Question Number 198933 by cherokeesay last updated on 25/Oct/23
Commented by mr W last updated on 26/Oct/23
Commented by Rasheed.Sindhi last updated on 26/Oct/23
$$\mathbb{T}\boldsymbol{\mathrm{han}}\Bbbk\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{\mathrm{help}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{diagram}}\:\boldsymbol{\mathrm{sir}}! \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/23
![Right vertex of yellow triangle=D (say) Intersection of DC^(→) & BA^(→) =E BE is diameter=2r [∵ BDE is right triangle] CE=CB=(√(3^2 +4^2 )) =5 DE=DC+CE=3+5=8 △BDE: BE^2 =BD^2 +DE^2 (2r)^2 =4^2 +8^2 =80 r^2 =20⇒r=AB=2(√5) △ABC: AC^2 =BC^2 −AB^2 =5^2 −(2(√5) )^2 =5 AC=(√5) ▲ABC=(1/2)∙AB.AC =(1/2)(2(√5) )((√5) )=5 cm^2](https://www.tinkutara.com/question/Q198947.png)
$${Right}\:{vertex}\:{of}\:{yellow}\:{triangle}={D}\:\left({say}\right) \\ $$$${Intersection}\:{of}\:\overset{\rightarrow} {{DC}}\:\&\:\overset{\rightarrow} {{BA}}\:={E} \\ $$$${BE}\:{is}\:{diameter}=\mathrm{2}{r}\:\left[\because\:{BDE}\:{is}\:{right}\:{triangle}\right] \\ $$$${CE}={CB}=\sqrt{\mathrm{3}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }\:=\mathrm{5} \\ $$$${DE}={DC}+{CE}=\mathrm{3}+\mathrm{5}=\mathrm{8} \\ $$$$\bigtriangleup{BDE}: \\ $$$${BE}^{\mathrm{2}} ={BD}^{\mathrm{2}} +{DE}^{\mathrm{2}} \\ $$$$\left(\mathrm{2}{r}\right)^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} =\mathrm{80} \\ $$$${r}^{\mathrm{2}} =\mathrm{20}\Rightarrow{r}={AB}=\mathrm{2}\sqrt{\mathrm{5}}\: \\ $$$$\bigtriangleup{ABC}: \\ $$$${AC}^{\mathrm{2}} ={BC}^{\mathrm{2}} −{AB}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} −\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)^{\mathrm{2}} =\mathrm{5} \\ $$$${AC}=\sqrt{\mathrm{5}}\: \\ $$$$\blacktriangle{ABC}=\frac{\mathrm{1}}{\mathrm{2}}\centerdot{AB}.{AC} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\sqrt{\mathrm{5}}\:\right)\left(\sqrt{\mathrm{5}}\:\right)=\mathrm{5}\:\mathrm{cm}^{\mathrm{2}} \\ $$
Commented by cherokeesay last updated on 26/Oct/23
$${So}\:{nice}\:! \\ $$$${thank}\:{you}\:{sir}\:! \\ $$