Question-43945

Question Number 43945 by ajfour last updated on 18/Sep/18

Commented by ajfour last updated on 18/Sep/18

T h e r e c t a n g u l a r p l a t e c a n s w i n g

f r e e l y a b o u t t h e d i a g o n a l b u t i s

a t b a l a n c e, n o t t u r n i n g a n y m o r e .

F i n d p, h i n t e r m s o f a, b, d .

Answered by MrW3 last updated on 19/Sep/18

G = \sqrt{a^{2} + b^{2}} = l e n g t h o f d i a g o n a l

H = \sqrt{G^{2} - d^{2}} = \sqrt{a^{2} + b^{2} - d^{2}} = h e i g h t o f t o p m o s t c o r n e r

\frac{e}{b} = \frac{b}{G} \Rightarrow e = \frac{b^{2}}{G}

\frac{f}{a} = \frac{a}{G} \Rightarrow f = \frac{a^{2}}{G}

p = \frac{e}{G} \times H = \frac{b^{2}}{a^{2} + b^{2}} \sqrt{a^{2} + b^{2} - d^{2}}

h = \frac{f}{G} \times H = \frac{a^{2}}{a^{2} + b^{2}} \sqrt{a^{2} + b^{2} - d^{2}}

Commented by MrW3 last updated on 19/Sep/18

Commented by ajfour last updated on 19/Sep/18

G r e a t s i m p l i f i c a t i o n S i r .

A w e s o m e s o l u t i o n .

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