find the area

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May 3rd 2009, 12:37 PM
beq!x

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find the area

In this regular pyramid find the area of http://alt1.mathlinks.ro/latexrender...3585b165f6.gif
i think that we have to solve this by using similar triangles and than continue with other things , and we have to solve it in terms of $a$ and angle http://alt2.mathlinks.ro/latexrender...80b4948e68.gif and http://alt1.mathlinks.ro/latexrender...8632e2f8c4.gif.

I'm having problems with this.
Please help!
May 3rd 2009, 11:41 PM
Grandad

Area
Hello beq!x

Quote:

Originally Posted by beq!x

In this regular pyramid find the area of http://alt1.mathlinks.ro/latexrender...3585b165f6.gif
i think that we have to solve this by using similar triangles and than continue with other things , and we have to solve it in terms of $a$ and angle http://alt2.mathlinks.ro/latexrender...80b4948e68.gif and http://alt1.mathlinks.ro/latexrender...8632e2f8c4.gif.

I'm having problems with this.
Please help!

You haven't told us much about the diagram, but I'll make the following assumptions:
- the sides of the base are a units long
- JK is parallel to AH
- M and N are the mid-points of the sides BG and AH
You need to know the formula:
- Area of a trapezium = $\tfrac12(a+b)h$
where the lengths of the parallel sides are $a$ and $b$ , and the height is $h$ .

So in BGKJ, the area we need is $\tfrac12(BG + JK)LM$ .

Now $BG = a$ (I'm assuming!), and you can find LM using the Sine Rule on $\triangle NML$ :

$\frac{LM}{\sin\alpha}= \frac{a}{\sin(180 - (\alpha+\beta))}$

$\Rightarrow LM = \frac{a\sin\alpha}{\sin(\alpha+\beta)}$

To find JK, you need to use the fact that triangles IJK and IAH are similar, together with the lengths of IL and IN. First IN: drop a perpendicular from I to the base, and then from the right-angled triangle thus formed:

$IN = \tfrac12a\tan\alpha$

From $\triangle LMN,\, NL = \frac{a\sin\beta}{\sin(\alpha+\beta)}$

$\Rightarrow IL = \tfrac12a\tan\alpha-\frac{a\sin\beta}{\sin(\alpha+\beta)}$

So from similar triangles $\frac{JK}{a}=\frac{IL}{IN} = \frac{\tfrac12a\tan\alpha-\dfrac{a\sin\beta}{\sin(\alpha+\beta)}}{\tfrac12a\ tan\alpha}$

Can you complete it from here?

Grandad
May 4th 2009, 12:22 AM
beq!x

no i can't (Crying)
is the solution available ?
May 4th 2009, 04:51 AM
Grandad

Hello beq!x

A correction to my first posting, and then the solution. I meant $\sec\alpha$ rather than $\tan\alpha$ . I.e.

$IN = \tfrac12a\sec\alpha$

$\Rightarrow IL = \tfrac12a\sec\alpha-\frac{a\sin\beta}{\sin(\alpha+\beta)}$

$\Rightarrow\frac{JK}{a}= \frac{IL}{IN}=\frac{\tfrac12a\sec\alpha-\dfrac{a\sin\beta}{\sin(\alpha+\beta)}}{\tfrac12a\ sec\alpha}$

$\Rightarrow JK = \frac{a\sec\alpha-\dfrac{2a\sin\beta}{\sin(\alpha+\beta)}}{\sec\alph a}$

= $a - \frac{2a\sin\beta\cos\alpha}{\sin(\alpha+\beta)}$

So the area = $\tfrac12\Big(2a -\frac{2a\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\B ig)\frac{a\sin\alpha}{\sin(\alpha+\beta)}$

= $\Big(1 -\frac{\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\Big )\frac{a^2\sin\alpha}{\sin(\alpha+\beta)}$

= $\Big(\frac{\sin(\alpha+\beta)-\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\Big)\frac {a^2\sin\alpha}{\sin(\alpha+\beta)}$

= $\Big(\frac{\sin\alpha\cos\beta}{\sin(\alpha+\beta) }\Big)\frac{a^2\sin\alpha}{\sin(\alpha+\beta)}$

= $\frac{a^2\sin^2\alpha\cos\beta}{\sin^2(\alpha+\bet a)}$

Grandad