Question Number 185178 by Rupesh123 last updated on 18/Jan/23
Commented by Rupesh123 last updated on 18/Jan/23
All are squares sitting on a horizontal base with areas given. Find Green Areas
Answered by HeferH last updated on 18/Jan/23
Commented by HeferH last updated on 18/Jan/23
$$\frac{\mathrm{2}\sqrt{\mathrm{5}}\centerdot\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}\:+\:\frac{\mathrm{2}\sqrt{\mathrm{5}}\centerdot\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{2}}=\:\mathrm{30} \\ $$
Answered by som(math1967) last updated on 18/Jan/23
$${cos}\angle{FBH}=\frac{\sqrt{\mathrm{45}}}{\:\sqrt{\mathrm{65}}}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}} \\ $$$$\angle{ABC}=\mathrm{180}−\angle{FBH} \\ $$$${sin}\angle{ABC}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$${sin}\angle{AED}=\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}} \\ $$$${green}\:{area}={area}\left(\bigtriangleup{ABC}+\bigtriangleup{DEA}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{AB}×{BC}×{sin}\angle{ABC} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}×{DE}×{AE}×{sin}\angle{AED} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{65}}×\sqrt{\mathrm{45}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{13}}} \\ $$$$+\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\mathrm{20}}×\sqrt{\mathrm{65}}×\frac{\mathrm{3}}{\:\sqrt{\mathrm{13}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2}×\mathrm{15}+\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{3}×\mathrm{10}=\mathrm{30}\boldsymbol{{sq}}\:\boldsymbol{{unit}} \\ $$
Commented by som(math1967) last updated on 18/Jan/23
Answered by mr W last updated on 20/Jan/23
Commented by mr W last updated on 20/Jan/23
$${a}=\sqrt{\mathrm{20}} \\ $$$${b}=\sqrt{\mathrm{45}} \\ $$$${A}_{\mathrm{1}} ={A}_{\mathrm{2}} =\frac{{a}×{b}}{\mathrm{2}}=\frac{\sqrt{\mathrm{20}}×\sqrt{\mathrm{45}}}{\mathrm{2}} \\ $$$${green}\:{area}={A}_{\mathrm{1}} +{A}_{\mathrm{2}} =\sqrt{\mathrm{20}}×\sqrt{\mathrm{45}}=\mathrm{30} \\ $$