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Question-221815

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Question Number 221815 by alvan545 last updated on 11/Jun/25
Answered by mr W last updated on 11/Jun/25
say AC=c BC^2 =CD^2 a^2 +c^2 −2ac cos θ=b^2 +c^2 −2bc cos θ 2(b−a)c cos θ=b^2 −a^2 2c cos θ=a+b ⇒c=((a+b)/(2 cos θ)) S_(ABCD) =S_(ABC) +S_(CDA) =((ac sin θ)/2)+((bc sin θ)/2) =(((a+b)c sin θ)/2) =(((a+b)(a+b) sin θ)/(4 cos θ)) =(((a+b)^2 tan θ)/4) ✓
$${say}\:{AC}={c} \\ $$$${BC}^{\mathrm{2}} ={CD}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ac}\:\mathrm{cos}\:\theta={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{2}\left({b}−{a}\right){c}\:\mathrm{cos}\:\theta={b}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{c}\:\mathrm{cos}\:\theta={a}+{b} \\ $$$$\Rightarrow{c}=\frac{{a}+{b}}{\mathrm{2}\:\mathrm{cos}\:\theta} \\ $$$${S}_{{ABCD}} ={S}_{{ABC}} +{S}_{{CDA}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{ac}\:\mathrm{sin}\:\theta}{\mathrm{2}}+\frac{{bc}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+{b}\right){c}\:\mathrm{sin}\:\theta}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+{b}\right)\left({a}+{b}\right)\:\mathrm{sin}\:\theta}{\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\left({a}+{b}\right)^{\mathrm{2}} \:\mathrm{tan}\:\theta}{\mathrm{4}}\:\checkmark \\ $$
Commented by alvan545 last updated on 11/Jun/25
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