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Question Number 172692 by mnjuly1970 last updated on 30/Jun/22

Answered by som(math1967) last updated on 30/Jun/22

Commented by som(math1967) last updated on 30/Jun/22

$$let\mathrm{\angle }ABC=\theta sideofsquare=a$$$$\therefore DC=asin\theta CE=acos\theta$$$$BF=acot\theta AG=atan\theta$$$$\therefore \frac{1}{2}\times a^{2}sin\theta cos\theta +\frac{1}{2}\times a^{2}cot\theta$$$$+\frac{1}{2}{a}^{2}tan\theta +a^2=\frac{1}{2}\times 3\times 4$$$$(\frac{1}{2}\times \frac{3}{5}\times \frac{4}{5}+\frac{1}{2}\times \frac{4}{3}+\frac{1}{2}\times \frac{3}{4}+1)a^2$$$$=6$$$$(\frac{6}{25}+\frac{2}{3}+\frac{3}{8}+1)a^2=6$$$$\frac{(144+400+225+600)a^2}{600}=6$$$$a^2=\frac{6\times 600}{1369}=2.63squnit\left(approx\right)$$

Commented by mnjuly1970 last updated on 30/Jun/22

$$\mathrm{tayeballah}\mathrm{sir}$$

Commented by Tawa11 last updated on 01/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by mr W last updated on 30/Jun/22

$$a=sidelengthofsquare$$$$4=\frac{a}{5}\times 4+\frac{a}{3}\times 5$$$$\Rightarrow a=\frac{60}{37}$$$$areaofsquare=a^2={\left(\frac{60}{37}\right)}^2=\frac{3600}{1369}$$

Commented by mr W last updated on 30/Jun/22

Commented by Tawa11 last updated on 01/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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