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Question Number 193686 by Rupesh123 last updated on 18/Jun/23

	Answered by mr W last updated on 18/Jun/23

	Commented by mr W last updated on 18/Jun/23

	$c = \sqrt{a^{2} + b^{2}}$ $R = a + b$ $R^{2} = b^{2} + c^{2} - 2 b c \cos (α + θ)$ $a^{2} + b^{2} + 2 a b = b^{2} + a^{2} + b^{2} - 2 b \sqrt{a^{2} + b^{2}} \cos (α + θ)$ $b - 2 a = 2 \sqrt{a^{2} + b^{2}} \cos (α + θ)$ $b - 2 a = 2 \sqrt{a^{2} + b^{2}} (\cos α \cos θ - \sin α \sin θ)$ $b - 2 a = 2 (b \cos θ - a \sin θ)$ $a \sin θ - \frac{2 a - b}{2} = b \cos θ$ $a^{2} \sin^{2} θ - (2 a - b) a \sin θ + \frac{{(2 a - b)}^{2}}{4} = b^{2} - b^{2} \sin^{2} θ$ $(a^{2} + b^{2}) \sin^{2} θ - (2 a - b) a \sin θ + \frac{4 a^{2} - 4 a b - 3 b^{2}}{4} = 0$ $\sin θ = \frac{1}{2 (a^{2} + b^{2})} [a (2 a - b) + \sqrt{a^{2} {(2 a - b)}^{2} - (a^{2} + b^{2}) (4 a^{2} - 4 a b - 3 b^{2}})]$ $A r e a = \frac{c^{2} \sin θ}{2} = \frac{1}{4} [a (2 a - b) + b \sqrt{4 a (2 a - b) - b^{2}}]$
	Commented by Rupesh123 last updated on 18/Jun/23
	[1] check last two lines (1/2 ? --> 1/4 ) [2] simplify last radical
	Commented by Mingma last updated on 18/Jun/23
	It's been a while sir! Neat work!
	Commented by Rupesh123 last updated on 18/Jun/23
	Perfect ��
	Commented by ajfour last updated on 20/Jun/23

	$P (r \cos ϕ, r \sin ϕ)$ $r = a + b$ $r^{2} \cos^{2} ϕ + {(r \sin ϕ - b)}^{2} = a^{2} + b^{2}$ $\Rightarrow 2 a b + b^{2} = 2 b r \sin ϕ$ $& a l t i t u d e o f △ i s ⊥ t o b a s e, s o$ $\frac{r \sin ϕ}{r \cos ϕ - a} = \frac{a + r \cos ϕ}{2 b - r \sin ϕ}$ $\Rightarrow 2 b r \sin ϕ = r^{2} - a^{2}$ $c o n s i d e r \frac{x}{a} + \frac{y}{b} = 1$ $o r b x + a y - a b = 0$ $l e t h = \frac{∣ b r \cos ϕ + a r \sin ϕ - a b ∣}{\sqrt{a^{2} + b^{2}}}$ $c = \sqrt{a^{2} + b^{2}}$ $r \sin ϕ = a + \frac{b}{2}$ $h = \frac{∣ b \sqrt{{(a + b)}^{2} - {(a + \frac{b}{2})}^{2}} + a^{2} - \frac{a b}{2} ∣}{\sqrt{a^{2} + b^{2}}}$ $△ = \frac{c h}{2}$ $= \frac{b^{2}}{2} ∣ \sqrt{\frac{3}{4} + \frac{a}{b}} + \frac{a^{2}}{b^{2}} - \frac{a}{2 b} ∣$
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