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Question Number 48522 by ajfour last updated on 25/Nov/18

Commented by ajfour last updated on 25/Nov/18

$A r e g u l a r h e x a g o n a l p y r a m i d$ $w i t h b a s e e d g e a a n d a l t i t u d e h .$ $F i n d t h e a r e a o f a s e c t i o n t h a t$ $p a s s e s t h r o u g h m i d p o i n t s P, Q$ $o f s i d e s A B a n d C D a n d a l s o$ $t h r o u g h M, t h e m i d p o i n t o f$ $a l t i t u d e .$
	Answered by mr W last updated on 25/Nov/18

	Commented by mr W last updated on 25/Nov/18

	$A D = 2 a$ $b_{1} = P Q = (a + 2 a) / 2 = 3 a / 2$ $b_{2} = U R = A D / 2 = a$ $c = \sqrt{{(\sqrt{3} a / 4)}^{2} + {(h / 2)}^{2}} = (\sqrt{3 a^{2} + 4 h^{2}}) / 4$ $\frac{e}{f - h / 2} = \frac{\sqrt{3} a / 4}{h / 2}$ $\frac{e}{h - f} = \frac{\sqrt{3} a / 2}{h}$ $\Rightarrow f - h / 2 = h - f$ $\Rightarrow f = 3 h / 4$ $b_{3} = S T$ $\frac{b_{3}}{a} = \frac{h - f}{h} = 1 / 4$ $\Rightarrow b_{3} = a / 4$ $\frac{d + c}{c} = \frac{f}{h / 2} = \frac{3}{2}$ $\frac{d}{c} = \frac{1}{2}$ $\Rightarrow d = c / 2 = (\sqrt{3 a^{2} + 4 h^{2}}) / 8$ $A r e a o f P Q R S T U P = A$ $A = (b_{1} + b_{2}) c / 2 + (b_{2} + b_{3}) d / 2 = [b_{1} c + b_{2} (c + d) + b_{3} d] / 2$ $= \frac{1}{2} [\frac{3 a}{2} \times \frac{\sqrt{3 a^{2} + 4 h^{2}}}{4} + a \times \frac{3 \sqrt{3 a^{2} + 4 h^{2}}}{8} + \frac{a}{4} \times \frac{\sqrt{3 a^{2} + 4 h^{2}}}{8}]$ $\Rightarrow A = \frac{25 a \sqrt{3 a^{2} + 4 h^{2}}}{64}$
	Commented by ajfour last updated on 25/Nov/18

	$e x c e l l e n t S i r, b u t a n s w e r i s b i t$ $d i f f e r e n t . .$
	Commented by mr W last updated on 25/Nov/18

	$I m a d e a l i t t l e m i s t a k e, n o w {i t}^{'} s f i x e d .$
	Commented by ajfour last updated on 25/Nov/18

	$R e a l l y C o r r e c t n o w, a n d b e y o n d$ $a l l a d m i r a t i o n a n d p r a i s e S i r;$ $\frac{b_{3}}{a} = \frac{h - f}{h} (p l e a s e h e l p; h o w ?)$
	Commented by mr W last updated on 26/Nov/18

	$S^{'} = m i d p o i n t o f S a n d T$ $E^{'} = m i d p o i n t o f E a n d F$ $\frac{S T}{E F} = \frac{{V S}^{'}}{{V E}^{'}} = \frac{h - f}{h}$ $\Rightarrow \frac{b_{3}}{a} = \frac{h - f}{h}$
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