Question Number 222533 by Tawa11 last updated on 29/Jun/25
Answered by fantastic last updated on 29/Jun/25
Commented by fantastic last updated on 29/Jun/25
![Total area of the triangle=54+42=96cm^2 So (1/2)×(4+α)×α=96 or α^2 +4α=192 or α^2 +4α−192=0 α=((−4±(√(4^2 −4×1×(−192))))/(2×1))=((−4±(√(16+768)))/2)=((−4±(√(784)))/2)=((−4±28)/2) So α=((−4+28)/2)=((24)/2)=12cm[∵α>0 but ((−4−28)/2)<0] Now (1/2)×α×β=54 So β=((54×2)/α)=((54×2)/(12))=9cm So β=9cm So γ=(√(α^2 +β^2 ))=(√(12^2 +9^2 ))=15cm So γ=15cm Length of x+γ=(√(α^2 +(α+4)^2 ))=(√(12^2 +(12+4)^2 ))=(√(12^2 +16^2 ))=20cm So x+γ=20cm ∴x=20−γ=20−15=5cm determinant (((Final Answer x=5cm)))](https://www.tinkutara.com/question/Q222535.png)
$${Total}\:{area}\:{of}\:{the}\:{triangle}=\mathrm{54}+\mathrm{42}=\mathrm{96}{cm}^{\mathrm{2}} \\ $$$${So}\:\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{4}+\alpha\right)×\alpha=\mathrm{96} \\ $$$${or}\:\alpha^{\mathrm{2}} +\mathrm{4}\alpha=\mathrm{192} \\ $$$${or}\:\alpha^{\mathrm{2}} +\mathrm{4}\alpha−\mathrm{192}=\mathrm{0} \\ $$$$\alpha=\frac{−\mathrm{4}\pm\sqrt{\mathrm{4}^{\mathrm{2}} −\mathrm{4}×\mathrm{1}×\left(−\mathrm{192}\right)}}{\mathrm{2}×\mathrm{1}}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{16}+\mathrm{768}}}{\mathrm{2}}=\frac{−\mathrm{4}\pm\sqrt{\mathrm{784}}}{\mathrm{2}}=\frac{−\mathrm{4}\pm\mathrm{28}}{\mathrm{2}} \\ $$$${So}\:\alpha=\frac{−\mathrm{4}+\mathrm{28}}{\mathrm{2}}=\frac{\mathrm{24}}{\mathrm{2}}=\mathrm{12}{cm}\left[\because\alpha>\mathrm{0}\:{but}\:\frac{−\mathrm{4}−\mathrm{28}}{\mathrm{2}}<\mathrm{0}\right] \\ $$$${Now}\:\frac{\mathrm{1}}{\mathrm{2}}×\alpha×\beta=\mathrm{54} \\ $$$${So}\:\beta=\frac{\mathrm{54}×\mathrm{2}}{\alpha}=\frac{\mathrm{54}×\mathrm{2}}{\mathrm{12}}=\mathrm{9}{cm} \\ $$$${So}\:\beta=\mathrm{9}{cm} \\ $$$${So}\:\gamma=\sqrt{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{9}^{\mathrm{2}} }=\mathrm{15}{cm} \\ $$$${So}\:\gamma=\mathrm{15}{cm} \\ $$$${Length}\:{of}\:{x}+\gamma=\sqrt{\alpha^{\mathrm{2}} +\left(\alpha+\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{12}^{\mathrm{2}} +\left(\mathrm{12}+\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\mathrm{12}^{\mathrm{2}} +\mathrm{16}^{\mathrm{2}} }=\mathrm{20}{cm} \\ $$$${So}\:{x}+\gamma=\mathrm{20}{cm} \\ $$$$\therefore{x}=\mathrm{20}−\gamma=\mathrm{20}−\mathrm{15}=\mathrm{5}{cm} \\ $$$$\begin{array}{|c|}{{Final}\:{Answer}\:{x}=\mathrm{5}{cm}}\\\hline\end{array} \\ $$
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by fantastic last updated on 29/Jun/25
Commented by fantastic last updated on 29/Jun/25
$$\alpha−\beta=\mathrm{12}−\mathrm{9}=\mathrm{3}{cm} \\ $$$${x}=\sqrt{\mathrm{4}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\mathrm{5}{cm} \\ $$$${So}\:{x}=\mathrm{5}{cm} \\ $$
Commented by fantastic last updated on 29/Jun/25
$${Another}\:{easy}\:{method} \\ $$$$ \\ $$
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by fantastic last updated on 29/Jun/25
Commented by fantastic last updated on 29/Jun/25
$$\frac{\gamma}{{x}}=\frac{\alpha}{\mathrm{4}} \\ $$$${x}=\frac{\mathrm{4}\gamma}{\alpha}=\frac{\mathrm{4}×\mathrm{15}}{\mathrm{12}}=\mathrm{5}{cm} \\ $$$${x}=\mathrm{5}{cm} \\ $$
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by mr W last updated on 29/Jun/25
Commented by mr W last updated on 29/Jun/25
$$\left(\frac{{y}+\mathrm{4}}{{y}}\right)^{\mathrm{2}} =\frac{\mathrm{54}+\mathrm{42}}{\mathrm{54}}=\frac{\mathrm{16}}{\mathrm{9}}\: \\ $$$$\Rightarrow{y}=\mathrm{3}×\mathrm{4}=\mathrm{12},\:{z}=\mathrm{3}{x} \\ $$$$\left(\frac{{x}+{z}}{{y}}\right)^{\mathrm{2}} =\frac{\mathrm{54}+\mathrm{42}+\mathrm{54}}{\mathrm{54}}=\frac{\mathrm{25}}{\mathrm{9}} \\ $$$$\mathrm{4}{x}=\frac{\mathrm{5}}{\mathrm{3}}×\mathrm{12}\:\Rightarrow{x}=\mathrm{5}\:\checkmark \\ $$
Commented by Tawa11 last updated on 29/Jun/25
$$\mathrm{Thanks}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$
Commented by fantastic last updated on 30/Jun/25
$${creative}! \\ $$