Question-46919

Question Number 46919 by behi83417@gmail.com last updated on 02/Nov/18

Commented by ajfour last updated on 02/Nov/18

Commented by ajfour last updated on 02/Nov/18

l e t C F = h; F M = F B = \sqrt{4 - h^{2}}

A B = \sqrt{9 - h^{2}} + \sqrt{4 - h^{2}} = 2 M B

\Rightarrow \sqrt{9 - h^{2}} + \sqrt{4 - h^{2}} = 4 \sqrt{4 - h^{2}}

\Rightarrow 9 - h^{2} = 9 (4 - h^{2})

o r 8 h^{2} = 27 \Rightarrow h = \frac{3 \sqrt{3}}{2 \sqrt{2}}

A B = 4 \sqrt{4 - \frac{27}{8}} = \sqrt{10}

A r e a (△ A B C) = \frac{h}{2} \times A B

= \frac{3 \sqrt{3}}{4 \sqrt{2}} \times \sqrt{10} = \frac{3 \sqrt{15}}{4} .

Commented by behi83417@gmail.com last updated on 02/Nov/18

t h a n k y o u s o m u c h s i r A j f o u r .

y o u r a n s w e r i s r i g h t w i t h 13 %

d e f f e r e n c e .

a r e a = 2.76 u n i t s .

You can't use 'macro parameter character #' in math mode

Commented by ajfour last updated on 02/Nov/18

B e h i S i r, i c h e c k e d, b u t {c o u l d n}^{'} t

f i n d m i s t a k e i n m y s o l u t i o n . .!

Commented by MJS last updated on 02/Nov/18

A M = 2 m

M F = F B = m

m^{2} + h^{2} = 4

9 m^{2} + h^{2} = 9

\Rightarrow m = \frac{\sqrt{10}}{4} \land h = \frac{3 \sqrt{6}}{4}

\Rightarrow c = A B = 4 m = \sqrt{10}

\Rightarrow area = \frac{3 \sqrt{15}}{4}

\dots so Sir Aifour is right

Commented by behi83417@gmail.com last updated on 02/Nov/18

t h a n k y o u v e r y m u c h s i r .

y o u a n d s i r A j f o u r, u s e :

[a = 2, m = 2] t o t a k e C \overset{△}{M B}, a s i s o s c a l e

t r i a n g l e . n o w i f a \neq 2, w h a t c a n

w e d o ? [s a y a = 5, f o r e x a m p l e]

(f o r : a = 5, b = 3, m = 2,

a r e a = \frac{8 \sqrt{161}}{15} = 6.77)

Commented by behi83417@gmail.com last updated on 03/Nov/18

2 (a^{2} + b^{2}) = c^{2} + 4 m_{c}^{2} \Rightarrow 2 (2^{2} + 3^{2}) = c^{2} + 4 \times 2^{2}

\Rightarrow c^{2} = 26 - 16 = 10 \Rightarrow c = \sqrt{10}

S = \frac{1}{4} \sqrt{(2 + 3 + \sqrt{10}) (2 + 3 - \sqrt{10}) (2 + \sqrt{10} - 3) (3 + \sqrt{10} - 2)} =

= \frac{1}{4} \sqrt{(5 + \sqrt{10}) (5 - \sqrt{10}) (\sqrt{10} + 1) (\sqrt{10} - 1)} =

= \frac{1}{4} \sqrt{(25 - 10) (10 - 1)} = \frac{3 \sqrt{15}}{4} .

t h e f o r m u l a g i v e n f o r i s :

S = \frac{m_{c} . (a + b)}{4 ab} \sqrt{4 a^{2} b^{2} - m_{c}^{2} . {(a + b)}^{2}}

but i {can}^{'} t p r o v e i t .

Commented by ajfour last updated on 03/Nov/18

l e t M (0, 0); A (- l, 0); B (l, 0);

C (x, y)

\Rightarrow S = \frac{1}{2} (2 l) y = l y

\Rightarrow x^{2} + y^{2} = m^{2}

{(x + l)}^{2} + y^{2} = b^{2}

{(x - l)}^{2} + y^{2} = a^{2}

\Rightarrow 2 l^{2} = a^{2} + b^{2} - 2 m^{2}

& 4 l x = b^{2} - a^{2}

\Rightarrow 16 l^{2} x^{2} = {(b^{2} - a^{2})}^{2}

\Rightarrow 16 l^{2} (m^{2} - y^{2}) = {(b^{2} - a^{2})}^{2}

o r 16 l^{2} m^{2} - 16 S^{2} = {(b^{2} - a^{2})}^{2}

S^{2} = \frac{16 l^{2} m^{2} - {(b^{2} - a^{2})}^{2}}{16}

= \frac{8 m^{2} (a^{2} + b^{2} - 2 m^{2}) - {(b^{2} - a^{2})}^{2}}{16}

= \frac{8 m^{2} (a^{2} + b^{2}) - {(a^{2} + b^{2})}^{2} + 4 a^{2} b^{2} - 16 m^{4}}{16}

= \frac{4 a^{2} b^{2} - {(a^{2} + b^{2} - 4 m^{2})}^{2}}{16}

\Rightarrow f o r a = 2, b = 3, m = 2

S^{2} = \frac{144 - 9}{16} = \frac{135}{16}

S = \frac{3 \sqrt{15}}{4} .

I f b = 3, a = 5, m = 2, t h e n

S^{2} = \frac{900 - 324}{16} = \frac{576}{16} = 36

S = 6 .

Commented by behi83417@gmail.com last updated on 03/Nov/18

s i r A j f o u r . y o u a n d y o u r s o l u t i o n

a r e a m a z i n g . t h a n k y o u s o m u c h .