold brackets..)
Likewise, |BS| = |CS| * (|BS|/|CS|) = ab * b = ab^2
and |AS| = |BS| * (|AS|/|BS|) = ab^2 * b = ab^3
Get it?

Quote:
 Originally Posted by [b [color=#0040FF]perwin[/color][/b]] Code: b² = a² - c² = 63
How do you get this?

November 11th, 2012, 08:29 AM   #10
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Quote:
 Originally Posted by Denis ...because diagonals cross at right angles, right?
Yes. The area is half the product of the diagonals multiplied by the sine of their angle of intersection.

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