Hello beq!xYou 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?
Grandad
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 = $\displaystyle \tfrac12\Big(2a -\frac{2a\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\B ig)\frac{a\sin\alpha}{\sin(\alpha+\beta)}$
= $\displaystyle \Big(1 -\frac{\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\Big )\frac{a^2\sin\alpha}{\sin(\alpha+\beta)}$
= $\displaystyle \Big(\frac{\sin(\alpha+\beta)-\sin\beta\cos\alpha}{\sin(\alpha+\beta)}\Big)\frac {a^2\sin\alpha}{\sin(\alpha+\beta)}$
= $\displaystyle \Big(\frac{\sin\alpha\cos\beta}{\sin(\alpha+\beta) }\Big)\frac{a^2\sin\alpha}{\sin(\alpha+\beta)}$
= $\displaystyle \frac{a^2\sin^2\alpha\cos\beta}{\sin^2(\alpha+\bet a)}$
Grandad