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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.
The task is:
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.)
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November 9th, 2012, 01:23 AM   #2
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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?
Hoempa is offline  
November 9th, 2012, 06:56 PM   #3
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I've corrected some slips in the above post.
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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) - b)
Due to Pythagoras:
Code:
a + b = 12 = 144
b + c = 9 = 81
Then
Code:
b = 144 - a = 81 - c
b = a - c = 63
So
Code:
a + b = 144
a + 63 = 144
a = 81
a = 9
And
Code:
b + c = 81
63 + c = 81
c = 18
... etc.
Next thing, if AS = x, CS = 12-x, DS = y and BS = 9-y
then due to Pythagoras
Code:
a = 12x
b = 9*(9-y) = 12*(12-x)
c = 9y
and
Code:
12-x = y(9-y)
9-y = x(12-x)
12x - 9y = x - y
so
Code:
c - y = a - x
x - y = a -c
... b = a -c
But it doesn't fit, where am I doing a mistake?
Thanks very much.
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November 10th, 2012, 12:42 PM   #5
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There's more than one error. If you try to justify everything you have done, you should be able to spot the errors.
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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.
The task is:
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 ?
Denis is offline  
November 10th, 2012, 01:46 PM   #7
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Both expressions are valid.
skipjack is offline  
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?
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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:
b = a - c = 63
How do you get this?
Hoempa is offline  
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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