given-isoscele-triangle-with-sides-10-and-inradius-3-how-find-base-

Question Number 212533 by emilagazade last updated on 16/Oct/24

g i v e n i s o s c e l e t r i a n g l e w i t h s i d e s 10 a n d i n r a d i u s 3 . h o w f i n d b a s e ?

Answered by A5T last updated on 16/Oct/24

L e t b a s e = b; a n g l e b e t w e e n n o n - e q u a l s i d e s = θ

[△] = \frac{10 \times 10 s i n (180 - 2 θ)}{2} = \frac{b \times 10 \times s i n θ}{2}

\Rightarrow 50 \times 2 c o s θ = 5 b \Rightarrow c o s θ = \frac{b}{20} \Rightarrow s i n θ = \frac{\sqrt{400 - b^{2}}}{20}

[△] = \frac{3 (10 + 10 + b)}{2} = \frac{10 b s i n θ}{2} = \frac{b \sqrt{400 - b^{2}}}{4}

\Rightarrow 120 + 6 b = b \sqrt{400 - b^{2}} \Rightarrow b = 12 o r 4 + 2 \sqrt{19}

Commented by emilagazade last updated on 16/Oct/24

t h a n k y o u a l o t

Answered by mr W last updated on 16/Oct/24

Commented by mr W last updated on 17/Oct/24

s a y x = \frac{b}{2}

\frac{a - x}{r} = \frac{\sqrt{a^{2} - x^{2}}}{x}

\frac{a - x}{r^{2}} = \frac{a + x}{x^{2}}

x^{3} - {a x}^{2} + r^{2} x + {a r}^{2} = 0

x^{3} - 10 x^{2} + 9 x + 90 = 0

(x - 6) (x^{2} - 4 x - 15) = 0

\Rightarrow x = 6, 2 \pm \sqrt{19}

\Rightarrow b = 12, 4 + 2 \sqrt{19}

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