Calculate-the-side-of-an-equilateral-triangle-whose-vertices-are-situated-on-three-parallel-coplanar-lines-knowing-that-a-and-b-are-the-distances-of-the-parallel-line-to-the-others-

Question Number 76974 by Maclaurin Stickker last updated on 02/Jan/20

C a l c u l a t e t h e s i d e o f a n e q u i l a t e r a l

t r i a n g l e w h o s e v e r t i c e s a r e s i t u a t e d

o n t h r e e p a r a l l e l c o p l a n a r l i n e s,

k n o w i n g t h a t \boldsymbol a a n d \boldsymbol b a r e t h e d i s t a n c e s

o f t h e p a r a l l e l l i n e t o t h e o t h e r s .

Answered by mr W last updated on 02/Jan/20

\sqrt{l^{2} - {(a + b)}^{2}} = \sqrt{l^{2} - b^{2}} - \sqrt{l^{2} - a^{2}}

l^{2} - a^{2} - b^{2} - 2 a b = l^{2} - b^{2} + l^{2} - a^{2} - 2 \sqrt{(l^{2} - b^{2}) (l^{2} - a^{2})}

l^{2} + 2 a b = 2 \sqrt{(l^{2} - b^{2}) (l^{2} - a^{2})}

l^{4} + 4 {a b l}^{2} + 4 a^{2} b^{2} = 4 (l^{2} - b^{2}) (l^{2} - a^{2})

3 l^{2} - 4 (a^{2} + b^{2} + a b) = 0

\Rightarrow l = 2 \sqrt{\frac{a^{2} + b^{2} + a b}{3}}

Commented by Maclaurin Stickker last updated on 02/Jan/20

h o w d i d y o u g e t t h e f i r s t e x p r e s s i o n ?

Commented by mr W last updated on 02/Jan/20

Commented by mr W last updated on 02/Jan/20

A B = \sqrt{l^{2} - b^{2}}

C D = \sqrt{l^{2} - {(a + b)}^{2}}

D E = \sqrt{l^{2} - a^{2}}

C D + D E = A B

\sqrt{l^{2} - {(a + b)}^{2}} + \sqrt{l^{2} - a^{2}} = \sqrt{l^{2} - b^{2}}

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