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Question Number 189123 by Rupesh123 last updated on 12/Mar/23

Answered by mr W last updated on 13/Mar/23

Commented by mr W last updated on 13/Mar/23

$$letAC=p,AD=q$$$$\frac{AE}{BF}=\frac{AC}{BC}=\frac{CE}{CF}$$$$\frac{AE}{1}=\frac{p}{1}=\frac{5}{CF}$$$$\Rightarrow AE=p,CF=\frac{5}{p}$$$$similarly$$$$\Rightarrow AB=\frac{3q}{2},DG=\frac{8}{q}$$$$\cos \alpha =\frac{2^2+p^2-q^2}{2\times 2\times p}=\frac{6^2+p^2-{\left(\frac{3q}{2}\right)}^2}{2\times 6\times p}$$$$\Rightarrow 8p^2-3q^2=96...\left(i\right)$$$$\cos \beta =\frac{1^2+{\left(\frac{3q}{2}\right)}^2-p^2}{2\times 1\times \frac{3q}{2}}=\frac{6^2+{\left(\frac{3q}{2}\right)}^2-p^2}{2\times 6\times \frac{3q}{2}}$$$$\Rightarrow 9q^2-4p^2=24...\left(ii\right)$$$$from\left(i\right)and\left(ii\right):$$$$\Rightarrow p^2=\frac{78}{5},q^2=\frac{48}{5}$$$$\frac{x^2-{(p+\frac{5}{p})}^2-{(q+\frac{8}{q})}^2}{2\times (p+\frac{5}{p})(q+\frac{8}{q})}=\frac{3^2-p^2-q^2}{2\times p\times q}$$$$x^2=26+p^2+\frac{25}{p^2}+q^2+\frac{64}{q^2}+(9-p^2-q^2)(1+\frac{5}{p^2})(1+\frac{8}{q^2})$$$$x^2=26+\frac{78}{5}+\frac{25\times 5}{78}+\frac{48}{5}+\frac{64\times 5}{48}+(9-\frac{78}{5}-\frac{48}{5})(1+\frac{5\times 5}{78})(1+\frac{8\times 5}{48})$$$$=\frac{81}{4}$$$$\Rightarrow x=\frac{9}{2}=4.5✓$$

Commented by Rupesh123 last updated on 14/Mar/23

Excellent!

Answered by HeferH last updated on 13/Mar/23

$$4.5$$

Commented by mr W last updated on 13/Mar/23

$$imadesomemistakes.{it}^{\prime }sfixed.$$

Commented by Rupesh123 last updated on 13/Mar/23

Can you show workings?

Commented by HeferH last updated on 13/Mar/23

Commented by HeferH last updated on 13/Mar/23

Commented by HeferH last updated on 13/Mar/23

$$\mathrm{Im}\mathrm{sorry}\mathrm{if}\mathrm{there}\mathrm{is}\mathrm{any}\mathrm{error}$$

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