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


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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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