Math Question and Answers


Question and Answers Forum

All Questions Topic List
Geometry Questions
Previous in All Question Next in All Question
Previous in Geometry Next in Geometry
Question Number 49830 by ajfour last updated on 11/Dec/18

Commented by ajfour last updated on 11/Dec/18

$F i n d c i r c u m r a d i u s r i n t e r m s o f a .$
	Answered by mr W last updated on 11/Dec/18

	$r = \frac{1}{4} \sqrt{\frac{(a b + c d) (a c + b d) (a d + b c)}{(s - a) (s - b) (s - c) (s - d)}}$ $s = \frac{1}{2} (a + b + c + d) = \frac{1}{2} (a + 2 a + 3 a + 4 a) = 5 a$ $r = \frac{1}{4} \sqrt{\frac{(2 a^{2} + 12 a^{2}) (3 a^{2} + 8 a^{2}) (4 a^{2} + 6 a^{2})}{(4 a) (3 a) (2 a) (a)}}$ $r = \frac{a}{4} \sqrt{\frac{14 \times 11 \times 10}{4 \times 3 \times 2 \times 1}}$ $r = \frac{\sqrt{2310} a}{24} \approx 2.0026 a$
	Commented by mr W last updated on 11/Dec/18

	Commented by ajfour last updated on 11/Dec/18

	$T h a n k y o u S i r! (d i n t k n o w o f t h e$ $f o r m u l a) .$
	Answered by behi83417@gmail.com last updated on 11/Dec/18

	$l e t : \frac{a^{2}}{2 r^{2}} = λ$ ${c o s}^{- 1} \frac{2 r^{2} - a^{2}}{2 r^{2}} + {c o s}^{- 1} \frac{2 r^{2} - 4 a^{2}}{2 r^{2}} + {c o s}^{- 1} \frac{2 r^{2} - 9 a^{2}}{2 r^{2}} +$ ${c o s}^{- 1} \frac{2 r^{2} - 16 a^{2}}{2 r^{2}} = 2 π$ ${c o s}^{- 1} (1 - λ) + {c o s}^{- 1} (1 - 4 λ) + {c o s}^{- 1} (1 - 9 λ) +$ $+ {c o s}^{- 1} (1 - 16 λ) = 2 π \Rightarrow$ $a + b + c + d = 2 π \Rightarrow c o s (a + c) = c o s (b + d)$ ${c o s}^{- 1} (1 - λ) + {c o s}^{- 1} (1 - 9 λ) = 2 π - [{c o s}^{- 1} (1 - 2 λ) + {c o s}^{- 1} (1 - 16 λ)]$ $c o s (a + c) = (1 - λ) . (1 - 9 λ) - [(1 - {(1 - λ)}^{2}] [[1 - {(1 - 9 λ)}^{2}] =$ $= 1 - 10 λ - 27 λ^{2} + 180 λ^{3} - 81 λ^{4}$ $c o s (b + d) = (1 - 4 λ) (1 - 16 λ) - [(1 - {(1 - 4 λ)}^{2}] [(1 - {(1 - 16 λ)}^{2}] =$ $= 1 - 20 λ - 192 λ^{2} + 2560 λ^{3} - 4096 λ^{4}$ $1 - 10 λ - 27 λ^{2} + 180 λ^{3} - 81 λ^{4} =$ $1 - 20 λ - 192 λ^{2} + 2560 λ^{3} - 4096 λ^{4}$ $\Rightarrow 4015 λ^{3} - 2380 λ^{2} + 165 λ + 10 = 0$ $\Rightarrow λ = \frac{a^{2}}{2 r^{2}} = 0.5008, 0.1302 \Rightarrow$ $\Rightarrow \frac{a}{r} = 1.007, 0.5103$
	Answered by mr W last updated on 11/Dec/18

	$a n o t h e r w a y w i t h o u t u s i n g f o r m u l a :$ $l e t α = ∠ A$ $\Rightarrow ∠ C = π - α$ $l e t B D = l$ $l^{2} = a^{2} + {(4 a)}^{2} - 2 (a) (4 a) \cos α$ $\Rightarrow l^{2} = 17 a^{2} - 8 a^{2} \cos α$ $l^{2} = {(2 a)}^{2} + {(3 a)}^{2} - 2 (2 a) (3 a) \cos (π - α)$ $\Rightarrow l^{2} = 13 a^{2} + 12 a^{2} \cos α$ $\Rightarrow 13 a^{2} + 12 a^{2} \cos α = 17 a^{2} - 8 a^{2} \cos α$ $\Rightarrow \cos α = \frac{1}{5}$ $\Rightarrow l^{2} = 17 a^{2} - 8 a^{2} \times \frac{1}{5} = \frac{77}{5} a^{2}$ $\Rightarrow l = \frac{\sqrt{385}}{5} a$ $r i s c i r c u m r a d i u s o f Δ A B D w i t h s i d e s$ $a, 4 a, \frac{\sqrt{385}}{5} a .$ $r = \frac{a \times 4 a \times \frac{\sqrt{385}}{5} a}{\sqrt{(a + 4 a + \frac{\sqrt{385}}{5} a) (- a + 4 a + \frac{\sqrt{385}}{5} a) (a - 4 a + \frac{\sqrt{385}}{5} a) (a + 4 a - \frac{\sqrt{385}}{5} a)}}$ $r = \frac{4 \times \frac{\sqrt{385}}{5} a}{\sqrt{(\frac{385}{25} - 9) (25 - \frac{385}{25})}} = \frac{20 \sqrt{385} a}{\sqrt{160 \times 240}}$ $r = \frac{\sqrt{385} a}{4 \sqrt{6}} = \frac{\sqrt{2310} a}{24} \approx 2.0026 a$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum