Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 194067 by Rupesh123 last updated on 27/Jun/23

Answered by som(math1967) last updated on 27/Jun/23

$$x^2+{10}^2=2{(10-x)}^2$$$$\Rightarrow x^2-40x+100=0$$$$x=10(2-\sqrt{3})$$$$sideoftriangle=\sqrt{2}\times 10(1-2+\sqrt{3})$$$$=10\sqrt{2}(\sqrt{3}-1)$$$$\frac{a}{10\sqrt{2}(\sqrt{3}-1)-a}=\frac{\sqrt{3}}{2}$$$$2a=30\sqrt{2}-10\sqrt{6}-\sqrt{3}a$$$$\therefore a=\frac{10(3\sqrt{2}-\sqrt{6})}{2+\sqrt{3}}$$$$a^2=\frac{100{(3\sqrt{2}-\sqrt{6})}^2}{{(2+\sqrt{3})}^2}{cm}^2$$

Commented by som(math1967) last updated on 27/Jun/23

Commented by Frix last updated on 27/Jun/23

$${\mathrm{I}}^{\prime }\mathrm{m}\mathrm{afraid}{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{a}\mathrm{mistake}\mathrm{calculating}a.$$$$\mathrm{I}\mathrm{get}a=10(9-5\sqrt{3})\sqrt{2}\approx 4.80$$$$\mathrm{while}\mathrm{your}a\approx 1.24$$$$\mathrm{which}\mathrm{seems}\mathrm{too}\mathrm{small}\mathrm{given}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}$$$$\mathrm{triangle}\mathrm{is}\approx 10.35$$

Commented by som(math1967) last updated on 27/Jun/23

$$Thankyousir,Icorrectit$$

Commented by Rupesh123 last updated on 27/Jun/23

Perfect ��

Commented by Rupesh123 last updated on 27/Jun/23

You're right!

Answered by mr W last updated on 27/Jun/23

$$s=sidelengthofequilateral$$$$s^2={10}^2+{(10-\frac{s}{\sqrt{2}})}^2$$$$-400+20\sqrt{2}s+s^2=0$$$$\Rightarrow s=10(\sqrt{6}-\sqrt{2})$$$$a=sidelengthofsmallsquare$$$$a+\frac{2a}{\sqrt{3}}=s$$$$\Rightarrow a=\frac{\sqrt{3}s}{\sqrt{3}+2}=\frac{10(3\sqrt{2}-\sqrt{6})}{\sqrt{3}+2}\approx 4.805$$

Commented by Rupesh123 last updated on 27/Jun/23

Perfect ��

Terms of Service

Privacy Policy

Contact: info@tinkutara.com