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 160014 by mr W last updated on 23/Nov/21

Commented by mr W last updated on 23/Nov/21

$Q 158675$
Commented by mr W last updated on 24/Nov/21

$e x a c t s o l u t i o n = \frac{118}{7}$
	Answered by mr W last updated on 23/Nov/21

	Commented by mr W last updated on 24/Nov/21

	$\frac{B F}{B A} = \frac{F G}{G A} = \frac{F C}{C A}$ $\frac{B F}{16} = \frac{F G}{p} = \frac{F C}{10} = k, s a y$ $\Rightarrow B F = 16 k, F C = 10 k, F G = k p$ $\frac{16^{2} k^{2} + {(1 + k)}^{2} p^{2} - 16^{2}}{2 \times 16 k \times (1 + k) p} = - \frac{10^{2} k^{2} + {(1 + k)}^{2} p^{2} - 10^{2}}{2 \times 10 k \times (1 + k) p}$ $\frac{{(1 + k)}^{2} p^{2} - 16^{2} (1 - k^{2})}{8} = - \frac{{(1 + k)}^{2} p^{2} - 10^{2} (1 - k^{2})}{5}$ $\frac{(1 + k) p^{2} - 16^{2} (1 - k)}{8} = - \frac{(1 + k) p^{2} - 10^{2} (1 - k)}{5}$ $5 (1 + k) p^{2} - 5 \times 16^{2} (1 - k) + 8 (1 + k) p^{2} - 8 \times 10^{2} (1 - k) = 0$ $13 (1 + k) p^{2} = (5 \times 16^{2} + 8 \times 10^{2}) (1 - k)$ $\Rightarrow p^{2} = \frac{160 (1 - k)}{1 + k}$ $\frac{8^{2} + p^{2} - s^{2}}{2 \times 8 \times p} = \frac{16^{2} + {(1 + k)}^{2} p^{2} - 16^{2} k^{2}}{2 \times 16 \times (1 + k) p}$ $8^{2} + p^{2} - s^{2} = \frac{(1 + k) p^{2} + 16^{2} (1 - k)}{2}$ $s^{2} = \frac{(1 - k) p^{2}}{2} + 64 (2 k - 1)$ $s^{2} = \frac{80 {(1 - k)}^{2}}{1 + k} + 64 (2 k - 1)$ $\Rightarrow s = 4 \sqrt{\frac{13 k^{2} - 6 k + 1}{1 + k}} . . . (i)$ $\frac{7^{2} + p^{2} - t^{2}}{2 \times 7 \times p} = \frac{10^{2} + {(1 + k)}^{2} p^{2} - 10^{2} k^{2}}{2 \times 10 \times (1 + k) p}$ $\frac{7^{2} + p^{2} - t^{2}}{7} = \frac{(1 + k) p^{2} + 10^{2} (1 - k)}{10}$ $t^{2} = \frac{(3 - 7 k) p^{2}}{10} + 70 k - 21$ $t^{2} = \frac{16 (3 - 7 k) (1 - k)}{1 + k} + 70 k - 21$ $\Rightarrow t = \sqrt{\frac{182 k^{2} - 111 k + 27}{1 + k}} . . . (i i)$ $\frac{8^{2} + 7^{2} - {(s + t)}^{2}}{2 \times 8 \times 7} = \frac{16^{2} + 10^{2} - 26^{2} k^{2}}{2 \times 16 \times 10}$ $\frac{113 - {(s + t)}^{2}}{7} = \frac{89 - 169 k^{2}}{5} e$ $113 \times 5 - 5 {(s + t)}^{2} = 7 \times 89 - 7 \times 169 k^{2}$ $\Rightarrow 1183 k^{2} - 5 {(s + t)}^{2} = 58 . . . (i i i)$ $i n s e r t i n g (i) a n d (i i) i n t o (i i i), w e g e t$ $k \approx 0.64835165 a n d k = 1 (r e j e c t e d)$ $\Rightarrow ? = B C = 26 k \approx 16.857143$ $====================$ $e x a c t s o l u t i o n :$ $5 (s^{2} + t^{2} + 2 s t) = 1183 k^{2} - 58$ $5 (390 k^{2} - 207 k + 43) + 40 \sqrt{(13 k^{2} - 6 k + 1) (182 k^{2} - 111 k + 27)} = (1183 k^{2} - 58) (1 + k)$ $40 \sqrt{(13 k^{2} - 6 k + 1) (182 k^{2} - 111 k + 27)} = 1183 k^{3} - 767 k^{2} + 977 k - 273$ $1600 (13 k^{2} - 6 k + 1) (182 k^{2} - 111 k + 27) = {(1183 k^{3} - 767 k^{2} + 977 k - 273)}^{2}$ $(k - 1) (k + 1) (13 k - 3) (13 k + 3) {(91 k - 59)}^{2} = 0$ $k = \pm 1 (r e j e c t e d, s i n c e k < \frac{16 + 10}{26} = 1)$ $k = \pm \frac{3}{13} (r e j e c t e d, s i n c e k > \frac{16 - 10}{26} = \frac{3}{13})$ $k = \frac{59}{91} ✓$ $\Rightarrow ? = B C = 26 k = \frac{26 \times 59}{91} = \frac{118}{7} \approx 16.857$
	Commented by Tawa11 last updated on 23/Nov/21

	$Great sir$
	Commented by Tawa11 last updated on 23/Nov/21

	$God bless you sir . I appreciate .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum