Question-45427

Question Number 45427 by ajfour last updated on 12/Oct/18

Commented by ajfour last updated on 12/Oct/18

D e t e r m i n e s i d e a o f s q u a r e,

w i t h o u t u s i n g c a l c u l a t o r .

Commented by ajfour last updated on 12/Oct/18

Commented by ajfour last updated on 12/Oct/18

\frac{A E}{E D} = \frac{D H}{C H} \Rightarrow \frac{A E}{a} = \frac{a}{1}

\Rightarrow A E = a^{2}

{E G}^{2} = 1 = x^{2} + {(a - x)}^{2} \dots . (i)

A n d \frac{a - x}{x} = \frac{a}{a + A E} = \frac{a}{a + a^{2}}

\Rightarrow (a - x) (1 + a) = x

o r a + a^{2} - x - a x = x

\Rightarrow x = \frac{a + a^{2}}{2 + a} \dots (i i)

u s i n g (i i) i n (i) :

{(\frac{a + a^{2}}{2 + a})}^{2} + {(a - \frac{a + a^{2}}{2 + a})}^{2} = 1

\Rightarrow a^{2} {(1 + a)}^{2} + a^{2} = {(2 + a)}^{2}

\Rightarrow a^{4} + 2 a^{3} + 2 a^{2} = 4 + 4 a + a^{2}

\Rightarrow a^{4} + 2 a^{3} + a^{2} - 4 a - 4 = 0

a^{3} (a + 1) + a^{2} (a + 1) - 4 (a + 1) = 0

a s a \neq - 1

\Rightarrow a^{3} + a^{2} - 4 = 0

l e t a = b - \frac{1}{3}

\Rightarrow b^{3} - b^{2} + \frac{b}{3} - \frac{1}{27} + b^{2} - \frac{2 b}{3} + \frac{1}{9} - 4 = 0

\Rightarrow b^{3} - \frac{b}{3} - \frac{106}{27} = 0

b = {(\frac{53}{27} + \sqrt{{(\frac{53}{27})}^{2} - {(\frac{1}{27})}^{2}})}^{1 / 3}

+ {(\frac{53}{27} - \sqrt{{(\frac{53}{27})}^{2} - {(\frac{1}{27})}^{2}})}^{1 / 3}

= {(\frac{53}{27} + \frac{\sqrt{54 \times 52}}{27})}^{1 / 3} + {(\frac{53}{27} - \frac{\sqrt{54 \times 52}}{27})}^{1 / 3}

a = - \frac{1}{3} + \frac{{(53 + \sqrt{54 \times 52})}^{1 / 3}}{3}

+ \frac{{(53 - \sqrt{54 \times 52})}^{1 / 3}}{3}

a \approx 1.3146

Commented by MrW3 last updated on 13/Oct/18

a n o t h e r n i c e a p p l i c a t i o n o f c u b i c e q n .!

Facebook Tweet Pin