# find the area

• May 3rd 2009, 12:37 PM
beq!x
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 $\displaystyle a$ and angle http://alt2.mathlinks.ro/latexrender...80b4948e68.gif and http://alt1.mathlinks.ro/latexrender...8632e2f8c4.gif.

I'm having problems with this.
• May 3rd 2009, 11:41 PM
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 $\displaystyle a$ and angle http://alt2.mathlinks.ro/latexrender...80b4948e68.gif and http://alt1.mathlinks.ro/latexrender...8632e2f8c4.gif.

I'm having problems with this.

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 = $\displaystyle \tfrac12(a+b)h$

where the lengths of the parallel sides are $\displaystyle a$ and $\displaystyle b$, and the height is $\displaystyle h$.

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

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

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

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

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:

$\displaystyle IN = \tfrac12a\tan\alpha$

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

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

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

Can you complete it from here?

• May 4th 2009, 12:22 AM
beq!x
no i can't (Crying)
is the solution available ?
• May 4th 2009, 04:51 AM
Hello beq!x

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

$\displaystyle IN = \tfrac12a\sec\alpha$

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

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

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

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

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

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

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

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

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